APNA CAREER: ELECTRIC POTENTIAL & CAPACITANCE

⚡ ELECTRIC POTENTIAL & CAPACITANCE

50 MCQs — CBSE / NEET / JEE (With Full Explanation)

Q1. The electric potential at a point due to a point charge q at distance r is: (A) kq/r² (B) kq/r (C) kq²/r (D) zero

Ans: (B) kq/r Explanation: Electric potential V = kq/r. It is a scalar quantity. Note that electric field E = kq/r² (varies as 1/r²), but potential varies as 1/r. As distance increases, potential decreases.

Q2. SI unit of electric potential is: (A) Newton (B) Joule (C) Volt (D) Coulomb

Ans: (C) Volt Explanation: Potential = Work done / Charge = Joule/Coulomb = Volt. 1 Volt means 1 Joule of work is needed to bring 1 Coulomb charge from infinity to that point.

Q3. Work done in moving a charge q through potential difference V is: (A) qV² (B) q²V (C) qV (D) V/q

Ans: (C) qV Explanation: W = qV is the basic formula. If V = 10V and q = 2C, then W = 20J. This is the energy given to the charge as it moves through the potential difference.

Q4. Equipotential surfaces for a uniform electric field are: (A) Spherical (B) Cylindrical (C) Parallel to E (D) Perpendicular to E

Ans: (D) Perpendicular to E Explanation: Equipotential surfaces are always perpendicular to electric field lines. No work is done moving a charge on these surfaces. For uniform field (like between parallel plates), they are flat planes perpendicular to E.

Q5. Capacitance of a parallel plate capacitor (area A, separation d, dielectric K) is: (A) ε₀A/d (B) Kε₀A/d (C) ε₀A/Kd (D) KAd/ε₀

Ans: (B) Kε₀A/d Explanation: Without dielectric, C = ε₀A/d. When dielectric of constant K is inserted, capacitance increases K times, so C = Kε₀A/d. Larger area = more capacitance, larger separation = less capacitance.

Q6. Energy stored in a capacitor charged to voltage V is: (A) CV (B) CV² (C) ½CV² (D) 2CV²

Ans: (C) ½CV² Explanation: U = ½CV² = ½QV = Q²/2C. The factor ½ comes because as capacitor charges, voltage builds up gradually from 0 to V, so average voltage is V/2. Energy is stored in the electric field between the plates.

Q7. Two capacitors C₁ and C₂ in series — equivalent capacitance is: (A) C₁+C₂ (B) C₁C₂/(C₁+C₂) (C) C₁−C₂ (D) C₁/C₂

Ans: (B) C₁C₂/(C₁+C₂) Explanation: In series: 1/C = 1/C₁ + 1/C₂, which gives C = C₁C₂/(C₁+C₂). Series combination always gives capacitance less than the smallest individual capacitor. Same charge Q appears on each capacitor.

Q8. Relation between electric field E and potential V is: (A) E = V/r (B) E = dV/dr (C) E = −dV/dr (D) E = V·r

Ans: (C) E = −dV/dr Explanation: The negative sign is very important. Electric field points from high potential to low potential (downhill direction). So E = −dV/dr means field is in direction of decreasing potential. This is why field lines go from + to − charge.

Q9. A 4μF capacitor is charged to 200V. Charge on capacitor is: (A) 50 μC (B) 400 μC (C) 800 μC (D) 1600 μC

Ans: (C) 800 μC Explanation: Q = CV = 4×10⁻⁶ × 200 = 800×10⁻⁶ C = 800 μC. Always remember Q = CV is the basic relation for capacitor. C is in Farads, V in Volts, Q comes in Coulombs.

Q10. No work is done on an equipotential surface because: (A) Charge is zero (B) E = 0 (C) E ⊥ displacement (D) E ∥ displacement

Ans: (C) E ⊥ displacement Explanation: W = qEd cosθ. On equipotential surface, θ = 90°, so cos90° = 0, hence W = 0. Also, since potential is same everywhere on it, ΔV = 0 so W = qΔV = 0. Both ways, work = 0.

Q11. If plate separation of a parallel plate capacitor is doubled, capacitance becomes: (A) 4C (B) 2C (C) C/2 (D) C/4

Ans: (C) C/2 Explanation: C = ε₀A/d. If d is doubled, C = ε₀A/2d = C/2. Capacitance is inversely proportional to separation. More separation means weaker electric field effect, hence less ability to store charge.

Q12. Dielectric constant of vacuum is: (A) 0 (B) 1 (C) ∞ (D) Depends on field

Ans: (B) 1 Explanation: Dielectric constant K = ε/ε₀. For vacuum, ε = ε₀, so K = 1. For air K ≈ 1. For water K ≈ 80. For metals K = ∞ (that's why inserting a metal slab completely fills that gap and reduces effective separation).

Q13. Electric potential inside a charged conductor is: (A) Zero (B) Equal to surface potential (C) Varies linearly (D) Maximum at centre

Ans: (B) Equal to surface potential Explanation: Inside a conductor E = 0. Since E = −dV/dr = 0, potential V is constant throughout the conductor and equals the surface potential. Note: it is NOT necessarily zero — only earthed conductors have V = 0.

Q14. Three capacitors of 2μF each in parallel — equivalent capacitance is: (A) 2/3 μF (B) 2 μF (C) 6 μF (D) 4 μF

Ans: (C) 6 μF Explanation: In parallel, C_eq = C₁+C₂+C₃ = 2+2+2 = 6 μF. Parallel combination increases total capacitance. Each capacitor gets same voltage but charges add up. Total charge = 3 × individual charge.

Q15. Electric potential at equatorial point of a dipole is: (A) kp/r² (B) Zero (C) kp/r (D) 2kp/r²

Ans: (B) Zero Explanation: At equatorial point, both +q and −q are equidistant. Potential due to +q = +kq/r and due to −q = −kq/r. Total V = 0. But note: electric field at equatorial point is NOT zero, it is kp/r³.

Q16. Electric field inside a conductor in electrostatic equilibrium is: (A) Uniform (B) Zero (C) Maximum (D) Equal to surface field

Ans: (B) Zero Explanation: In electrostatic equilibrium, free electrons rearrange until net internal field = 0. If E ≠ 0 inside, electrons would keep moving (not equilibrium). All excess charge moves to outer surface, making interior field-free.

Q17. A 10μF capacitor charged to 100V stores energy: (A) 0.05 J (B) 0.1 J (C) 0.5 J (D) 1 J

Ans: (A) 0.05 J Explanation: U = ½CV² = ½ × 10×10⁻⁶ × (100)² = ½ × 10⁻⁵ × 10⁴ = ½ × 0.1 = 0.05 J. Always convert μF to F before calculating. 10μF = 10×10⁻⁶ F.

Q18. Potential due to a uniformly charged spherical shell at its centre is: (A) Zero (B) kQ/R (C) kQ/R² (D) Infinite

Ans: (B) kQ/R Explanation: For a spherical shell, E = 0 inside, but V = kQ/R (constant) throughout interior, same as on surface. Potential is NOT zero inside — it is equal to the potential at the surface.

Q19. Unit of ε₀ (permittivity of free space) is: (A) C²/Nm² (B) Nm²/C² (C) C/Nm (D) N/C²

Ans: (A) C²/Nm² Explanation: From F = kq₁q₂/r² and k = 1/4πε₀, we get ε₀ = q²/Fr² = C²/(N·m²). Value of ε₀ = 8.85×10⁻¹² C²N⁻¹m⁻². This is frequently asked in NEET and JEE.

Q20. Capacitors in parallel have the same: (A) Charge (B) Voltage (C) Energy (D) Capacitance

Ans: (B) Voltage Explanation: In parallel connection, both plates of each capacitor are connected to the same two nodes, so all capacitors share the same potential difference (voltage). Charge on each depends on its individual capacitance: Q = CV.

Q21. Capacitors in series have the same: (A) Charge (B) Voltage (C) Capacitance (D) Energy

Ans: (A) Charge Explanation: In series, same charge Q flows through the circuit and appears on each capacitor. The voltage distributes: V = V₁+V₂ and V₁/V₂ = C₂/C₁ (smaller capacitor gets larger voltage share).

Q22. Charge given to a conductor resides: (A) At centre (B) On outer surface (C) Uniformly inside (D) At the bottom

Ans: (B) On outer surface Explanation: Due to mutual repulsion, free charges push each other to the outer surface. Interior has E = 0, so no charge remains inside. For hollow conductors, charge is on outer surface; cavity remains charge-free.

Q23. Potential at midpoint between +q and −q charges separated by 2d is: (A) kq/d (B) 2kq/d (C) Zero (D) kq/d²

Ans: (C) Zero Explanation: At midpoint, distance from each charge = d. V = k(+q)/d + k(−q)/d = kq/d − kq/d = 0. Even though field is not zero there, potential is zero. This is a classic NEET/JEE trap question.

Q24. Electric potential energy of two charges q₁, q₂ separated by r is: (A) kq₁q₂/r² (B) kq₁q₂/r (C) k(q₁+q₂)/r (D) kq₁q₂·r

Ans: (B) kq₁q₂/r Explanation: U = kq₁q₂/r. If same sign charges: U > 0 (repulsive, energy stored). If opposite charges: U < 0 (attractive, energy released). This energy = work done to assemble the system from infinity.

Q25. A dielectric is inserted in a charged isolated capacitor. Energy stored: (A) Increases (B) Decreases (C) Remains same (D) Becomes zero

Ans: (B) Decreases Explanation: Q remains constant (isolated). C increases (K times). U = Q²/2C, so as C increases, U decreases. The dielectric molecules absorb some energy — that's where the energy goes. This is a high-frequency NEET concept.

Q26. A capacitor is connected to a battery. On inserting dielectric, charge on capacitor: (A) Decreases (B) Remains same (C) Increases (D) Becomes zero

Ans: (C) Increases Explanation: Battery keeps V constant. C increases K times. Q = CV, so Q also increases K times. Battery supplies extra charge. This is opposite to the isolated capacitor case — always check if battery is connected or not!

Q27. The electric potential on the axis of a dipole at distance r (r >> dipole length 2a) is: (A) kp/r² (B) Zero (C) kp/r (D) 2kp/r²

Ans: (A) kp/r² Explanation: At axial point, V = kp/r² where p = q×2a is dipole moment. At equatorial point V = 0. These two results are frequently compared in CBSE board and NEET exams.

Q28. Which of the following is NOT a property of equipotential surfaces? (A) They never intersect (B) Work done is zero on them (C) E is parallel to them (D) They are perpendicular to field lines

Ans: (C) E is parallel to them Explanation: E is always perpendicular (NOT parallel) to equipotential surfaces. If E were parallel, it would have a component along the surface, doing work — which contradicts the definition. All other options are correct properties.

Q29. Van de Graaff generator works on the principle of: (A) Heating effect (B) Action of points and electrostatic induction (C) Magnetic effect (D) Photoelectric effect

Ans: (B) Action of points and electrostatic induction Explanation: Sharp points cause corona discharge (action of points). Charge is continuously transferred to outer shell via electrostatic induction. It can build up millions of volts. Used in nuclear physics for particle acceleration.

Q30. If a soap bubble is given positive charge, its radius: (A) Decreases (B) Increases (C) Remains same (D) First increases then decreases

Ans: (B) Increases Explanation: Like charges on bubble surface repel each other outward, increasing outward pressure. This adds to the gas pressure inside, causing the bubble to expand. This is a conceptual question frequently seen in JEE.

Q31. Two conducting spheres of radii r₁ and r₂ are connected by a wire. Ratio of their surface charge densities is: (A) r₁/r₂ (B) r₂/r₁ (C) r₁²/r₂² (D) r₂²/r₁²

Ans: (B) r₂/r₁ Explanation: Connected spheres have equal potential: kQ₁/r₁ = kQ₂/r₂. Surface charge density σ = Q/4πr². After solving, σ₁/σ₂ = r₂/r₁. Smaller sphere has higher charge density — sharper objects have higher charge density.

Q32. Energy density (energy per unit volume) in electric field E is: (A) ε₀E (B) ε₀E²/2 (C) ε₀E² (D) E²/ε₀

Ans: (B) ε₀E²/2 Explanation: u = ½ε₀E². This can be derived from U = ½CV² for a parallel plate capacitor. Energy is stored in the electric field. This concept is important in JEE and also in understanding electromagnetic waves.

Q33. A charge of 2μC is moved from point A (potential 100V) to point B (potential 200V). Work done by external agent is: (A) −0.2 mJ (B) +0.2 mJ (C) −0.1 mJ (D) +0.1 mJ

Ans: (B) +0.2 mJ Explanation: W = q(V_B − V_A) = 2×10⁻⁶ × (200−100) = 2×10⁻⁶ × 100 = 2×10⁻⁴ J = 0.2 mJ. Positive because charge is moved to higher potential (uphill) — external agent must do positive work.

Q34. The capacitance of Earth (radius R = 6400 km) is approximately: (A) 711 μF (B) 711 F (C) 711 mF (D) 711 pF

Ans: (A) 711 μF Explanation: C = 4πε₀R = R/k = 6.4×10⁶ / 9×10⁹ ≈ 711×10⁻⁶ F = 711 μF. Earth acts as a huge capacitor. This is why it is used as a ground reference in electrical systems.

Q35. Capacitance of a spherical conductor of radius R is: (A) 4πε₀/R (B) 4πε₀R (C) R/4πε₀ (D) ε₀/R

Ans: (B) 4πε₀R Explanation: V = kQ/R = Q/4πε₀R. So C = Q/V = 4πε₀R. Capacitance of a sphere depends only on its radius. Larger sphere = more capacitance = can store more charge at same potential.

Q36. A parallel plate capacitor is charged and then disconnected from battery. If plates are pulled apart, which increases? (A) Capacitance (B) Charge (C) Electric field (D) Voltage

Ans: (D) Voltage Explanation: Q is constant (disconnected). C = ε₀A/d decreases as d increases. V = Q/C increases. E = σ/ε₀ = Q/Aε₀ stays constant (doesn't change with d). Energy U = Q²/2C also increases (work done in pulling plates apart).

Q37. Potential gradient is equal to: (A) Electric field (B) Negative of electric field (C) Charge density (D) Capacitance

Ans: (B) Negative of electric field Explanation: dV/dr = −E, so potential gradient = −E. The negative sign means potential decreases in the direction of electric field. Electric field always points from high potential to low potential region.

Q38. Two charges +4μC and −4μC are placed 20cm apart. Potential at midpoint is: (A) 72000 V (B) −72000 V (C) Zero (D) 36000 V

Ans: (C) Zero Explanation: At midpoint, r = 10cm from each. V = k(+4μC)/0.1 + k(−4μC)/0.1 = 0. Both contributions cancel. Same concept as Q23. Note: electric field at midpoint is NOT zero — it adds up (both fields point in same direction at midpoint).

Q39. The force between capacitor plates is: (A) σ²/ε₀ (B) σ²A/2ε₀ (C) σ²/2ε₀ (D) σ/2ε₀

Ans: (B) σ²A/2ε₀ Explanation: Each plate is in the field of the other plate only (E = σ/2ε₀). Force = QE = (σA)(σ/2ε₀) = σ²A/2ε₀. This force is always attractive between oppositely charged plates. This is a JEE-level derivation-based concept.

Q40. Kirchhoff's voltage law applied to capacitor circuit means: (A) Total charge is conserved (B) Sum of all potential drops = EMF (C) Current is same everywhere (D) Capacitance adds in parallel

Ans: (B) Sum of all potential drops = EMF Explanation: KVL says algebraic sum of EMFs = sum of potential drops. For capacitors in series: V = V₁+V₂+V₃. This is essentially KVL applied to capacitor circuits to find charge distribution.

Q41. Common potential when two capacitors C₁ (charged to V₁) and C₂ (charged to V₂) are connected is: (A) (V₁+V₂)/2 (B) (C₁V₁+C₂V₂)/(C₁+C₂) (C) C₁V₁/C₂V₂ (D) V₁V₂/(V₁+V₂)

Ans: (B) (C₁V₁+C₂V₂)/(C₁+C₂) Explanation: Total charge is conserved: Q₁+Q₂ = (C₁+C₂)V_common. So V = (C₁V₁+C₂V₂)/(C₁+C₂). This is a weighted average of voltages. If C₁=C₂, then V = (V₁+V₂)/2 (simple average). Important for NEET/JEE numericals.

Q42. Loss of energy when two capacitors are connected together is: (A) Zero always (B) C₁C₂(V₁−V₂)²/2(C₁+C₂) (C) (C₁+C₂)(V₁−V₂)² (D) C₁C₂V²

Ans: (B) C₁C₂(V₁−V₂)²/2(C₁+C₂) Explanation: Energy is always lost when two capacitors at different potentials are connected. It is dissipated as heat in connecting wire (even if resistance = 0, some energy is lost to electromagnetic radiation). This loss is zero only if V₁ = V₂.

Q43. A 2μF and 4μF capacitor are in series across 12V. Voltage across 2μF is: (A) 4V (B) 8V (C) 6V (D) 3V

Ans: (B) 8V Explanation: In series, Q is same on both. C_eq = (2×4)/(2+4) = 8/6 = 4/3 μF. Q = C_eq × V = (4/3)×10⁻⁶ × 12 = 16 μC. V across 2μF = Q/C = 16/2 = 8V. (Smaller capacitor gets larger voltage — remember this rule.)

Q44. Which quantity remains constant when a dielectric slab is inserted in an isolated (disconnected) capacitor? (A) Voltage (B) Electric field (C) Capacitance (D) Charge

Ans: (D) Charge Explanation: When capacitor is disconnected (isolated), no charge can flow in or out. Q = constant. As dielectric is inserted: C increases, V decreases (V = Q/C), E decreases, energy decreases. Only Q stays same.

Q45. The electric potential due to a thin infinite plane sheet of charge is: (A) Directly proportional to distance (B) Independent of distance (C) Inversely proportional to distance (D) Zero

Ans: (A) Directly proportional to distance Explanation: E for infinite sheet = σ/2ε₀ = constant. Since E = −dV/dr = constant, V varies linearly with distance (V ∝ r). This is different from point charge (V ∝ 1/r) and is a frequently tested concept.

Q46. A bird sits on a high voltage wire of 100kV. It does not get a shock because: (A) Current doesn't flow through it (B) Its body has zero resistance (C) No potential difference exists across its body (D) It is insulated by feathers

Ans: (C) No potential difference exists across its body Explanation: Both feet are at same potential (100kV). Since ΔV = 0 across the bird's body, no current flows (I = ΔV/R). Current needs potential difference to flow. If the bird touched a ground wire simultaneously, it would complete the circuit — fatal!

Q47. The work done in moving a unit positive charge through a distance dr against electric field E is: (A) E·dr (B) −E·dr (C) E/dr (D) Zero

Ans: (A) E·dr Explanation: Work done by external agent against field = −(work done by field) = −(−E·dr) = E·dr. This work increases the potential energy of the charge. This is also how potential difference is defined: dV = E·dr.

Q48. Two capacitors 6μF and 3μF are in series with a 9V battery. Energy stored in 6μF capacitor is: (A) 6 μJ (B) 12 μJ (C) 3 μJ (D) 18 μJ

Ans: (B) 12 μJ Explanation: C_eq = (6×3)/(6+3) = 2μF. Q = C_eq × V = 2×10⁻⁶ × 9 = 18μC (same on both). V across 6μF = Q/C = 18/6 = 3V. U = ½CV² = ½ × 6×10⁻⁶ × 9 = 27 μJ. Wait — recalculating: ½ × 6×10⁻⁶ × 3² = ½ × 6×10⁻⁶ × 9 = 27 μJ. Ans: 27μJ. (Rechecked — answer is 27 μJ, a common numerical in NEET.)

Q49. Electric flux through a closed surface depends on: (A) Size of surface (B) Shape of surface (C) Charges outside surface (D) Charges enclosed inside surface

Ans: (D) Charges enclosed inside surface Explanation: By Gauss's Law: φ = Q_enclosed/ε₀. Flux depends ONLY on total charge inside the closed surface. Shape, size, charges outside — none of these affect the total flux. This is the most powerful result in electrostatics.

Q50. A capacitor of 5μF charged to 400V is connected to an uncharged 5μF capacitor. Final voltage is: (A) 400V (B) 300V (C) 200V (D) 100V

Ans: (C) 200V Explanation: Initial Q = CV = 5×10⁻⁶ × 400 = 2000μC. After connection, total capacitance = 5+5 = 10μF (parallel). V_final = Q_total/C_total = 2000μC/10μF = 200V. Charge is conserved but voltage halves. Energy is also halved — lost as heat/radiation.

Quick Revision Summary

Topic	Key Formula
Electric Potential	V = kq/r
Work Done	W = qV
Capacitance	C = Q/V = Kε₀A/d
Series	1/C = 1/C₁ + 1/C₂
Parallel	C = C₁ + C₂
Energy	U = ½CV²
E and V	E = −dV/dr
Energy density	u = ½ε₀E²
Common Potential	V = (C₁V₁+C₂V₂)/(C₁+C₂)
Gauss's Law	φ = Q/ε₀

⚡ Electric Potential & Capacitance — Q51 to Q100

CBSE / NEET / JEE Level | With Answer & Explanation

Q51. The potential difference between two points is 50V. Work done in moving 5C charge between them is:

(A) 10 J — (B) 250 J — (C) 55 J — (D) 45 J

Ans: (B) 250 J

Explanation: W = qV = 5 × 50 = 250 J. Simple and direct application of W = qΔV. Always remember: if charge moves from low to high potential, external agent does positive work.

Q52. Which of the following has highest capacitance?

(A) 3 capacitors of 3μF in series — (B) 3 capacitors of 3μF in parallel — (C) 2 capacitors of 3μF in series — (D) Single 3μF capacitor

Ans: (B) 3 capacitors of 3μF in parallel

Explanation: Parallel gives C = 3+3+3 = 9μF. Series gives C = 1μF. Parallel combination always gives maximum capacitance. This is a concept-based comparison question common in NEET.

Q53. The potential at a point is 20V. The work done in bringing a charge of −2C from infinity to that point is:

(A) +40 J — (B) −40 J — (C) +10 J — (D) −10 J

Ans: (B) −40 J

Explanation: W = qV = (−2) × 20 = −40 J. Negative work means the electric field itself does the work — no external effort needed. Negative charge naturally moves toward higher potential region.

Q54. A capacitor blocks:

(A) AC only — (B) DC only — (C) Both AC and DC — (D) Neither AC nor DC

Ans: (B) DC only

Explanation: A fully charged capacitor blocks DC (steady current) because once charged, no more current flows. It allows AC to pass as it continuously charges and discharges with alternating voltage. This is why capacitors are used as DC blockers in circuits.

Q55. Electric potential is a:

(A) Vector quantity — (B) Scalar quantity — (C) Tensor — (D) Dimensionless quantity

Ans: (B) Scalar quantity

Explanation: Electric potential has magnitude but no direction. It is a scalar. Electric field is a vector. Potential can be added algebraically (with signs), not vectorially. This makes calculation of potential easier than field in complex systems.

Q56. Dimension of electric potential is:

(A) ML²T⁻³A⁻¹ — (B) ML²T⁻²A⁻¹ — (C) MLT⁻²A⁻¹ — (D) ML²T⁻²A

Ans: (A) ML²T⁻³A⁻¹

Explanation: V = W/q. Dimension of W = ML²T⁻². Dimension of q = AT. So V = ML²T⁻²/AT = ML²T⁻³A⁻¹. This is same as Watt/Ampere = Volt. Dimensional analysis is frequently tested in JEE Mains.

Q57. Two equal positive charges are placed at two corners of an equilateral triangle of side a. Potential at third corner is:

(A) kq/a — (B) 2kq/a — (C) kq/2a — (D) 4kq/a

Ans: (B) 2kq/a

Explanation: Each charge is at distance a from the third corner. V = kq/a + kq/a = 2kq/a. Since potential is scalar, simply add the contributions from each charge algebraically. No angle or direction needed.

Q58. Farad is equivalent to:

(A) Coulomb/Volt — (B) Volt/Coulomb — (C) Joule/Coulomb — (D) Coulomb²/Joule

Ans: (A) Coulomb/Volt

Explanation: C = Q/V, so Farad = Coulomb/Volt. Also, 1F = 1C²/J (from U = Q²/2C). 1 Farad is a very large unit — most practical capacitors are in μF, nF, or pF range.

Q59. A parallel plate capacitor with air between plates has capacitance 8pF. If gap is filled with dielectric of K=6, new capacitance is:

(A) 8 pF — (B) 48 pF — (C) 16 pF — (D) 4 pF

Ans: (B) 48 pF

Explanation: C_new = K × C_old = 6 × 8 = 48 pF. When a dielectric completely fills the gap, capacitance increases exactly K times. Dielectric reduces the effective electric field, allowing more charge to be stored at same voltage.

Q60. Potential energy of a charge q in a potential V is:

(A) qV — (B) q/V — (C) q²V — (D) V/q

Ans: (A) qV

Explanation: Potential energy U = qV. This is the energy stored by the charge due to its position in the electric field. For a capacitor, total energy = ½QV because voltage builds from 0 to V during charging (average = V/2).

Q61. When a charged capacitor is disconnected and plate area is doubled, capacitance:

(A) Doubles — (B) Halves — (C) Remains same — (D) Becomes 4 times

Ans: (A) Doubles

Explanation: C = ε₀A/d. If A → 2A, then C → 2C (doubles). Since capacitor is disconnected, Q is constant. As C doubles, V = Q/C halves. Electric field E = V/d also halves. Energy U = Q²/2C halves.

Q62. Three capacitors 1μF, 2μF, 3μF are in series. Equivalent capacitance is:

(A) 6 μF — (B) 11/6 μF — (C) 6/11 μF — (D) 1 μF

Ans: (C) 6/11 μF

Explanation: 1/C = 1/1 + 1/2 + 1/3 = 6/6 + 3/6 + 2/6 = 11/6. So C = 6/11 ≈ 0.545 μF. In series, equivalent is always less than smallest (here 1μF). Finding LCM is the key step in these numericals.

Q63. What is the electric potential at infinity?

(A) Maximum — (B) Minimum — (C) Zero — (D) Depends on charge

Ans: (C) Zero

Explanation: By convention, electric potential at infinity is taken as zero reference point. V = kq/r → 0 as r → ∞. All potential values are measured relative to this zero. This is why potential of a negative charge is negative (less than zero).

Q64. A proton is moved from lower potential to higher potential. Its potential energy:

(A) Increases — (B) Decreases — (C) Remains same — (D) First increases then decreases

Ans: (A) Increases

Explanation: For positive charge, U = qV. Moving to higher V means higher U. It's like pushing a ball uphill — potential energy increases. The proton would naturally move the other way (from high to low potential) if released freely.

Q65. An electron is moved from lower potential to higher potential. Its kinetic energy:

(A) Increases — (B) Decreases — (C) Remains same — (D) Becomes zero

Ans: (A) Increases

Explanation: Electron has negative charge. When moved to higher potential, electric force actually does positive work on it (force is opposite to field for negative charge). By work-energy theorem, KE increases. This seems counterintuitive but is correct.

Q66. In a region where electric potential is constant, the electric field is:

(A) Maximum — (B) Uniform — (C) Zero — (D) Infinite

Ans: (C) Zero

Explanation: E = −dV/dr. If V = constant, dV = 0, so E = 0. Inside a conductor, V is constant → E = 0. This also confirms that inside a conductor there is no electric field in electrostatic equilibrium.

Q67. Which of following is correct for a conductor surface?

(A) E = 0, V = constant — (B) E ≠ 0, V = constant — (C) E = 0, V ≠ constant — (D) Both E and V vary

Ans: (B) E ≠ 0, V = constant

Explanation: On the surface of a conductor, potential is constant (same as interior) BUT electric field is NOT zero. Surface field E = σ/ε₀ and is perpendicular to the surface. This distinguishes the surface from the interior.

Q68. A point charge +q is placed at center of a hollow conducting sphere. Charge on inner surface is:

(A) +q — (B) −q — (C) Zero — (D) +2q

Ans: (B) −q

Explanation: By electrostatic induction, the inner surface acquires −q (attracted toward +q inside). Since total conductor charge = 0, outer surface gets +q. Electric field inside conductor wall = 0 always. This is the basis of electrostatic shielding.

Q69. Electrostatic shielding means:

(A) Shielding electric potential — (B) Protecting interior from external electric field — (C) Blocking magnetic field — (D) Preventing charge leakage

Ans: (B) Protecting interior from external electric field

Explanation: A hollow conductor completely shields its interior from external electric fields. Whatever field exists outside, E = 0 inside cavity. Used in sensitive electronic instruments, Faraday cages, and microwave ovens (mesh door). Not effective for magnetic shielding.

Q70. Capacitance depends on:

(A) Charge on capacitor — (B) Voltage applied — (C) Geometry of conductors and medium — (D) Material of plates

Ans: (C) Geometry of conductors and medium

Explanation: C = Kε₀A/d — it depends on area A, separation d, and dielectric K. NOT on Q or V individually (though C = Q/V, the ratio is constant). Changing Q or V does not change C — only changing geometry or dielectric does.

Q71. The electric field between plates of a charged parallel plate capacitor:

(A) Is zero — (B) Is uniform — (C) Varies linearly — (D) Is maximum at centre

Ans: (B) Is uniform

Explanation: E = σ/ε₀ = V/d between parallel plates — completely uniform. This is why parallel plate capacitors are used to study uniform electric fields. At edges, field becomes non-uniform (fringing effect), but this is ignored in standard problems.

Q72. A 100V battery charges a 2μF capacitor. Energy supplied by battery is:

(A) 0.01 J — (B) 0.02 J — (C) 0.04 J — (D) 0.005 J

Ans: (B) 0.02 J

Explanation: Energy supplied by battery = QV = CV² = 2×10⁻⁶ × 100² = 0.02 J. But energy stored in capacitor = ½CV² = 0.01 J. Half the energy (0.01J) is lost as heat in connecting wires. This 50% loss always occurs during charging — a classic JEE concept.

Q73. When two identical capacitors charged to different voltages are connected in parallel, total energy:

(A) Is conserved — (B) Increases — (C) Decreases — (D) Remains same as larger one

Ans: (C) Decreases

Explanation: Energy is always lost when capacitors at different potentials are connected. Loss = C₁C₂(V₁−V₂)²/2(C₁+C₂). This energy is dissipated as heat. Only if V₁ = V₂ is energy conserved. This is analogous to inelastic collision where KE is lost.

Q74. The ratio of charge to potential (Q/V) gives:

(A) Electric field — (B) Capacitance — (C) Resistance — (D) Energy

Ans: (B) Capacitance

Explanation: By definition, C = Q/V. Capacitance is the ability to store charge per unit voltage. Higher capacitance means more charge stored for same voltage. It is always positive and depends only on geometry.

Q75. Potential due to a line charge at perpendicular distance r is proportional to:

(A) 1/r — (B) 1/r² — (C) ln(r) — (D) r

Ans: (C) ln(r)

Explanation: For infinite line charge, E ∝ 1/r. Integrating E, V = −∫E dr ∝ −ln(r) + constant. So V varies logarithmically with distance. This is different from point charge (V ∝ 1/r) and is important for JEE Advanced.

Q76. A metal plate is inserted between plates of a capacitor (without touching). Effective capacitance:

(A) Decreases — (B) Increases — (C) Remains same — (D) Becomes zero

Ans: (B) Increases

Explanation: Metal plate (thickness t) reduces effective separation to (d−t). New C = ε₀A/(d−t) > original C. Metal has K = ∞ effectively. So inserting metal always increases capacitance. The plate gets induced charges and acts like part of both main plates.

Q77. In a parallel plate capacitor, if both plate area and separation are doubled, capacitance:

(A) Doubles — (B) Halves — (C) Remains same — (D) Becomes 4 times

Ans: (C) Remains same

Explanation: C = ε₀A/d. New C = ε₀(2A)/(2d) = ε₀A/d = same. Area and separation both double → effects cancel out. This is a tricky question — many students choose "doubles." Always substitute and check.

Q78. The capacitance of a parallel plate capacitor is 2μF with air. With dielectric K=5 filling half the gap (in series), new capacitance is:

(A) 10/6 μF — (B) 10/3 μF — (C) 5 μF — (D) 2 μF

Ans: (B) 10/3 μF

Explanation: Half gap: two capacitors in series — C₁ = Kε₀A/(d/2) = 2KC and C₂ = ε₀A/(d/2) = 2C (in terms of original C = ε₀A/d = 2μF). C_eq = (2KC × 2C)/(2KC + 2C) = 2KC/(K+1) = 2×5×2/(5+1) = 20/6 = 10/3 μF.

Q79. Potential at any point on the surface of a charged conductor is:

(A) Zero — (B) Same everywhere on surface — (C) Maximum at sharp points — (D) Minimum at centre

Ans: (B) Same everywhere on surface

Explanation: Entire surface of a conductor is an equipotential surface. If different points had different potentials, charge would flow along surface until equilibrium — which means electrostatic condition requires uniform surface potential.

Q80. The electric field just outside a conductor surface is:

(A) σ/ε₀ perpendicular to surface — (B) σ/2ε₀ parallel to surface — (C) Zero — (D) σ/ε₀ parallel to surface

Ans: (A) σ/ε₀ perpendicular to surface

Explanation: E = σ/ε₀, directed perpendicular (normal) to surface. If it had a component along surface, surface charges would redistribute — contradicting equilibrium. Note: for infinite sheet, field is σ/2ε₀. For conductor surface it is σ/ε₀ (double) because charges exist only on one side.

Q81. Charge distribution on a conductor's outer surface is:

(A) Uniform always — (B) Higher at flat portions — (C) Higher at sharp pointed regions — (D) Zero

Ans: (C) Higher at sharp pointed regions

Explanation: Surface charge density σ ∝ 1/R (curvature). Sharp points (small R) have very high σ and hence very high E. This causes corona discharge (air ionization) at sharp points. Lightning rods work on this principle — they have pointed tips.

Q82. The energy stored in a capacitor is doubled when charge is increased by factor:

(A) 2 — (B) √2 — (C) 4 — (D) 1/√2

Ans: (B) √2

Explanation: U = Q²/2C. If U → 2U, then Q² → 2Q², so Q → Q√2. Charge must increase by √2 times to double the energy. This is because energy is proportional to Q² (not Q linearly). Very common in NEET objective questions.

Q83. Work done in moving a charge q around a closed path in an electric field is:

(A) qE — (B) qV — (C) Zero — (D) 2qV

Ans: (C) Zero

Explanation: Electric force is conservative. Work done over any closed path is always zero: ∮E·dl = 0. This means electric potential energy depends only on start and end points, not on the path taken. This is a fundamental property of electrostatic fields.

Q84. The potential gradient at a point where electric field is 500 V/m is:

(A) 500 V/m — (B) −500 V/m — (C) 250 V/m — (D) 1000 V/m

Ans: (B) −500 V/m

Explanation: E = −dV/dr, so dV/dr = −E = −500 V/m. Potential gradient is negative of electric field. The negative sign shows potential decreases in direction of field. Magnitude of potential gradient = magnitude of E = 500 V/m.

Q85. A 6μF capacitor is charged by 12V battery. Charge stored is:

(A) 2 μC — (B) 18 μC — (C) 72 μC — (D) 6 μC

Ans: (C) 72 μC

Explanation: Q = CV = 6×10⁻⁶ × 12 = 72×10⁻⁶ C = 72 μC. Direct application of Q = CV. Energy stored = ½ × 6×10⁻⁶ × 144 = 432 μJ. Battery supplies double this (864 μJ), half is lost as heat.

Q86. If a charge is placed at centre of a cube, electric flux through one face is:

(A) q/ε₀ — (B) q/6ε₀ — (C) q/8ε₀ — (D) 6q/ε₀

Ans: (B) q/6ε₀

Explanation: By Gauss's Law, total flux through cube = q/ε₀. Cube has 6 equal faces. By symmetry, flux through each face = q/6ε₀. This elegant result comes from symmetry — a top favourite in CBSE board and NEET exams.

Q87. Which statement about electric potential lines is WRONG?

(A) They are always perpendicular to field lines — (B) They can intersect — (C) Work done along them is zero — (D) They are closer in stronger field region

Ans: (B) They can intersect

Explanation: Equipotential surfaces (or lines) can NEVER intersect. If they did, a point would have two different potentials — contradiction. Field lines also never intersect (same reason). This is a common negative-marking trap in MCQs.

Q88. A parallel plate capacitor has charge Q. If a dielectric slab completely fills the gap, the electric field between plates:

(A) Increases K times — (B) Decreases K times — (C) Remains same — (D) Becomes zero

Ans: (B) Decreases K times

Explanation: E = σ/Kε₀ = Q/KAε₀. Since capacitor is isolated (Q constant), field decreases K times. The dielectric's polarization creates an opposing field that reduces net E. This is why dielectrics are used — they reduce field and allow more charge storage.

Q89. Two point charges +q and +q are separated by distance d. The point where potential is minimum between them is:

(A) Closer to smaller charge — (B) At midpoint — (C) No such point exists between them — (D) Closer to larger charge

Ans: (C) No such point exists between them

Explanation: Both charges are positive, so potential is positive everywhere between them. Potential has a minimum somewhere between them (at midpoint for equal charges) but it is NOT zero or negative — it's just the lowest positive value. The question asks for "minimum" — that is at the midpoint.

Corrected Ans: (B) At midpoint — potential is minimum (but still positive) at midpoint for equal charges.

Q90. Capacitance of a capacitor does NOT depend on:

(A) Area of plates — (B) Separation of plates — (C) Charge on plates — (D) Dielectric between plates

Ans: (C) Charge on plates

Explanation: C = Kε₀A/d — depends on K, A, d only. C does NOT depend on Q or V. Doubling Q doubles V too, keeping C = Q/V constant. This is the definition: C is constant for given geometry regardless of how much charge is placed.

Q91. The work done by electric field when a charge moves between two equipotential surfaces of potentials V₁ and V₂ is:

(A) q(V₂−V₁) — (B) q(V₁−V₂) — (C) q(V₁+V₂) — (D) Zero

Ans: (B) q(V₁−V₂)

Explanation: W by field = q(V_initial − V_final) = q(V₁−V₂). If charge moves from high to low potential, field does positive work (like ball rolling downhill). External agent does W = q(V₂−V₁) which is opposite in sign.

Q92. An uncharged capacitor is connected to a battery of EMF E and internal resistance r. At steady state, current through circuit is:

(A) E/r — (B) Zero — (C) E/2r — (D) Infinite

Ans: (B) Zero

Explanation: At steady state, capacitor is fully charged. No more current flows because capacitor blocks DC. Initial current = E/r, but it exponentially decays to zero. At steady state: I = 0, V across capacitor = E. This is fundamental RC circuit behaviour.

Q93. Which of the following cannot be the value of dielectric constant?

(A) 1 — (B) 0.5 — (C) 80 — (D) 2.3

Ans: (B) 0.5

Explanation: K = ε/ε₀ ≥ 1 always. For vacuum K = 1 (minimum). No material can have K < 1 because dielectric always reduces E (increases ε). K = 80 is water, K = 2.3 is paper, K = 1 is vacuum/air. K = 0.5 is physically impossible.

Q94. Potential energy of a system of three charges q each at corners of equilateral triangle of side a is:

(A) kq²/a — (B) 2kq²/a — (C) 3kq²/a — (D) kq²/3a

Ans: (C) 3kq²/a

Explanation: Three pairs of charges: (1,2), (2,3), (1,3). Each pair has U = kq²/a. Total U = 3 × kq²/a = 3kq²/a. In general, for n charges, count all pairs: n(n−1)/2 pairs. For 3 charges = 3 pairs. Important assembly energy problem in JEE.

Q95. If the distance between two charges is halved, force between them becomes:

(A) Half — (B) Double — (C) 4 times — (D) 8 times

Ans: (C) 4 times

Explanation: F = kq₁q₂/r². If r → r/2, then F → kq₁q₂/(r/2)² = 4kq₁q₂/r² = 4F. Force increases 4 times. Potential energy U = kq₁q₂/r also doubles when distance is halved. Note: force changes as 1/r² but potential as 1/r.

Q96. Two capacitors 4μF and 6μF are connected in series to 100V supply. Charge on each capacitor is:

(A) 400 μC — (B) 240 μC — (C) 600 μC — (D) 100 μC

Ans: (B) 240 μC

Explanation: C_eq = (4×6)/(4+6) = 24/10 = 2.4 μF. Q = C_eq × V = 2.4×10⁻⁶ × 100 = 240 μC. In series, same charge (240μC) appears on both capacitors. Voltage splits: V across 4μF = 60V, V across 6μF = 40V. Check: 60+40 = 100V ✓

Q97. The electric potential due to a uniformly charged ring at its centre is:

(A) Zero — (B) kQ/R — (C) kQ/R² — (D) kQR

Ans: (B) kQ/R

Explanation: Every element of ring is at same distance R from centre. V = kQ/R (scalar sum). But electric field at centre = 0 by symmetry (field contributions from opposite elements cancel). Classic example: V ≠ 0 but E = 0.

Q98. On moving from A (V = 30V) to B (V = 20V), the kinetic energy of a positive charge:

(A) Increases — (B) Decreases — (C) Remains same — (D) First increases then decreases

Ans: (A) Increases

Explanation: Positive charge moves from high (30V) to low (20V) potential — this is the natural direction of electric force. Work by field = q(30−20) = 10q > 0. By work-energy theorem, KE increases. It's like the charge rolling downhill naturally.

Q99. A capacitor of capacitance C is given charge Q. The energy stored if only half the charge leaks away is:

(A) Q²/8C — (B) Q²/4C — (C) Q²/2C — (D) Q²/C

Ans: (A) Q²/8C

Explanation: Remaining charge = Q/2. Energy = (Q/2)²/2C = Q²/(4×2C) = Q²/8C. Original energy was Q²/2C, so energy is now 1/4 of original. Energy reduces as square of charge — losing half the charge loses 3/4 of the energy. Important insight!

Q100. The principle of superposition of potentials states:

(A) Potential vectors add up — (B) Potentials add algebraically with signs — (C) Only magnitudes add — (D) Potentials cancel each other

Ans: (B) Potentials add algebraically with signs

Explanation: Since potential is a scalar, total V = V₁ + V₂ + V₃ + ... (simple algebraic addition with + and − signs). This is much simpler than electric field superposition (which requires vector addition). This is the biggest advantage of working with potential instead of field.

Quick Revision Table Q51–Q100

Concept	Key Point
W = qΔV	Always check sign of q and ΔV
Series C	Always less than smallest C
Parallel C	Always more than largest C
Isolated capacitor	Q = constant
Battery connected	V = constant
E inside conductor	Always zero
V inside conductor	Equals surface potential
K (dielectric)	Always ≥ 1
Energy loss on sharing	Always occurs if V₁ ≠ V₂
Conservative field	Work in closed loop = 0

⚡ Electric Potential & Capacitance

Theory Questions 101–150 | CBSE / NEET / JEE Level | With Answer & Explanation

Q101. What is electric potential? Write its SI unit.

Ans: Electric potential at a point is defined as the work done per unit positive charge in bringing a test charge from infinity to that point.

V = W/q

SI unit is Volt (V) = Joule/Coulomb.

Explanation: Potential tells us about the "electric level" at a point — just like height tells us gravitational level. A point at higher potential means more work was needed to bring positive charge there. Potential is a scalar quantity, so it has no direction.

Q102. What is the difference between electric potential and electric potential energy?

Ans:

Electric Potential (V) = Work done per unit charge = W/q (unit: Volt)
Electric Potential Energy (U) = Work done to bring charge q to that point = qV (unit: Joule)

Explanation: Potential is a property of the point in space (independent of test charge). Potential energy is a property of the charge-field system. Example: A point may have V = 100V. A 2C charge placed there has U = 200J, while a 3C charge has U = 300J. V remains 100V in both cases.

Q103. Define equipotential surface. Give two examples.

Ans: A surface on which electric potential is the same at every point is called an equipotential surface. No work is done in moving a charge along this surface.

Examples:

Sphere around a point charge (spherical equipotential surfaces)
Flat planes perpendicular to uniform electric field

Explanation: Equipotential surfaces and field lines are always perpendicular to each other. They are closer together where field is stronger (like contour lines on a map). No work is done on them because E ⊥ displacement, so W = qEd cos90° = 0.

Q104. Why do electric field lines never intersect?

Ans: If two field lines intersected, there would be two different directions of electric field at that point — which is physically impossible. At any point, the electric field has only one unique direction (tangent to field line).

Explanation: Same logic applies to equipotential surfaces — they never intersect. If two equipotential surfaces intersected, a point would have two different potential values, which is a contradiction. Both field lines and equipotential surfaces are unique at every point.

Q105. State the relation between electric field and electric potential. Explain the negative sign.

Ans: E = −dV/dr

Electric field is the negative gradient (rate of change) of electric potential.

Explanation: The negative sign means electric field points in the direction of decreasing potential — from high potential to low potential. Just like water flows downhill (decreasing height), positive charges naturally move in direction of decreasing potential. This is why field lines go from + to − charge (+ is high potential, − is low potential).

Q106. What is an equipotential surface for a dipole? Draw and explain.

Ans: For an electric dipole, the equipotential surfaces are not simple spheres. They are distorted and asymmetric shapes:

Near + charge: surfaces crowd around + charge
Near − charge: surfaces crowd around − charge
At equatorial plane: V = 0 (this entire plane is one equipotential surface)
The equipotential surfaces are always perpendicular to the field lines of the dipole

Explanation: The V = 0 equipotential plane passes through the middle of the dipole, perpendicular to the dipole axis. This is why potential at equatorial point is always zero. The field lines and equipotential surfaces together give a complete picture of the dipole field.

Q107. What is a capacitor? On what principle does it work?

Ans: A capacitor is a device used to store electric charge and electrical energy. It consists of two conductors (plates) separated by an insulating medium (dielectric).

Principle: When a grounded conductor is placed near a charged conductor, the capacitance of the charged conductor increases enormously. This is because the induced negative charges on the nearby conductor reduce the potential of the charged conductor, allowing more charge to be stored at the same potential.

Explanation: C = Q/V. If V decreases (due to induced charges), more Q can be stored for same applied voltage. This is the entire working principle of a capacitor — using nearby conductors to increase charge storage capacity.

Q108. Derive the expression for capacitance of a parallel plate capacitor.

Ans: Consider two parallel plates of area A, separation d, with surface charge density σ.

Step 1: Charge on each plate = Q = σA

Step 2: Electric field between plates = E = σ/ε₀ = Q/ε₀A

Step 3: Potential difference = V = Ed = Qd/ε₀A

Step 4: Capacitance = C = Q/V = Q/(Qd/ε₀A)

C = ε₀A/d

With dielectric: C = Kε₀A/d

Explanation: Capacitance increases with larger plate area (more surface to store charge) and decreases with larger separation (weaker induction effect). Dielectric K increases capacitance by reducing effective field between plates.

Q109. What happens to capacitance, charge, voltage and energy when a dielectric is inserted in: (a) Isolated capacitor (b) Battery-connected capacitor?

Ans:

(a) Isolated Capacitor (Q = constant):

Capacitance C → increases (KC)
Charge Q → constant
Voltage V = Q/C → decreases (V/K)
Electric field E → decreases (E/K)
Energy U = Q²/2C → decreases (U/K)

(b) Battery Connected (V = constant):

Capacitance C → increases (KC)
Voltage V → constant
Charge Q = CV → increases (KQ)
Electric field E = V/d → constant
Energy U = ½CV² → increases (KU)

Explanation: The key difference is what stays constant. Battery fixes V, isolation fixes Q. Always identify this first before solving any dielectric problem in NEET/JEE.

Q110. Derive energy stored in a capacitor.

Ans: As a capacitor charges from 0 to final charge Q, at any intermediate stage when charge is q and potential is v = q/C:

Small work done = dW = v·dq = (q/C)dq

Total work done (energy stored):

U = ∫₀Q (q/C)dq = Q²/2C

Since Q = CV:

U = Q²/2C = ½CV² = ½QV

Explanation: The factor ½ appears because voltage builds gradually from 0 to V as capacitor charges. Average voltage = V/2, so average work per unit charge = V/2. Energy is stored in the electric field between plates, not in the plates themselves.

Q111. What is energy density of electric field? Derive its expression.

Ans: Energy density is energy stored per unit volume in the electric field.

For parallel plate capacitor:

U = ½CV² = ½(ε₀A/d)(Ed)² = ½ε₀E²·Ad
Volume between plates = Ad

Energy density u = U/Volume = ½ε₀E²

With dielectric: u = ½Kε₀E² = ½εE²

Explanation: This result is universal — not just for capacitors. Wherever there is an electric field in space, energy is stored with density ½ε₀E². This concept is fundamental to electromagnetic waves, where energy is stored in both E and B fields.

Q112. What is dielectric? Explain polar and non-polar dielectrics.

Ans: A dielectric is an insulating material that can be polarized by an electric field.

Polar Dielectrics:

Molecules have permanent dipole moment even without external field
Example: Water (H₂O), HCl, NH₃
Dipoles are randomly oriented normally
External field aligns them partially

Non-polar Dielectrics:

Molecules have zero dipole moment normally
Example: O₂, N₂, CO₂, benzene
External field induces dipole moment (distorts electron cloud)
Induced dipoles align with field

Explanation: In both cases, when placed in external field E₀, the dielectric develops surface charges that create an opposing field E_p. Net field = E₀ − E_p = E₀/K. This reduction in field is why K > 1 always for dielectrics.

Q113. What is dielectric polarization? Define polarization vector P.

Ans: Dielectric polarization is the process by which electric dipoles in a dielectric align themselves along an external electric field, creating a net dipole moment per unit volume.

Polarization vector P = Net dipole moment per unit volume = p/V

Unit: C/m²

P = ε₀(K−1)E = χₑε₀E

where χₑ = (K−1) is called electric susceptibility.

Explanation: Larger K means stronger polarization. For K = 1 (vacuum), P = 0 (no polarization). For water (K = 80), polarization is very strong. The surface charge density due to polarization = P (bound charges). These bound charges reduce the net electric field inside dielectric.

Q114. Explain the effect of inserting a conducting slab in a parallel plate capacitor.

Ans: When a conducting slab of thickness t is inserted between plates (separation d):

Slab is an equipotential volume
Effective separation reduces to (d − t)
New capacitance: C' = ε₀A/(d−t)
Since (d−t) < d, C' > C (capacitance increases)

Special case: If t = d (slab fills entire gap), C → ∞ (plates are effectively connected — not a capacitor anymore)

Explanation: A conducting slab is equivalent to a dielectric with K = ∞. The slab develops induced charges on its faces which exactly cancel the field inside it. The electric field exists only in the remaining air gap (d−t). This is a frequently tested JEE concept.

Q115. What is Van de Graaff generator? Explain its working.

Ans: Van de Graaff generator is a device that generates very high electrostatic potential (millions of volts) using the following principles:

Action of sharp points (corona discharge)
Electrostatic induction
Property that charge on inner surface transfers to outer surface

Working:

A large hollow metallic sphere is mounted on insulating pillars
A long insulating belt runs between two pulleys
At lower end, a comb (spray brush) at high potential sprays charge onto belt
Belt carries charge up to inner surface of sphere
Upper comb transfers charge from belt to inner surface
By induction, charge moves to outer surface of sphere
Process repeats — potential builds up to millions of volts

Use: Accelerating charged particles in nuclear physics experiments.

Q116. State and prove that work done in moving a charge in a closed loop in electric field is zero.

Ans: Electric force is a conservative force.

For any closed path: W = q∮E·dl = 0

Proof: Consider moving charge from A to B via path 1 and returning via path 2.

Work from A to B (path 1) = W₁

Work from B to A (path 2) = W₂

For conservative field: W₁ + W₂ = 0

This means W₁ = −W₂ → work depends only on endpoints, not path.

This gives us: ∮E·dl = 0 (Electrostatic field is irrotational)

Explanation: This is why we can define potential — if work depended on path, potential would not be well-defined. Magnetic force is NOT conservative (that's why we cannot define magnetic potential easily). This property is used in Kirchhoff's Voltage Law.

Q117. Explain the principle of electrostatic shielding.

Ans: Electrostatic shielding is the phenomenon of protecting a region from external electric fields by enclosing it in a hollow conductor.

Principle:

E = 0 inside any conductor in electrostatic equilibrium
A hollow conductor completely shields its interior from external fields
Whatever the external field, no field penetrates inside the cavity

Applications:

Sensitive electronic instruments are enclosed in metallic cases
Faraday cage — protects from lightning
Coaxial cables — outer conductor shields inner wire from external interference
Microwave oven door mesh — prevents leakage of microwaves

Limitation: Cannot shield from gravitational field or magnetic field (magnetic shielding needs special materials like mu-metal).

Explanation: When external field is applied to a conductor, charges rearrange on surface such that internal field is cancelled. This happens almost instantaneously. The shielding is perfect for electrostatic fields but not for rapidly changing fields.

Q118. Derive expression for potential on the axis of an electric dipole.

Ans: Let dipole have charge +q and −q separated by 2a. Consider point P on axis at distance r from centre.

Distance of P from +q = (r − a) Distance of P from −q = (r + a)

V = k[q/(r−a) + (−q)/(r+a)]

V = kq[(r+a−r+a)/(r²−a²)]

V = kq[2a/(r²−a²)]

Since p = q×2a and for r >> a, r²−a² ≈ r²:

V = kp/r²

Explanation: Potential on axis is kp/r² (not zero). It varies as 1/r². Compare: at equatorial point V = 0. At general point: V = kp cosθ/r². The axial point has θ = 0°, giving maximum potential. Equatorial has θ = 90°, giving V = 0.

Q119. Derive expression for potential at equatorial point of electric dipole.

Ans: At equatorial point P at distance r from centre:

Both +q and −q are at equal distance = √(r² + a²) from P.

V₊ = kq/√(r²+a²)

V₋ = −kq/√(r²+a²)

V_total = V₊ + V₋ = 0

Explanation: At equatorial point, both charges are equidistant, so their potential contributions are equal and opposite. They cancel exactly. But note: electric field at equatorial point is NOT zero — it equals kp/r³ directed opposite to dipole moment. This is a very important distinction — V = 0 does not mean E = 0.

Q120. What is Gauss's Law? State it mathematically.

Ans: Gauss's Law states that the total electric flux through any closed surface (Gaussian surface) is equal to the total charge enclosed divided by ε₀.

φ = ∮E·dA = Q_enclosed/ε₀

Key Points:

Closed surface is called Gaussian surface
Shape and size of surface do not matter
Charges outside the surface contribute zero net flux
Works for any charge distribution

Explanation: This is one of the four Maxwell's equations. It is the electrostatic equivalent of saying "what goes in must come out" — field lines that start inside the surface must come out through it. A negative charge "sucks in" field lines. Gauss's Law is most useful for symmetric charge distributions.

Q121. What is the potential of Earth? Why is Earth taken as zero potential reference?

Ans: Earth is taken as the reference zero potential (V_Earth = 0V).

Reason:

Earth is a huge conductor
Its capacitance is very large (≈ 711μF)
Adding or removing a small amount of charge does not change its potential significantly
It acts as an infinite reservoir of charge

Explanation: Just like sea level is taken as zero height reference, Earth is zero potential reference. Any conductor connected to Earth (earthed/grounded) acquires zero potential. In practice, Earth's actual charge is negative (about −500,000C) but its potential is defined as 0V by convention.

Q122. Explain the concept of common potential when two capacitors are connected.

Ans: When two charged capacitors C₁ (voltage V₁) and C₂ (voltage V₂) are connected together:

Conservation of charge: Q_total = C₁V₁ + C₂V₂

Total capacitance = C₁ + C₂

Common potential: V_common = (C₁V₁ + C₂V₂)/(C₁ + C₂)

Energy lost: ΔU = C₁C₂(V₁−V₂)²/[2(C₁+C₂)]

This energy is always positive and is lost as heat.

Explanation: This is similar to mixing two liquids at different temperatures — final temperature is the weighted average. Energy is always lost (like in inelastic collision). The only way to avoid loss is if V₁ = V₂ (already at same potential).

Q123. Why is the electric field inside a conductor zero in electrostatic equilibrium?

Ans: If electric field E were non-zero inside a conductor, the free electrons would experience force F = eE and keep moving. This would mean current flows, which contradicts electrostatic (static) condition.

In equilibrium: no current → no net force on electrons → E = 0 inside.

Mathematical reasoning:

Conductor has free electrons
Any external field causes electron movement
Electrons rearrange until they create equal and opposite field
Net internal field = External field − Field by rearranged charges = 0

Explanation: This takes some time (relaxation time ≈ 10⁻¹⁷ seconds for metals — essentially instantaneous). After this, all charges are on surface, interior is field-free. This is not true for semiconductors and insulators.

Q124. What is a Faraday cage? Give examples of its application.

Ans: A Faraday cage is a hollow conducting enclosure that blocks external electric fields from entering the interior.

Working: When external field is applied, charges redistribute on outer surface to cancel the field inside perfectly.

Applications:

Cars protect passengers during lightning strike
Elevators and metal buildings reduce mobile signals
Microwave oven metal mesh door
MRI rooms are Faraday caged
Sensitive laboratory equipment cases
Submarine cables use conducting sheath
Airplane fuselage protects avionics from lightning

Note: Faraday cage does NOT protect from magnetic fields — only electric fields.

Q125. Explain how a capacitor stores energy. Where is the energy actually stored?

Ans: When a capacitor is charged, work is done against the repulsion of already-stored charges. This work is stored as potential energy.

Storage location: Energy is stored in the electric field between the plates (not in the plates or charges themselves).

Energy density = ½ε₀E² (energy per unit volume in field)

Total energy = ½ε₀E² × (volume between plates) = ½ε₀E² × (Ad) = ½CV²

Explanation: This understanding is critical for electromagnetic waves — energy is carried by oscillating E and B fields through empty space. Capacitor energy is purely in E field. Inductor energy is purely in B field. EM wave energy is shared equally between E and B fields.

Q126. Explain series combination of capacitors and derive equivalent capacitance.

Ans: In series combination, capacitors are connected end-to-end. Same charge Q appears on each capacitor.

V = V₁ + V₂ + V₃

Q/C_eq = Q/C₁ + Q/C₂ + Q/C₃

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃

Properties:

Same charge on each: Q₁ = Q₂ = Q₃ = Q
Voltage divides: V₁:V₂:V₃ = 1/C₁ : 1/C₂ : 1/C₃
Equivalent C < smallest individual C
Smaller capacitor gets larger voltage share

Use: Used when higher voltage tolerance is needed (voltage distributes across capacitors).

Q127. Explain parallel combination of capacitors and derive equivalent capacitance.

Ans: In parallel, all capacitors share same potential V. Charges add up.

Q_total = Q₁ + Q₂ + Q₃

C_eq × V = C₁V + C₂V + C₃V

C_eq = C₁ + C₂ + C₃

Properties:

Same voltage across each: V₁ = V₂ = V₃ = V
Charge distributes: Q₁:Q₂:Q₃ = C₁:C₂:C₃
Equivalent C > largest individual C
More capacitance means more charge stored

Use: Used when more charge storage (capacitance) is needed at same voltage.

Q128. What is the significance of negative electric potential?

Ans: Negative electric potential at a point means that a negative charge creates that potential, or the point is at lower potential than the reference (infinity).

V = kq/r. If q is negative, V is negative.
Negative V means work done by external agent is negative (field does positive work) in bringing positive charge there.
A positive charge placed at negative potential has negative potential energy — meaning it is in a bound state (attracted).

Explanation: An electron near a proton is at negative potential — this is why it is bound (needs energy to escape). Ionization energy is the energy needed to bring electron from negative potential to zero (infinity). This concept is fundamental to atomic physics and chemistry.

Q129. Explain the concept of bound charges and free charges in a dielectric.

Ans:

Free charges: Charges that can move freely through material (electrons in metals, ions in solution). They cause conduction current.

Bound charges: Charges that are bound to atoms/molecules and cannot move freely. They appear on the surface of a dielectric when polarized.

When dielectric is polarized in external field E₀:

Surface bound charge density = σ_b = P (polarization)
These bound charges create field E_p opposing E₀
Net field inside = E = E₀ − E_p = E₀/K

Explanation: Bound charges are like prisoners — they shift position slightly when field is applied (creating dipoles) but cannot leave the material. Their surface accumulation is what creates the opposing field and increases capacitance.

Q130. What is electric susceptibility? How is it related to dielectric constant?

Ans: Electric susceptibility (χₑ) measures how easily a dielectric gets polarized in an external electric field.

P = ε₀χₑE

Relation with dielectric constant K:

K = 1 + χₑ

or χₑ = K − 1

Explanation:

For vacuum: K = 1, so χₑ = 0 (cannot be polarized — no molecules)
For water: K = 80, so χₑ = 79 (highly polarizable)
Higher χₑ means material polarizes more easily
χₑ is dimensionless and always positive (K ≥ 1)

Q131. State the superposition principle for electric potential.

Ans: The total electric potential at a point due to a system of charges is the algebraic sum of potentials due to individual charges.

V_total = V₁ + V₂ + V₃ + ... = Σ kqᵢ/rᵢ

Advantage over field superposition:

Potential is scalar → simple algebraic addition
Field is vector → requires vector addition (components)
Potential calculation is much simpler for complex charge distributions

Explanation: This is why potential is often preferred over field in complex problems. For a system of 10 charges, finding field at a point needs 10 vector additions. Finding potential needs only 10 scalar additions. NEET and JEE frequently test this advantage.

Q132. What is the potential at the centre of a square having charges at its corners?

Ans: Let square have side a. Charges q₁, q₂, q₃, q₄ at corners. Distance from each corner to centre = a√2/2 = a/√2.

V_centre = k(q₁ + q₂ + q₃ + q₄)/(a/√2) = k√2(q₁+q₂+q₃+q₄)/a

Special cases:

All charges equal +q: V = 4kq√2/a
Alternate +q and −q: V = 0 (equal positive and negative cancel)
Three +q and one −q: V = 2kq√2/a

Explanation: Since potential is scalar, just add all contributions. All corners are equidistant from centre, so distance factor is same for all. Only the charge values matter. This is a very common NEET/JEE board question.

Q133. Explain why a bird sitting on a high voltage wire is safe but a man touching it gets a shock.

Ans: Bird on wire:

Bird's both feet are at same potential (say 11000V)
No potential difference exists across bird's body
No current flows (I = ΔV/R, if ΔV = 0, I = 0)
Bird is safe

Man touching wire:

One hand touches wire (11000V)
Feet touch ground (0V)
Potential difference = 11000V across body
Huge current flows through body
Fatal electric shock

Explanation: It is always the potential difference (not absolute potential) that drives current and causes shock. Even 11000V is harmless if there's no path for current to flow. This is why electricians work with one hand in their pocket — to avoid creating a path across their body.

Q134. What is the effect of temperature on dielectric constant?

Ans:

For polar dielectrics (like water):

K decreases with increasing temperature
At higher T, thermal agitation randomizes dipole alignment
Less net polarization → smaller K

For non-polar dielectrics:

K is relatively independent of temperature
Induced dipoles don't depend on thermal motion

Practical implication:

Capacitors using polar dielectrics are temperature-sensitive
This is why ceramic (non-polar) capacitors are more stable
Electrolytic capacitors (polar) should not overheat

Explanation: This is why the K of water drops from 80 at 20°C to about 55 at 100°C. The random thermal motion fights against the aligning effect of the electric field. More heat = more randomness = less net alignment = less polarization = lower K.

Q135. Derive the expression for potential energy of a system of two point charges.

Ans: Let charges q₁ and q₂ be separated by distance r.

Step 1: q₁ is brought from infinity to its position. Work done = 0 (no field yet). U₁ = 0.

Step 2: q₂ is brought from infinity to distance r from q₁. It moves against (or with) field of q₁.

Work done = W = q₂ × V₁ = q₂ × (kq₁/r)

U = kq₁q₂/r

If q₁q₂ > 0 (same sign): U > 0 (need to do work to assemble — repulsive system) If q₁q₂ < 0 (opposite sign): U < 0 (energy is released — attractive, bound system)

Explanation: Negative potential energy means the system is bound — you need to supply energy to separate the charges. This is exactly like gravitational PE being negative (bound system). The more negative U is, the more tightly the system is bound.

Q136. What is meant by "1 electron volt" (eV)? How is it related to Joule?

Ans: One electron volt (eV) is the kinetic energy gained by one electron when accelerated through a potential difference of 1 volt.

1 eV = e × V = 1.6×10⁻¹⁹ C × 1V = 1.6×10⁻¹⁹ J

Other units:

1 keV = 10³ eV = 1.6×10⁻¹⁶ J
1 MeV = 10⁶ eV = 1.6×10⁻¹³ J
1 GeV = 10⁹ eV = 1.6×10⁻¹⁰ J

Explanation: eV is used in atomic and nuclear physics because Joule is too large for atomic-scale energies. X-ray photons have energies of keV. Nuclear reactions involve MeV. The rest mass energy of a proton = 938.3 MeV. This unit is essential for NEET (biology + physics overlap) and JEE Advanced nuclear physics.

Q137. Explain why capacitance of a conductor increases when another grounded conductor is brought near it.

Ans: When grounded conductor B is brought near charged conductor A:

Step 1: Positive charge on A induces negative charge on near side of B and positive on far side.

Step 2: Far side positive charge flows to Earth (B is grounded).

Step 3: Only negative induced charge remains on B.

Step 4: This negative charge on B reduces the potential of A.

Step 5: Lower V with same Q means C = Q/V increases.

Explanation: This is the exact principle of a capacitor! One plate is never grounded in practice, but the presence of the second plate (even ungrounded) dramatically increases capacitance. Without the second plate, a conductor of 1cm radius has capacitance ≈ 1pF. With a nearby plate, same conductor can have μF capacitance.

Q138. What is the difference between capacitance and capacity of a capacitor?

Ans: In everyday language, both words are used interchangeably, but technically:

Capacitance: The ratio C = Q/V. It measures the ability to store charge per unit voltage. It depends only on geometry and medium (not on Q or V). Unit: Farad.

Capacity (charge capacity): The maximum charge a capacitor can hold before the dielectric breaks down (dielectric breakdown). Related to dielectric strength.

Dielectric strength: Maximum electric field a dielectric can withstand before it breaks down and becomes conducting. Unit: V/m.

Explanation: A capacitor may have high capacitance (geometrically) but low charge capacity (weak dielectric). For example, air has K = 1 (low capacitance) but high dielectric strength. Designing capacitors requires balancing both.

Q139. What happens to a capacitor if the dielectric between its plates breaks down?

Ans: Dielectric breakdown occurs when the electric field exceeds the dielectric strength of the material.

What happens:

Dielectric becomes conducting (loses insulating property)
Charges flow from one plate to other through dielectric
Capacitor discharges suddenly
Generates heat, spark, or permanent damage
Dielectric material may be permanently damaged (charred, punctured)

Prevention:

Use capacitor below its rated voltage
Choose dielectric with high dielectric strength
Maintain operating temperature within limits

Explanation: This is why capacitors have a maximum voltage rating (printed as e.g., "100V, 10μF"). Exceeding this causes breakdown. The dielectric strength of air is 3×10⁶ V/m. For mica it is 10⁸ V/m — much safer. This is why mica capacitors are used in high-voltage applications.

Q140. Derive the expression for loss of energy when two capacitors are connected.

Ans: Let C₁ be at V₁ and C₂ at V₂ (V₁ > V₂).

Initial energy: Uᵢ = ½C₁V₁² + ½C₂V₂²

After connection, common potential: V = (C₁V₁ + C₂V₂)/(C₁+C₂)

Final energy: Uf = ½(C₁+C₂)V²

Energy lost: ΔU = Uᵢ − Uf = C₁C₂(V₁−V₂)²/[2(C₁+C₂)]

This is always positive (since it's a perfect square numerator) so energy is ALWAYS lost.

Explanation: Energy is lost as heat in connecting wires. Even for zero resistance wire, energy is lost through electromagnetic radiation. This is analogous to perfectly inelastic collision where KE is always lost. The only way ΔU = 0 is if V₁ = V₂ (like elastic collision requires equal masses and velocities).

Q141. What is meant by charge sharing between capacitors?

Ans: When two capacitors at different potentials are connected, charge flows from higher potential to lower potential until both reach the same common potential. This redistribution is called charge sharing.

Key results:

Total charge is conserved: Q₁ + Q₂ = Q₁' + Q₂'
Final common potential: V = (C₁V₁ + C₂V₂)/(C₁+C₂)
Energy is NOT conserved (some is always lost as heat)
Larger capacitor retains more charge after sharing

Explanation: Charge sharing is like mixing two liquids — total volume conserved but energy (temperature) is partially lost. The analogy: C = thermal capacity, V = temperature, Q = heat content. Final temperature = weighted average. This analogy is perfect and helps remember the formula.

Q142. What is the potential of a charged soap bubble?

Ans: A soap bubble is a thin conducting shell (two surfaces). When charge Q is given to bubble of radius R:

All charge resides on outer surface:

V = kQ/R

When bubble expands (radius increases to R'):

Since charge Q is conserved and V = kQ/R, as R increases, V decreases.

Energy analysis:

When charged bubble expands: C increases, Q constant, V decreases, Energy decreases
Extra energy goes into surface energy (expanding the bubble)

Explanation: Charging a soap bubble makes it expand because like charges on surface repel outward. This is a favourite NEET concept question. The bubble reaches new equilibrium when electric repulsion pressure balances excess atmospheric + surface tension pressure.

Q143. What is meant by electric potential due to a continuous charge distribution?

Ans: For a continuous charge distribution (line, surface, or volume), we cannot use point charge formula directly. Instead:

Divide distribution into small elements dq, each treated as point charge:

dV = k·dq/r

Total potential: V = ∫k·dq/r = k∫dq/r

For different distributions:

Line charge: dq = λ·dl (λ = linear charge density, C/m)
Surface charge: dq = σ·dA (σ = surface charge density, C/m²)
Volume charge: dq = ρ·dV (ρ = volume charge density, C/m³)

Explanation: Integration is needed because each element is at different distance r from the field point. Since V is scalar, no direction needs to be tracked — much simpler than finding E by integration (which requires vector components). This is a JEE-level concept.

Q144. What is the physical significance of negative potential energy in electrostatics?

Ans: Negative potential energy means the system is in a bound state — energy must be supplied from outside to separate the charges to infinity.

Examples:

Electron around proton in hydrogen atom: U = −ke²/r (negative — bound state). Energy needed to ionize = 13.6 eV
Two opposite charges −q and +q: U = −kq²/r (bound, attractive system)
Like charges: U = +kq²/r (positive — not bound, repulsive, charges fly apart if released)

Explanation: By convention, U = 0 at infinity. If two charges attract, they naturally come together (releasing energy → U decreases → becomes negative). If they repel, you must push them together (U increases → positive). The more negative U is, the more stable (more tightly bound) the system is.

Q145. Explain the working of a capacitor in an AC circuit.

Ans: In AC circuit, voltage alternates direction sinusoidally. The capacitor responds as follows:

Half cycle 1 (positive voltage): Capacitor charges — positive plate on top, negative on bottom.

Half cycle 2 (negative voltage): Capacitor reverses — charges and discharges in opposite direction.

Effect: Capacitor continuously charges and discharges, causing current to flow back and forth — AC passes through.

Capacitive reactance: Xc = 1/(2πfC) — opposition to AC

Higher frequency → lower Xc → easier passage
Lower frequency → higher Xc → harder passage
f = 0 (DC) → Xc = ∞ → completely blocked

Explanation: This is why capacitors are used as high-pass filters — they pass high frequency AC easily but block low frequency and DC. Coupling capacitors in amplifiers, bypass capacitors in power supplies — all use this principle.

Q146. What is Kelvin's method of successive induction? (Used in Van de Graaff generator)

Ans: Kelvin's method states that if a charged body is placed inside a hollow conductor connected to ground:

Inner surface gets induced charge (−Q if body has +Q)
Outer surface gets +Q (since conductor is neutral and inner has −Q)
If outer is grounded: +Q flows to earth
If inner ground connection is removed and body removed: inner surface has −Q permanently

By repeating this process, charge builds up on the hollow sphere.

Application in Van de Graaff generator:

Belt continuously brings charge inside sphere
Inner comb neutralizes belt charge (takes it to inner surface)
Charge automatically appears on outer surface
Potential builds up progressively to millions of volts

Q147. What is the difference between electrostatic potential and gravitational potential?

Ans:

Property	Electric Potential	Gravitational Potential
Formula	V = kq/r	Vg = −GM/r
Sign	Can be + or −	Always negative
Nature	Scalar	Scalar
Unit	Volt (J/C)	J/kg
Cause	Charge	Mass
Reference	V = 0 at infinity	Vg = 0 at infinity
Field relation	E = −dV/dr	g = −dVg/dr

Key difference: Gravitational potential is always negative (gravity is always attractive). Electric potential can be positive (due to +ve charge) or negative (due to −ve charge). Mass is always positive, but charge can be positive or negative.

Q148. Explain why electric potential inside a hollow conductor is non-zero even though field is zero.

Ans: This is a common point of confusion.

Inside conductor: E = 0 (always in electrostatic equilibrium)

But V ≠ 0 (it equals the surface potential)

Why?

E = −dV/dr = 0 means V is constant, not zero
V = constant inside (same as surface value)
V = 0 only if conductor is grounded (earthed)

Analogy: Imagine a flat plateau (constant height). The slope (gradient) is zero everywhere on it — but the height is not zero. Similarly, E (gradient of V) = 0, but V itself is not zero.

Explanation: The potential everywhere inside and on the surface of an isolated charged conductor is the same non-zero value. Only an earthed conductor has V = 0. This confusion between E = 0 and V = 0 is the most common mistake in CBSE and NEET exams.

Q149. What are the practical applications of capacitors?

Ans:

1. Energy Storage:

Flash cameras store energy in capacitor, release it instantly
Defibrillators charge capacitor and discharge through patient's heart

2. Filter circuits:

Smooth ripples in DC power supplies (filtering AC component)
Bypass capacitors in amplifier circuits

3. Timing circuits:

RC circuits determine time constants in oscillators and timers

4. Coupling:

Pass AC signal while blocking DC between amplifier stages

5. Tuning:

Variable capacitors tune radio receivers to different frequencies

6. Power factor correction:

Improve power factor in industrial AC systems

7. Memory:

DRAM (Dynamic RAM) stores each bit as charge in a capacitor

Explanation: Capacitors are one of the three fundamental passive components (R, L, C). Their ability to store and release energy quickly makes them invaluable. A camera flash needs huge power (high energy in short time) — capacitor provides this burst that a battery cannot deliver directly.

Q150. Write a note on the limitations of electrostatics.

Ans: Electrostatics deals with charges at rest. Its limitations are:

1. Static charges only:

Cannot describe moving charges (that needs electrodynamics)
As soon as charges move, magnetic effects appear (Maxwell's equations needed)

2. Point charge idealization:

Real charges have finite size — purely point charges don't exist
Leads to infinite self-energy problem for point charges

3. Dielectric breakdown:

Capacitors fail when field exceeds dielectric strength
Electrostatic theory doesn't predict breakdown — needs quantum mechanics

4. No radiation:

Static charges don't radiate energy
Accelerating charges do radiate (Larmor formula) — needs classical electrodynamics

5. Quantum effects ignored:

At atomic scales, quantum mechanics (not classical electrostatics) governs electron behavior
Wave nature of electrons, tunneling, etc. are beyond electrostatics

Explanation: Electrostatics is an excellent approximation for slowly varying or static systems. It is the foundation of all electronics, capacitor technology, and electrostatic applications. But for a complete picture, it must be extended to Maxwell's electrodynamics and eventually quantum electrodynamics (QED).

📋 Complete Theory Summary Table (Q101–Q150)

Topic	Key Formula / Concept
Electric Potential	V = kq/r, unit = Volt
Potential Energy	U = qV = kq₁q₂/r
Field-Potential	E = −dV/dr
Equipotential	Always ⊥ to field lines
Capacitance	C = Q/V = Kε₀A/d
Energy in capacitor	U = ½CV² = Q²/2C
Energy density	u = ½ε₀E²
Dielectric	K = 1+χₑ, K ≥ 1 always
Series	1/C = 1/C₁+1/C₂+...
Parallel	C = C₁+C₂+...
Common potential	V = (C₁V₁+C₂V₂)/(C₁+C₂)
Energy lost on sharing	ΔU = C₁C₂(V₁−V₂)²/2(C₁+C₂)
Inside conductor	E = 0, V = constant
Axial dipole potential	V = kp/r²
Equatorial potential	V = 0
Gauss's Law	φ = Q_enc/ε₀

⚡ Electric Potential & Capacitance

Questions 151–200 | Mixed Theory + Numericals + Application

CBSE / NEET / JEE Level | With Full Explanation

Q151. Two point charges +3μC and −3μC are placed 20cm apart. Find the potential at the midpoint.

Ans: Distance from each charge to midpoint = 10cm = 0.1m

V₁ = k(+3×10⁻⁶)/0.1 = 9×10⁹ × 3×10⁻⁶/0.1 = +270,000 V

V₂ = k(−3×10⁻⁶)/0.1 = −270,000 V

V_total = V₁ + V₂ = 270,000 − 270,000 = 0 V

Explanation: At the midpoint between equal and opposite charges, potential is always zero. Both charges are equidistant, so their contributions are equal and opposite and cancel. However, electric field at this point is NOT zero — it is maximum here and points from +q to −q. This is a classic NEET trap — V = 0 does not mean E = 0.

Q152. A charge of 5μC is moved from point A (V = 400V) to point B (V = 100V). Find work done by electric field.

Ans: Work done by electric field = q(V_A − V_B)

W = 5×10⁻⁶ × (400 − 100)

W = 5×10⁻⁶ × 300

W = 1.5×10⁻³ J = 1.5 mJ

Explanation: Positive charge moves from higher (400V) to lower (100V) potential — this is the natural direction of electric force. So field does positive work. Work done by external agent would be −1.5mJ (opposite sign). Always remember: W_field = q(V_initial − V_final). W_external = q(V_final − V_initial).

Q153. Find the equivalent capacitance of the combination: 3μF and 6μF in parallel, and this combination in series with 2μF.

Ans:

Step 1: 3μF and 6μF in parallel C₁₂ = 3 + 6 = 9μF

Step 2: 9μF in series with 2μF 1/C_eq = 1/9 + 1/2 = 2/18 + 9/18 = 11/18

C_eq = 18/11 ≈ 1.636 μF

Explanation: Always solve from innermost/simplest combination outward. First simplify parallel, then series (or vice versa as required). Parallel always adds directly. Series uses reciprocal formula. Final answer is less than smallest series capacitor (2μF) which confirms series part is correct.

Q154. A parallel plate capacitor has plate area 100cm² and separation 2mm. Find capacitance with air and with glass (K=5).

Ans: A = 100cm² = 100×10⁻⁴ m² = 10⁻² m² d = 2mm = 2×10⁻³ m ε₀ = 8.85×10⁻¹² C²/Nm²

With air: C = ε₀A/d = (8.85×10⁻¹²×10⁻²)/(2×10⁻³) C = 8.85×10⁻¹⁴/2×10⁻³ C = 4.425×10⁻¹¹ F = 44.25 pF

With glass (K=5): C' = KC = 5 × 44.25 = 221.25 pF

Explanation: Unit conversions are critical here. cm² → m² (divide by 10⁴), mm → m (divide by 10³). Always convert to SI before substituting. Dielectric increases capacitance exactly K times (5 times here). This type of numerical is standard in CBSE boards.

Q155. Energy stored in a 20μF capacitor charged to 500V. Also find charge stored.

Ans: C = 20μF = 20×10⁻⁶ F, V = 500V

Charge: Q = CV = 20×10⁻⁶ × 500 = 10⁻² C = 10 mC

Energy: U = ½CV² U = ½ × 20×10⁻⁶ × (500)² U = ½ × 20×10⁻⁶ × 250000 U = ½ × 5 U = 2.5 J

Explanation: Note how much energy 2.5J is — enough to lift a 250g weight by 1 metre. Capacitors can store significant energy and release it very quickly (unlike batteries which release slowly). This makes them useful in defibrillators and camera flashes where burst power is needed.

Q156. What is the potential at the centre of a square of side 1m with charges +1μC, +1μC, −1μC, −1μC at its four corners?

Ans: Distance from each corner to centre = (diagonal/2) = (√2/2) × 1 = 1/√2 m

V = k(q₁+q₂+q₃+q₄)/r

Sum of charges = 1+1−1−1 = 0

V = k × 0 / r = 0 V

Explanation: When total charge is zero and all charges are equidistant from centre, potential is zero. This happens for charge configurations with equal positive and negative charges symmetrically arranged. Note: If arrangement was not symmetric (unequal distances), we would need to calculate each contribution separately even if sum is zero.

Q157. A proton is accelerated through potential difference of 1000V. Find its final kinetic energy and speed.

Ans: Charge of proton = 1.6×10⁻¹⁹ C Mass of proton = 1.67×10⁻²⁷ kg

Kinetic energy gained: KE = qV = 1.6×10⁻¹⁹ × 1000 KE = 1.6×10⁻¹⁶ J = 1000 eV = 1 keV

Speed: ½mv² = 1.6×10⁻¹⁶ v² = 2×1.6×10⁻¹⁶/(1.67×10⁻²⁷) v² = 3.2×10⁻¹⁶/1.67×10⁻²⁷ v² ≈ 1.916×10¹¹ v ≈ 4.38×10⁵ m/s

Explanation: Every charge accelerated through potential difference V gains kinetic energy = qV. This is the working principle of particle accelerators — cyclotrons, synchrotrons, and the Large Hadron Collider all use electric potential difference to accelerate particles. An electron through 1V gains 1eV = 1.6×10⁻¹⁹ J.

Q158. A 4μF capacitor is charged to 100V and a 2μF capacitor is charged to 200V. They are connected with positive plate to positive plate. Find common potential and energy lost.

Ans: Q₁ = C₁V₁ = 4×100 = 400μC Q₂ = C₂V₂ = 2×200 = 400μC

Common potential: V = (C₁V₁+C₂V₂)/(C₁+C₂) = (400+400)/(4+2) = 800/6

V = 133.3 V

Initial energy: Uᵢ = ½C₁V₁² + ½C₂V₂² = ½×4×10⁻⁶×10000 + ½×2×10⁻⁶×40000 = 0.02 + 0.04 = 0.06 J

Final energy: Uf = ½(C₁+C₂)V² = ½×6×10⁻⁶×(133.3)² = 3×10⁻⁶×17778 = 0.0533 J

Energy lost = 0.06 − 0.0533 = 0.0067 J = 6.7 mJ

Explanation: Positive to positive plate connection means charges add (both push in same direction). Common potential formula uses algebraic sum of charges. Energy is always lost when capacitors share charge — here 6.7mJ becomes heat in connecting wire.

Q159. A capacitor of capacitance C₁ = 6μF is charged to V₁ = 100V. It is then connected to uncharged capacitor C₂ = 3μF. Find final charge on each capacitor.

Ans: Initial charge Q = C₁V₁ = 6×100 = 600μC C₂ is uncharged, so V₂ = 0.

Common potential: V = (C₁V₁+C₂V₂)/(C₁+C₂) = (600+0)/(6+3) = 600/9

V = 66.67 V

Charge on C₁: Q₁ = C₁×V = 6×10⁻⁶×66.67 = 400 μC

Charge on C₂: Q₂ = C₂×V = 3×10⁻⁶×66.67 = 200 μC

Verification: Q₁+Q₂ = 400+200 = 600μC = initial charge ✓

Explanation: Charge is conserved — total 600μC is redistributed between the two capacitors. The larger capacitor (6μF) gets more charge (400μC) while smaller (3μF) gets less (200μC). Charge distributes in ratio of capacitances (6:3 = 2:1 → 400:200 ✓). This verification step is important in JEE.

Q160. Find the electric field between plates of a capacitor if potential difference is 240V and plate separation is 0.4cm.

Ans: V = 240V, d = 0.4cm = 4×10⁻³ m

For uniform field between parallel plates: E = V/d = 240/(4×10⁻³)

E = 60,000 V/m = 6×10⁴ V/m

Explanation: Between parallel plates, field is perfectly uniform (ignoring edge effects). E = V/d is a simpler form of E = −dV/dr for the case of uniform field. This field is same everywhere between plates — same near positive plate, middle, or near negative plate. Air breaks down at about 3×10⁶ V/m, so this field is well within safe range.

Q161. What is the work done in rotating an electric dipole from θ₁ to θ₂ in a uniform electric field E?

Ans: Potential energy of dipole in field: U = −pE cosθ

Work done by external agent in rotating from θ₁ to θ₂:

W = U₂ − U₁ = −pE cosθ₂ − (−pE cosθ₁)

W = pE(cosθ₁ − cosθ₂)

Special cases:

θ₁ = 90° to θ₂ = 0° (equatorial to axial): W = pE(0−1) = −pE (field does this work)
θ₁ = 0° to θ₂ = 180° (flip dipole): W = pE(1−(−1)) = 2pE (maximum work needed)
θ₁ = 0° to θ₂ = 90°: W = pE(1−0) = pE

Explanation: At θ = 0° (parallel to E), dipole has minimum energy (−pE) — most stable position. At θ = 180° (antiparallel), maximum energy (+pE) — most unstable. At θ = 90°, U = 0. These three positions are important reference points in NEET and JEE.

Q162. Three capacitors of 2μF, 3μF, and 6μF are connected in series to a 12V battery. Find: (a) equivalent capacitance (b) charge on each (c) voltage across each.

Ans:

(a) Equivalent capacitance: 1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 C = 1μF

(b) Charge on each (same in series): Q = CV = 1×10⁻⁶×12 = 12μC (same on all three)

(c) Voltage across each: V₁ (across 2μF) = Q/C₁ = 12/2 = 6V V₂ (across 3μF) = Q/C₂ = 12/3 = 4V V₃ (across 6μF) = Q/C₃ = 12/6 = 2V

Verification: V₁+V₂+V₃ = 6+4+2 = 12V ✓

Explanation: In series: smallest capacitor (2μF) gets largest voltage (6V). Largest capacitor (6μF) gets smallest voltage (2V). Voltage is inversely proportional to capacitance in series. This result (V ∝ 1/C) is exactly opposite to parallel where all get same V. Always verify by checking voltages sum to supply voltage.

Q163. Two metal spheres of radii r₁ = 3cm and r₂ = 6cm are charged to same potential V = 1200V. Find charge on each sphere.

Ans: C₁ = 4πε₀r₁ = r₁/k = 0.03/(9×10⁹) = 3.33×10⁻¹² F = 3.33pF

C₂ = 4πε₀r₂ = r₂/k = 0.06/(9×10⁹) = 6.67×10⁻¹² F = 6.67pF

Q₁ = C₁V = 3.33×10⁻¹² × 1200 = 4×10⁻⁹ C = 4nC

Q₂ = C₂V = 6.67×10⁻¹² × 1200 = 8×10⁻⁹ C = 8nC

Explanation: Capacitance of sphere = 4πε₀R = R/k. Larger sphere has more capacitance and stores more charge at same potential. Ratio Q₁:Q₂ = r₁:r₂ = 3:6 = 1:2 (confirmed: 4nC:8nC = 1:2). This is why large conductors can store more charge — they have larger capacitance.

Q164. Find the potential energy of a system of four charges +q at corners of a square of side a.

Ans: Number of pairs = 4C₂ = 6 pairs total

Side pairs (4 pairs, distance = a): U_side = 4 × kq²/a

Diagonal pairs (2 pairs, distance = a√2): U_diag = 2 × kq²/(a√2) = 2kq²/(a√2) = √2kq²/a

Total energy: U = 4kq²/a + √2kq²/a

U = kq²/a(4 + √2) = kq²(4+1.414)/a ≈ 5.414 kq²/a

Explanation: Count all unique pairs carefully — for 4 charges there are 6 pairs: 4 sides and 2 diagonals. Each pair contributes kq₁q₂/r. Side length = a, diagonal length = a√2. All charges are positive so all contributions are positive (energy stored against repulsion). This is a standard JEE assembly energy problem.

Q165. A parallel plate capacitor is charged by battery to 200V. Battery is removed. Plate separation is then doubled. Find new values of V, E, C, Q, U.

Ans: Initial: V₀=200V, d₀=d, C₀=ε₀A/d, Q₀=C₀×200, E₀=200/d, U₀=½C₀×200²

After doubling separation (d→2d), Q remains constant (battery removed):

Capacitance: C = ε₀A/2d = C₀/2 (halved)

Charge: Q = Q₀ (unchanged)

Voltage: V = Q/C = Q₀/(C₀/2) = 2×(Q₀/C₀) = 2×200 = 400V (doubled)

Electric field: E = Q/ε₀A = Q₀/ε₀A = E₀ (unchanged)

Energy: U = Q²/2C = Q₀²/(2×C₀/2) = Q₀²/C₀ = 2U₀ (doubled)

Explanation: After battery removal, Q is fixed. Doubling d halves C. Since V = Q/C, V doubles. Surprisingly, E = σ/ε₀ = Q/Aε₀ stays same (doesn't depend on d). Energy doubles — the extra energy comes from work done in pulling the plates apart against attraction.

Q166. An electron and proton are at distance 1Å (10⁻¹⁰m) apart. Find the potential energy of the system.

Ans: q₁ = +e = 1.6×10⁻¹⁹C (proton) q₂ = −e = −1.6×10⁻¹⁹C (electron) r = 10⁻¹⁰m

U = kq₁q₂/r = (9×10⁹×1.6×10⁻¹⁹×(−1.6×10⁻¹⁹))/10⁻¹⁰

U = 9×10⁹ × (−2.56×10⁻³⁸)/10⁻¹⁰

U = −9×10⁹ × 2.56×10⁻²⁸

U = −23.04×10⁻¹⁹ J = −14.4 eV

Explanation: Negative energy means bound state — electron is attracted to proton. To ionize hydrogen atom, you must supply 13.6eV (slightly less than 14.4eV because this was calculated for 1Å, the actual Bohr radius is 0.529Å). This negative potential energy is what keeps electrons bound in atoms — the foundation of all chemistry.

Q167. What is the capacitance of Earth? (R = 6.4×10⁶m)

Ans: Earth behaves as a spherical conductor.

C = 4πε₀R = R/k = 6.4×10⁶/(9×10⁹)

C = 0.711×10⁻³ F

C ≈ 711 μF

Explanation: 711μF is enormous for a single conductor (most capacitors are nF or pF). This is why Earth is used as ground reference — its huge capacitance means adding or removing small charges barely changes its potential. In circuit terms, Earth is like an infinite charge reservoir. Grounding a conductor means connecting it to this reservoir — its potential immediately becomes 0V.

Q168. Find electric field and potential at point P due to two charges +4μC at origin and −2μC at (3,0) m. Point P is at (0,4) m.

Ans: Distance r₁ (from origin to P) = 4m Distance r₂ (from (3,0) to P) = √(3²+4²) = √25 = 5m

Potential at P (scalar addition): V₁ = k(+4×10⁻⁶)/4 = 9×10⁹×4×10⁻⁶/4 = 9000V V₂ = k(−2×10⁻⁶)/5 = 9×10⁹×(−2×10⁻⁶)/5 = −3600V

V_total = 9000 − 3600 = 5400 V

Explanation: Potential calculation only needs distances (scalar) — no direction needed. Electric field would require finding components along x and y directions from each charge and then adding vectorially — much more work. This illustrates why potential is preferred for problems where only potential (not field) is asked. The 3-4-5 triangle is a common trick in JEE coordinate problems.

Q169. Derive the expression for potential at a general point due to an electric dipole.

Ans: Let dipole have charges +q and −q separated by 2a. Point P at distance r from centre at angle θ from dipole axis.

Distance from +q to P ≈ r − a cosθ Distance from −q to P ≈ r + a cosθ (for r >> a)

V = kq[1/(r−a cosθ) − 1/(r+a cosθ)]

V = kq[(r+a cosθ − r+a cosθ)/(r²−a²cos²θ)]

V = kq[2a cosθ/(r²−a²cos²θ)]

For r >> a: r²−a²cos²θ ≈ r²

V = kp cosθ/r²

where p = q×2a = dipole moment

Special cases:

θ = 0° (axial): V = kp/r²
θ = 90° (equatorial): V = 0
θ = 180°: V = −kp/r²

Q170. A capacitor of 8μF is charged to 250V. It is then connected to another uncharged capacitor of 4μF. Find energy lost.

Ans: Q = 8×10⁻⁶×250 = 2000μC = 2×10⁻³ C

Common potential: V = Q/(C₁+C₂) = 2000/(8+4) = 2000/12 = 500/3 V

Initial energy: Uᵢ = ½C₁V₁² = ½×8×10⁻⁶×250² = ½×8×10⁻⁶×62500 = 0.25 J

Final energy: Uf = ½(C₁+C₂)V² = ½×12×10⁻⁶×(500/3)² = 6×10⁻⁶×27778 = 0.1667 J

Energy lost = 0.25 − 0.1667 = 0.0833 J ≈ 83.3 mJ

Using formula directly: ΔU = C₁C₂(V₁−V₂)²/2(C₁+C₂) = (8×4×10⁻¹²×250²)/(2×12×10⁻⁶) = (32×10⁻¹²×62500)/(24×10⁻⁶) = 2×10⁻⁶/24×10⁻⁶ × ... = 83.3 mJ ✓

Q171. What is the charge on a 6μF capacitor in a circuit where it is in series with 3μF across 18V?

Ans: C_eq = (6×3)/(6+3) = 18/9 = 2μF

Charge (same on both in series): Q = C_eq × V = 2×10⁻⁶ × 18

Q = 36μC

Voltage across 6μF = Q/C = 36/6 = 6V Voltage across 3μF = Q/C = 36/3 = 12V

Verification: 6+12 = 18V ✓

Explanation: In series, Q is same on both capacitors. The charge on any series combination always equals C_eq × V_total. Smaller capacitor (3μF) gets larger voltage (12V) and larger capacitor (6μF) gets smaller voltage (6V) — voltage is inversely proportional to capacitance in series.

Q172. An infinite plane sheet has surface charge density σ = 2×10⁻⁶ C/m². Find electric field and potential difference between two points 10cm apart on the same side.

Ans: Electric field: E = σ/2ε₀ = (2×10⁻⁶)/(2×8.85×10⁻¹²) = 10⁻⁶/8.85×10⁻¹² = 1.13×10⁵ V/m

Potential difference (between points at d₁=0.05m and d₂=0.15m from sheet): If the two points are 10cm apart, the potential difference: V = E × d = 1.13×10⁵ × 0.10

ΔV = 1.13×10⁴ V = 11,300 V

Explanation: For infinite plane sheet, E is uniform and constant (does not decrease with distance). This is completely different from a point charge where E decreases as 1/r². Constant E means equal potential difference for equal distances — like parallel plate capacitor. The field is directed away from positive sheet on both sides.

Q173. A 5μF capacitor is connected to a 100V battery. Then the battery is disconnected and the capacitor is connected to a 20μF uncharged capacitor. Find final voltage.

Ans: Initial charge: Q = 5×10⁻⁶×100 = 500μC

After connecting to 20μF (battery disconnected → Q conserved):

V_final = Q/(C₁+C₂) = 500μC/(5+20)μF = 500/25

V_final = 20V

Explanation: Total capacitance becomes 5+20 = 25μF. Same charge 500μC spreads over larger capacitance, so voltage drops dramatically from 100V to just 20V. This is why connecting a small charged capacitor to a large uncharged one nearly discharges the small one. Energy drops from ½×5×10⁻⁶×10000 = 25mJ to ½×25×10⁻⁶×400 = 5mJ. 20mJ is lost as heat.

Q174. The electric potential in a region is given by V = 5x² + 3y − 2z. Find the electric field components at point (1,2,3).

Ans: E = −∇V = −(∂V/∂x î + ∂V/∂y ĵ + ∂V/∂z k̂)

∂V/∂x = 10x → at (1,2,3): = 10 ∂V/∂y = 3 → at (1,2,3): = 3 ∂V/∂z = −2 → at (1,2,3): = −2

Eₓ = −∂V/∂x = −10 V/m Ey = −∂V/∂y = −3 V/m Ez = −∂V/∂z = +2 V/m

E = −10î − 3ĵ + 2k̂ V/m

|E| = √(100+9+4) = √113 ≈ 10.6 V/m

Explanation: This is a JEE Advanced level problem using gradient of potential to find field. E = −∇V means each component of E is the negative partial derivative of V with respect to that coordinate. This is the 3D generalisation of E = −dV/dr. The point (1,2,3) is substituted only in the derivative (not V itself) since we need field at that specific point.

Q175. What is the potential difference required to accelerate an alpha particle from rest to kinetic energy 5MeV?

Ans: Charge of alpha particle = 2e = 2×1.6×10⁻¹⁹ = 3.2×10⁻¹⁹ C

KE = 5 MeV = 5×10⁶×1.6×10⁻¹⁹ = 8×10⁻¹³ J

From: KE = qV

V = KE/q = 8×10⁻¹³/(3.2×10⁻¹⁹)

V = 2.5×10⁶ V = 2.5 MV

Explanation: Alpha particle has charge 2e (not e). This is important — many students use e instead of 2e and get wrong answer. Alpha particle (₂⁴He) has 2 protons, so charge = 2e. Cyclotrons in nuclear physics must provide millions of volts to accelerate alpha particles to energies needed for nuclear reactions.

Q176. In a parallel plate capacitor, plate area is 0.02m², separation is 1mm. Calculate the force between the plates when charged to 200V.

Ans: C = ε₀A/d = 8.85×10⁻¹²×0.02/10⁻³ = 1.77×10⁻¹⁰ F

Q = CV = 1.77×10⁻¹⁰×200 = 3.54×10⁻⁸ C

σ = Q/A = 3.54×10⁻⁸/0.02 = 1.77×10⁻⁶ C/m²

Field due to one plate = σ/2ε₀ (only one plate acts on other)

Force = Q × E_one_plate = Q × σ/2ε₀

F = σ²A/2ε₀ = (1.77×10⁻⁶)²×0.02/(2×8.85×10⁻¹²)

F = 3.13×10⁻¹²×0.02/(1.77×10⁻¹¹) = 6.26×10⁻¹⁴/1.77×10⁻¹¹

F ≈ 3.54×10⁻³ N = 3.54 mN

Explanation: The force uses field of ONE plate (σ/2ε₀) not both plates. Plate doesn't exert force on itself — it only feels the field of the other plate. The force is always attractive (opposite charges on opposite plates). This force can be used to measure charge (electrostatic microbalance) or in MEMS devices.

Q177. What is meant by the term "1 Farad is a very large unit"? Explain with calculation.

Ans: A sphere of capacitance 1F would need radius:

C = 4πε₀R = R/k

1 = R/(9×10⁹)

R = 9×10⁹ m = 9,000,000 km

This is about 64 times the radius of Earth (Earth radius = 6400km)!

Practical capacitors:

Typical electronic capacitor: 1pF to 1000μF
Supercapacitors (modern): up to few Farads
1F capacitor at 1V stores only ½×1×1² = 0.5J (vs a AA battery = 3000J)

Explanation: 1 Farad is indeed enormous. To build a 1F parallel plate capacitor with 1mm separation and air: A = Cd/ε₀ = 1×10⁻³/8.85×10⁻¹² ≈ 10⁸ m² — larger than a small country! This is why electrolytic capacitors use very thin dielectric (oxide layer, few nm) and large surface area (rolled foil) to achieve μF in small packages.

Q178. Two capacitors C₁ = 4μF and C₂ = 2μF are connected in series. A dielectric slab of K = 3 is inserted in C₁. Find new equivalent capacitance.

Ans: Original: C₁ = 4μF, C₂ = 2μF

After dielectric in C₁: C₁' = K × C₁ = 3 × 4 = 12μF

New series combination: 1/C_eq = 1/C₁' + 1/C₂ = 1/12 + 1/2 = 1/12 + 6/12 = 7/12

C_eq = 12/7 ≈ 1.71 μF

Original C_eq = (4×2)/(4+2) = 8/6 = 1.33μF

So capacitance increased from 1.33μF to 1.71μF.

Explanation: Inserting dielectric in only one capacitor of a series combination increases overall capacitance, but not as dramatically as if it were in parallel. In series, the smallest capacitor dominates — here C₂ = 2μF limits the equivalent. If dielectric were inserted in C₂ instead (making it 6μF), C_eq = (4×6)/(4+6) = 2.4μF — an even bigger increase.

Q179. A 10μF capacitor is charged to 100V. Find the electric field energy stored if plate area is 200cm² and separation is 1cm.

Ans: Method 1 (Direct): U = ½CV² = ½×10×10⁻⁶×100² = ½×10⁻⁵×10⁴ = 0.05 J

Method 2 (Energy density): E = V/d = 100/(10⁻²) = 10⁴ V/m

u = ½ε₀E² = ½×8.85×10⁻¹²×(10⁴)² = ½×8.85×10⁻¹²×10⁸ = 4.425×10⁻⁴ J/m³

Volume = A×d = 200×10⁻⁴×10⁻² = 2×10⁻⁴ m³

U = u×Volume = 4.425×10⁻⁴×2×10⁻⁴ = 8.85×10⁻⁸ J

Note: Discrepancy because real capacitance = ε₀A/d = 8.85×10⁻¹²×0.02/0.01 = 17.7×10⁻¹² F ≠ 10μF given. Method 1 uses given C=10μF. Both methods are correct within their own assumptions.

Q180. Three equal charges q are placed at vertices of equilateral triangle of side a. What is the electric potential at the centroid?

Ans: Distance from each vertex to centroid of equilateral triangle: r = a/√3

Potential at centroid (scalar sum of three equal contributions): V = 3 × kq/r = 3 × kq/(a/√3) = 3kq√3/a

V = 3√3 kq/a

Explanation: The centroid is equidistant from all three vertices (distance = a/√3). Since potential is scalar, just multiply one contribution by 3. The electric field at centroid = 0 (by symmetry, three equal forces cancel). This gives another example where V ≠ 0 but E = 0. This problem is frequently asked in JEE Mains.

Q181. A capacitor of 100pF is connected to 50V battery and then disconnected. The charge is then shared with another capacitor. If final voltage is 20V, find the capacitance of second capacitor.

Ans: Initial charge: Q = 100×10⁻¹²×50 = 5000 pC = 5×10⁻⁹ C

After sharing, V_final = 20V, total charge = 5000 pC

Total capacitance = Q/V_final = 5000pC/20V = 250 pF

C₁ + C₂ = 250 pF

100 + C₂ = 250

C₂ = 150 pF

Explanation: This is a reverse problem — given final voltage, find unknown capacitance. Key insight: total charge is conserved. Total charge = 5000pC. If final V = 20V, total C must be 5000/20 = 250pF. Subtract known C₁ = 100pF to get C₂ = 150pF. This type of reverse-working question is common in JEE Mains.

Q182. What is corona discharge? Where is it useful and where is it harmful?

Ans: Corona discharge is the ionization of air around a sharp pointed conductor due to extremely high electric field at the tip.

Why it happens:

Sharp points have very small radius of curvature
σ ∝ 1/R → high charge density at sharp points
E = σ/ε₀ → very high field at tips (can exceed 3×10⁶ V/m, breakdown of air)
Air molecules ionize → glow (corona) + charge leakage

Useful applications:

Lightning conductors (rods): safely discharge cloud charge through pointed rod
Van de Graaff generator: spray comb uses corona to charge belt
Electrostatic precipitators: remove pollution particles from chimney smoke
Photocopiers and laser printers: corona wire charges the drum

Harmful effects:

Power transmission lines: corona causes energy loss
Creates radio frequency interference (noise)
Produces ozone (O₃) which can be harmful at high concentrations
Damages high-voltage equipment insulation over time

Q183. Explain the term "Dielectric Strength". Give values for common materials.

Ans: Dielectric strength is the maximum electric field a dielectric material can withstand without breaking down (becoming conducting). It is measured in V/m or kV/mm.

Formula connecting to capacitor: Maximum voltage = Dielectric strength × plate separation V_max = E_max × d

Values of dielectric strength:

Material	Dielectric Strength
Air	3×10⁶ V/m = 3 kV/mm
Glass	9×10⁶ V/m
Rubber	12×10⁶ V/m
Mica	1.5×10⁸ V/m
Paper	1.6×10⁷ V/m
Porcelain	1.2×10⁷ V/m

Explanation: Mica has very high dielectric strength — this is why mica capacitors are used in high-voltage applications. Air is weakest — lightning occurs because storm clouds create field exceeding 3×10⁶ V/m in air. Capacitor failure happens when applied voltage creates field exceeding dielectric strength. Always check V_max = E_max × d when designing capacitors.

Q184. A charge Q is distributed on two concentric hollow spheres of radii r and R (R > r). Find potential at centre, on inner sphere surface, and on outer sphere surface.

Ans: Let charge on inner sphere = q₁, outer sphere = q₂, q₁+q₂ = Q.

Potential at centre (r = 0): Both spheres contribute. Point is inside both spheres. V_centre = kq₁/r + kq₂/R

Potential on inner sphere surface (distance r): V_inner = kq₁/r + kq₂/R (inner sphere contributes at its own surface, outer sphere contributes as if all charge at centre)

Potential on outer sphere surface (distance R): V_outer = kq₁/R + kq₂/R = k(q₁+q₂)/R = kQ/R

Note: V_centre = V_inner (potential is constant inside and on inner sphere since E = 0 inside)

Explanation: For a sphere, potential inside equals surface potential (constant). The outer sphere contributes kq₂/R at all points inside it (including centre and inner sphere surface). This concept is used in problems involving concentric spherical capacitors which appear frequently in JEE Advanced.

Q185. Explain the working of a capacitor as a charge reservoir or voltage stabilizer.

Ans: As charge reservoir:

Capacitor stores charge when voltage is high
Releases charge when voltage drops
Acts like a water tank — fills when pressure (voltage) is high, empties when low
Used in power supply filters to smooth voltage fluctuations

Mathematical basis: I = C × dV/dt

If V is constant: I = 0 (no current) If V changes rapidly: large current flows (charges or discharges)

Applications:

Power supply smoothing — capacitor across DC output smooths ripple from rectifier
Flash photography — capacitor charges slowly (seconds) from battery, discharges instantly (<1ms) through bulb
Computer RAM (DRAM) — each memory bit stored as charge state of tiny capacitor
Heart defibrillator — charges to high voltage, discharges controlled pulse through patient
Audio systems — bypass capacitors prevent voltage fluctuations when speakers draw sudden current

Q186. Two capacitors are in series across 100V. The voltages are 40V and 60V. Find their capacitance ratio.

Ans: In series, Q is same on both. V ∝ 1/C.

V₁/V₂ = C₂/C₁

40/60 = C₂/C₁

C₂/C₁ = 2/3

C₁:C₂ = 3:2

Verification: Larger capacitor (C₁) gets smaller voltage (40V) ✓ Smaller capacitor (C₂) gets larger voltage (60V) ✓

Explanation: In series, voltage across a capacitor is inversely proportional to its capacitance — V ∝ 1/C (since V = Q/C and Q is same for all). This is the reverse of parallel combination where all share same voltage. This ratio problem avoids finding actual capacitances — just the ratio from voltage ratio. Very common in NEET.

Q187. A charge q is placed at corner of a cube of side a. Find electric flux through one face not touching the charge.

Ans: The charge at corner is shared by 8 identical cubes (since a corner is shared by 8 cubes).

Total flux from charge q = q/ε₀

Flux through entire cube = q/8ε₀

The cube has 6 faces, but 3 faces touch the charge (no flux through them — field lines are parallel to these faces).

Only 3 faces receive flux:

Flux through one face (not touching charge) = (q/8ε₀)/3 = q/24ε₀

Explanation: This is a classic Gauss's Law application using symmetry. The trick is recognising that corner charge is shared by 8 cubes, and 3 of 6 faces have zero flux. This cannot be solved by direct integration easily — symmetry argument is the elegant JEE approach. The answer q/24ε₀ is a standard result to memorise.

Q188. Calculate the potential due to a ring of charge Q, radius R, at a point on its axis at distance x.

Ans: Every element dq of ring is at same distance from axial point P:

Distance from dq to P = √(R²+x²)

Since all elements are equidistant:

V = k∫dq/√(R²+x²) = k/√(R²+x²) × ∫dq

V = kQ/√(R²+x²)

Special cases:

At centre (x=0): V = kQ/R (maximum potential on axis)
Far away (x>>R): V ≈ kQ/x (behaves like point charge) ✓

Electric field on axis: E = kQx/(R²+x²)^(3/2)

E = 0 at centre (x=0) — symmetric cancellation

Explanation: The power of potential over field: potential needs only distance (scalar) so the integral is simple. Finding axial field by direct integration requires knowing vector directions. Finding potential first, then E = −dV/dx is the smarter JEE approach for ring/disc problems.

Q189. Find the work done in assembling three charges q₁=1μC, q₂=2μC, q₃=3μC at vertices of equilateral triangle of side 10cm.

Ans: Work done = Total potential energy of system

Pair 1 (q₁ and q₂): U₁₂ = kq₁q₂/r = 9×10⁹×1×2×10⁻¹²/0.1 = 0.18 J

Pair 2 (q₂ and q₃): U₂₃ = kq₂q₃/r = 9×10⁹×2×3×10⁻¹²/0.1 = 0.54 J

Pair 3 (q₁ and q₃): U₁₃ = kq₁q₃/r = 9×10⁹×1×3×10⁻¹²/0.1 = 0.27 J

Total work = U₁₂+U₂₃+U₁₃ = 0.18+0.54+0.27 = 0.99 J ≈ 1 J

Explanation: Work done to assemble = total potential energy stored. All pairs have same separation (equilateral triangle), but different charge products. Since all charges are positive, all pair energies are positive — you must do positive work against repulsion. Had some charges been negative, some pairs would release energy (negative work). Always identify all n(n−1)/2 = 3 pairs for 3 charges.

Q190. What is meant by polar molecule? Give examples and explain behaviour in electric field.

Ans: A polar molecule has a permanent electric dipole moment — its centre of positive charge and centre of negative charge do not coincide, even without an external field.

Examples: H₂O, HCl, NH₃, SO₂, CO

Structure of water (H₂O):

Oxygen is more electronegative
Electron cloud shifts toward oxygen
Positive centre (H side) and negative centre (O side) are separated
Permanent dipole moment ≈ 6.2×10⁻³⁰ C·m

Behaviour in external electric field:

Without field: Dipoles point in random directions — no net polarization.

With field: Field exerts torque τ = p×E = pE sinθ on each dipole, trying to align them along field.

Thermal agitation opposes perfect alignment → partial alignment → net polarization P.

Higher temperature → more thermal agitation → less alignment → lower K.

Explanation: This is why water has very high K = 80. Its large permanent dipole moment aligns strongly with field, creating large opposing field → great reduction in net E → large K. Non-polar molecules (H₂, N₂, O₂) have K ≈ 1 as they need induced dipoles.

Q191. A parallel plate capacitor with plate separation 4mm is connected to 48V battery. Find surface charge density on plates.

Ans: E = V/d = 48/(4×10⁻³) = 12000 V/m

For parallel plate capacitor: E = σ/ε₀

σ = ε₀×E = 8.85×10⁻¹²×12000

σ = 1.062×10⁻⁷ C/m² = 106.2 nC/m²

Explanation: The connection between electric field, voltage, and surface charge density for parallel plate capacitor: E = V/d = σ/ε₀. So σ = ε₀V/d. This is consistent with C = Q/V = σA/V = ε₀A/d ✓ (since σ = Q/A). Surface charge density directly gives the field — no other information needed.

Q192. Explain why two equipotential surfaces cannot intersect each other.

Ans: Suppose two equipotential surfaces S₁ (V = 10V) and S₂ (V = 20V) intersect at point P.

Then point P would have two values of potential simultaneously: V = 10V AND V = 20V.

This is a mathematical and physical contradiction — a point in space can have only one value of electric potential at any given time.

Therefore: Two equipotential surfaces can never intersect.

Additional argument using field: At the intersection point, the electric field would need to be perpendicular to both surfaces simultaneously — in two different directions. A vector cannot point in two directions at once. Contradiction. Hence no intersection is possible.

Similarly for field lines: If two field lines intersected, field would have two directions at that point — impossible.

Explanation: Both field lines and equipotential surfaces obey the rule of uniqueness — at any point, there is one unique value of potential and one unique direction of field. This uniqueness comes from the mathematics of solving Laplace's equation (which V must satisfy in charge-free regions).

Q193. A charge of 10μC is uniformly distributed on a thin shell of radius 10cm. Find potential (a) outside at 20cm (b) on surface (c) inside at 5cm.

Ans:

(a) Outside at r = 20cm = 0.2m: V = kQ/r = 9×10⁹×10×10⁻⁶/0.2 = 90000/0.2

V = 450,000 V = 450 kV

(b) On surface at r = 10cm = 0.1m: V = kQ/R = 9×10⁹×10×10⁻⁶/0.1

V = 900,000 V = 900 kV

(c) Inside at r = 5cm: E = 0 inside shell → V = constant = surface potential

V = 900 kV (same as surface)

Explanation: Outside: behaves like point charge — V = kQ/r (decreases as 1/r). On and inside: V = kQ/R = constant. The potential does not drop inside the shell even though we move closer to where there is "no charge." This is because the shell's charge, though distributed on surface, still creates potential inside. Field is zero inside but potential is not — classic exam distinction.

Q194. What is the significance of V = kQ/R for a charged sphere in high voltage technology?

Ans: For a sphere: V = kQ/R and E_surface = kQ/R²

Therefore: V = E_surface × R

This has major implications:

High voltage generators:

To achieve high voltage V, need either large E or large R
Large R (large sphere) at moderate E → high voltage
Van de Graaff uses large dome (R large) to achieve 10⁶V at moderate σ

Power line safety:

Round conductors chosen for minimum surface field
Sharp edges on power lines would cause corona and energy loss
Bundled conductors (multiple wires) effectively increase R

Medical equipment:

High voltage X-ray tubes use smooth electrodes
Any sharp point creates dangerous corona and field emission

Material science:

Dielectric breakdown: if E_surface > dielectric strength, breakdown occurs
V_max = E_max × R — maximum safe voltage proportional to sphere radius

Q195. A 10μF capacitor charged to 200V is connected in parallel with 5μF capacitor charged to 100V (positive plates connected together). Find final voltage and energy lost.

Ans: Initial charges: Q₁ = 10×200 = 2000μC Q₂ = 5×100 = 500μC

Positive to positive (same polarity) → charges ADD: Q_total = 2000+500 = 2500μC

C_total = 10+5 = 15μF

Final voltage: V = Q_total/C_total = 2500/15 = 166.7 V

Initial energy: Uᵢ = ½×10×10⁻⁶×200² + ½×5×10⁻⁶×100² = 0.2 + 0.025 = 0.225 J

Final energy: Uf = ½×15×10⁻⁶×(166.7)² = 7.5×10⁻⁶×27789 = 0.2084 J

Energy lost = 0.225−0.2084 = 0.0166 J = 16.6 mJ

Q196. Derive the expression for electric field using potential for a point charge and verify it.

Ans: Given: V = kq/r

E = −dV/dr = −d(kq/r)/dr = −kq×(−1/r²) = kq/r²

E = kq/r² ✓

This matches Coulomb's law result exactly.

Direction: Since E = −dV/dr and V decreases as r increases for positive q, dV/dr is negative, so E = −(negative) = positive (outward direction). ✓

For negative charge: V = k(−q)/r is negative. dV/dr is positive (less negative as r increases). E = −(positive) = negative (inward direction). ✓

Explanation: This derivation confirms that E = −dV/dr is consistent with Coulomb's law. The potential approach and force approach give identical results. This gives confidence that both frameworks (E field and V potential) are completely equivalent descriptions of the same physics. JEE often asks to "find E from V" using this technique.

Q197. What are the differences between a conductor and a dielectric in an electric field?

Ans:

Property	Conductor	Dielectric
Free charges	Present (electrons)	Absent
E inside	Zero	E₀/K (reduced)
V inside	Constant (= surface V)	Varies
Response to field	Charge redistribution on surface	Polarization (bound charges)
Induced effect	Free charge movement	Bound charge shift
Current flow	Yes (if connected)	No (insulator)
K value	Effectively ∞	1 to ~80
Time to respond	~10⁻¹⁷ s	Time for polarization
Field inside material	0	E₀/K
Example	Copper, Silver	Glass, Mica, Paper

Explanation: Conductor eliminates field completely inside (free charges completely cancel external field). Dielectric only reduces field by factor K (bound charges partially cancel field). This is why inserting a conductor increases C more than inserting dielectric — conductor effectively reduces gap (K=∞), dielectric reduces field by K.

Q198. Solve: A network has C₁=C₂=C₃=C₄=C each. C₁ and C₂ in series form one branch. C₃ and C₄ in series form another. Both branches in parallel. Find equivalent capacitance.

Ans: Branch 1: C₁ and C₂ in series C_branch1 = C×C/(C+C) = C²/2C = C/2

Branch 2: C₃ and C₄ in series C_branch2 = C×C/(C+C) = C/2

Both branches in parallel: C_eq = C/2 + C/2 = C

So equivalent capacitance = C (same as one capacitor!)

Explanation: Interesting result — this combination of four identical capacitors gives same equivalent capacitance as one. This happens because two series (giving C/2 each) in parallel (giving C) exactly cancels. This type of combination network problem is a favourite in JEE Mains — always systematically reduce step by step, never try to guess.

Q199. What are supercapacitors (ultracapacitors)? How are they different from normal capacitors?

Ans: Supercapacitors are electrochemical capacitors with extremely high capacitance (1F to 3000F) compared to conventional capacitors.

Working principle:

Use electric double layer (EDL) at electrode-electrolyte interface
Separation between charges is molecular (few Angstroms = 10⁻¹⁰m)
Huge plate area (porous carbon electrode, surface area ~1000 m²/g)
C = ε₀A/d → tiny d + huge A = enormous C

Comparison:

Property	Normal Capacitor	Supercapacitor	Battery
Capacitance	pF to mF	1F to 3000F	N/A
Energy density	Low	Medium	High
Power density	Very High	High	Low
Charge time	Microseconds	Seconds	Hours
Cycle life	Unlimited	500,000+	500-2000
Voltage	High (kV)	Low (2.7V)	Medium

Applications: Regenerative braking in electric vehicles, backup power, burst power in cameras, grid stabilization.

Q200. Write a comprehensive summary of all key formulas and concepts in Electric Potential and Capacitance.

Ans:

ELECTRIC POTENTIAL:

Formula	Meaning
V = kq/r	Potential due to point charge
V = kp cosθ/r²	Potential due to dipole
V_axial = kp/r²	Axial dipole potential
V_equatorial = 0	Equatorial dipole potential
E = −dV/dr	Field from potential
W = q(V_A−V_B)	Work by field
W = q(V_B−V_A)	Work by external agent
KE = qV	Kinetic energy gained
U = kq₁q₂/r	Potential energy of 2 charges

CAPACITANCE:

Formula	Meaning
C = Q/V	Definition
C = Kε₀A/d	Parallel plate capacitor
C = 4πε₀R	Spherical conductor
1/C = Σ1/Cᵢ	Series combination
C = ΣCᵢ	Parallel combination
U = ½CV² = Q²/2C	Energy stored
u = ½ε₀E²	Energy density
V_common = (C₁V₁+C₂V₂)/(C₁+C₂)	Charge sharing
ΔU = C₁C₂(V₁−V₂)²/2(C₁+C₂)	Energy lost on sharing

KEY CONCEPTS TO REMEMBER:

V is scalar — add algebraically with signs
E is vector — add as vectors
E = 0 inside conductor, V = constant (not zero unless earthed)
On surface: V = constant, E = σ/ε₀ (not zero)
Equipotential surfaces are always ⊥ to field lines
K ≥ 1 always (K = 1 for vacuum)
Battery disconnected → Q constant; Battery connected → V constant
Energy always lost when capacitors at different V share charge
Corona at sharp points (high σ = small R)
Faraday cage — interior shielded from external E field

🏆 Final Master Revision Table

Concept	Formula	Unit
Electric Potential	V = kq/r	Volt (V)
Potential Difference	ΔV = W/q	Volt
Electric Field	E = −dV/dr	V/m
Capacitance	C = Kε₀A/d	Farad (F)
Charge	Q = CV	Coulomb (C)
Energy	U = ½CV²	Joule (J)
Energy density	u = ½ε₀E²	J/m³
Dipole potential	V = kp cosθ/r²	Volt
Permittivity	ε₀ = 8.85×10⁻¹²	C²/Nm²
Coulomb constant	k = 9×10⁹	Nm²/C²
1 eV	= 1.6×10⁻¹⁹ J	Energy
Earth capacitance	≈ 711 μF	Farad

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Thursday, April 9, 2026

ELECTRIC POTENTIAL & CAPACITANCE

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