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Coulomb's Law - 2 Marks Questions with Answers
NEET Level Questions
Q1. Two point charges of +16 μC and -9 μC are placed 8 cm apart in air. At what point on the line joining the two charges is the electric potential zero?
Answer: Let the point be at distance x from +16 μC charge where potential is zero.
V = V₁ + V₂ = 0
k(16×10⁻⁶)/x + k(-9×10⁻⁶)/(0.08-x) = 0
16/x = 9/(0.08-x)
16(0.08-x) = 9x
1.28 - 16x = 9x
1.28 = 25x
x = 0.0512 m = 5.12 cm from +16 μC charge
Q2. Two charges of 1 μC and -4 μC are kept 30 cm apart. Find the point on the line joining the charges where electric field is zero.
Answer: Electric field will be zero at a point closer to smaller charge (1 μC), outside the charges.
Let distance from 1 μC = x
E₁ = E₂
k(1×10⁻⁶)/x² = k(4×10⁻⁶)/(x+0.30)²
1/x² = 4/(x+0.30)²
(x+0.30)² = 4x²
x+0.30 = 2x
x = 0.30 m = 30 cm from 1 μC charge
Q3. Calculate the electrostatic force between two alpha particles separated by 3.2 × 10⁻¹⁵ m. (Charge on alpha particle = 3.2 × 10⁻¹⁹ C)
Answer: Given: q₁ = q₂ = 3.2 × 10⁻¹⁹ C, r = 3.2 × 10⁻¹⁵ m
F = kq₁q₂/r²
F = (9×10⁹ × (3.2×10⁻¹⁹)² )/(3.2×10⁻¹⁵)²
F = (9×10⁹ × 10.24×10⁻³⁸)/(10.24×10⁻³⁰)
F = 9 N (repulsive)
Q4. Two identical spheres having charges +90 μC and -10 μC attract each other with a force of 0.4 N when separated by certain distance. If they are touched and kept at same distance, what will be the new force?
Answer: Initial: F₁ = k(90×10⁻⁶)(10×10⁻⁶)/r² = 0.4 N
After touching: Total charge = 90 - 10 = 80 μC Each sphere gets = 40 μC
F₂ = k(40×10⁻⁶)(40×10⁻⁶)/r²
F₂/F₁ = (40×40)/(90×10) = 1600/900 = 16/9
F₂ = 0.4 × 16/9 = 0.71 N (repulsive)
Q5. Three charges q, 2q and 8q are placed on a straight line. Where should 2q be placed between q and 8q so that the entire system is in equilibrium?
Answer: Let q be at origin, 8q at distance d, and 2q at distance x from q.
For 2q to be in equilibrium: Force from q = Force from 8q
kq(2q)/x² = k(8q)(2q)/(d-x)²
1/x² = 8/(d-x)²
(d-x)² = 8x²
d-x = 2√2 x
d = x(1 + 2√2)
x = d/(1 + 2√2) = d/3.828 ≈ 0.26d from charge q
CUET Level Questions
Q6. Two point charges +4 μC and +9 μC are separated by 1.5 m. Where should a third charge be placed on the line joining them so that it experiences no net force?
Answer: Let the point be at distance x from 4 μC charge.
For zero force on third charge: E₁ = E₂
k(4)/x² = k(9)/(1.5-x)²
4(1.5-x)² = 9x²
2(1.5-x) = 3x
3 - 2x = 3x
x = 0.6 m = 60 cm from 4 μC charge
Q7. If the distance between two equal charges is doubled and their individual charges are also doubled, what would happen to the force between them?
Answer: Initial force: F = kq²/r²
After changes: q' = 2q, r' = 2r
F' = k(2q)²/(2r)²
F' = k(4q²)/(4r²)
F' = kq²/r²
F' = F (Force remains same)
Q8. Four charges +q, +q, +q and -q are placed at corners of a square of side 'a'. Calculate the force on charge +q placed at the center.
Answer: At the center of square, distance from each corner = a/√2
Forces from three +q charges and one -q charge:
- Three +q → repulsive forces
- One -q → attractive force
Due to symmetry:
- Two opposite +q charges cancel each other
- Remaining +q and -q are also opposite
Vector sum of all forces = Zero
Net force = 0
Q9. Two small spheres each carrying a charge q are placed r meter apart. If one of the spheres is taken around the other one in a circular path of radius r, the work done will be:
Answer: Electrostatic force is a conservative force.
Work done in moving a charge in a closed path in electrostatic field = 0
Also, force is always perpendicular to displacement (centripetal direction).
W = F·s·cosθ = F·s·cos90° = 0
Work done = Zero
Q10. Calculate the distance between two electrons if the electrostatic force between them is equal to the weight of an electron. (Mass of electron = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C)
Answer: Weight of electron = mg = 9.1×10⁻³¹ × 10 = 9.1×10⁻³⁰ N
Electrostatic force = ke²/r²
ke²/r² = mg
r² = ke²/mg
r² = (9×10⁹ × (1.6×10⁻¹⁹)²)/(9.1×10⁻³⁰ × 10)
r² = (9×10⁹ × 2.56×10⁻³⁸)/(9.1×10⁻²⁹)
r = 5.08 m
Basic Level Questions
Q11. State Coulomb's law in electrostatics. Write its mathematical form.
Answer: Coulomb's Law: The electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematical form:
F = k(q₁q₂)/r²
where k = 9 × 10⁹ Nm²/C² (electrostatic constant)
or F = (1/4πε₀)(q₁q₂)/r²
where ε₀ = 8.85 × 10⁻¹² C²/Nm²
Q12. Two charges of 5 μC each are placed at two corners of an equilateral triangle of side 10 cm. Find the force on a charge of 1 μC placed at the third corner.
Answer: Force from each 5 μC charge:
F₁ = F₂ = k(5×10⁻⁶)(1×10⁻⁶)/(0.1)²
F₁ = (9×10⁹ × 5×10⁻¹²)/(0.01)
F₁ = 4.5 N
Angle between F₁ and F₂ = 60°
Resultant force: F = √(F₁² + F₂² + 2F₁F₂cos60°)
F = √(4.5² + 4.5² + 2×4.5×4.5×0.5)
F = √(20.25 + 20.25 + 20.25)
F = 7.79 N
Q13. What is the ratio of electrostatic force to gravitational force between a proton and an electron? (Given: e = 1.6 × 10⁻¹⁹ C, mₚ = 1.67 × 10⁻²⁷ kg, mₑ = 9.1 × 10⁻³¹ kg)
Answer: Fₑ/Fₘ = (ke²/r²)/(Gmₑmₚ/r²)
Fₑ/Fₘ = ke²/(Gmₑmₚ)
Fₑ/Fₘ = (9×10⁹ × (1.6×10⁻¹⁹)²)/(6.67×10⁻¹¹ × 9.1×10⁻³¹ × 1.67×10⁻²⁷)
Fₑ/Fₘ = (2.304×10⁻²⁸)/(1.014×10⁻⁶⁷)
Fₑ/Fₘ ≈ 2.27 × 10³⁹
Q14. Two identical metal balls with charges +6 μC and -2 μC are kept in contact and then separated. What will be charge on each ball?
Answer: When identical conductors touch, charge distributes equally.
Total charge = +6 + (-2) = +4 μC
After separation: Charge on each ball = Total charge/2
Charge on each = +4/2 = +2 μC
Q15. Define one coulomb of charge.
Answer: One Coulomb is the amount of charge which when placed at a distance of 1 meter from an equal and similar charge in vacuum, repels it with a force of 9 × 10⁹ N.
OR
One Coulomb is the charge flowing through a conductor when a current of 1 ampere flows for 1 second.
1 C = 1 A × 1 s
Coulomb's Law - 2 Marks Questions (Q16-Q50)
NEET Level Questions
Q16. Two point charges +3 μC and +12 μC repel each other with a force of 0.4 N. If each is given an additional charge of -4 μC, the new force of attraction will be:
Answer: After adding -4 μC to each:
- q₁' = 3 - 4 = -1 μC
- q₂' = 12 - 4 = +8 μC
Original: F = k(3)(12)/r² = 0.4 N So, k(36)/r² = 0.4 N
New: F' = k(1)(8)/r² = k(8)/r²
F'/F = 8/36 = 2/9
F' = 0.4 × 2/9
F' = 0.089 N (attractive)
Q17. Calculate the number of electrons that must be removed from a neutral metal sphere to give it a positive charge of 6.4 × 10⁻¹⁸ C.
Answer: Charge on one electron = e = 1.6 × 10⁻¹⁹ C
Total charge Q = ne
n = Q/e = (6.4 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹)
n = 40 electrons
Q18. Two charges -q and +q are placed at a distance d apart. Where should a third charge +2q be placed on the line joining them so that the net force on it is zero?
Answer: The point must lie outside the charges, on the side of -q.
Let distance from -q = x
For equilibrium: kq(2q)/x² = k(q)(2q)/(d+x)²
1/x² = 1/(d+x)²
This is impossible unless we place it inside.
Actually, for opposite charges, equilibrium point is between them.
Let distance from -q = x, then from +q = (d-x)
kq(2q)/x² = k(q)(2q)/(d-x)²
1/x² = 1/(d-x)²
x = d-x
x = d/2 (at midpoint)
Q19. Two small spheres each of mass 0.1 g carry equal charges and are suspended by silk threads of length 20 cm from a common point. If the distance between them is 6 cm, calculate the charge on each sphere. (g = 10 m/s²)
Answer: Given: m = 0.1 g = 10⁻⁴ kg, l = 20 cm = 0.2 m, r = 6 cm = 0.06 m
From geometry: sinθ = r/(2l) = 0.06/0.4 = 0.15 tanθ = sinθ/cosθ ≈ 0.15/0.989 ≈ 0.152
For equilibrium: T sinθ = F = kq²/r² T cosθ = mg
Dividing: tanθ = kq²/(mgr²)
q² = (mgr² tanθ)/k
q² = (10⁻⁴ × 10 × (0.06)² × 0.152)/(9×10⁹)
q² = (5.47 × 10⁻⁶)/(9×10⁹)
q = 2.47 × 10⁻⁸ C = 24.7 nC
Q20. The force between two charges separated by a distance 1 m is 1.8 N. If the distance is increased to 2 m, what will be the force?
Answer: F ∝ 1/r²
F₁/F₂ = r₂²/r₁²
1.8/F₂ = (2)²/(1)²
F₂ = 1.8/4
F₂ = 0.45 N
Q21. Four point charges -Q, -q, +2q and +Q are placed at the corners of a square. What is the relation between Q and q so that the resultant force on each charge is zero?
Answer: Place charges as: -Q, -q, +2q, +Q at corners.
For force on -Q to be zero: Forces from three other charges must balance.
Due to diagonal symmetry: Force from +Q (diagonal) + Forces from -q and +2q (adjacent) = 0
kQ²/(a√2)² = kQq/a² + kQ(2q)/a²
Q²/(2a²) = 3Qq/a²
Q = 6q
Q = 6q or Q/q = 6
Q22. Two charges +0.2 μC and -0.2 μC are placed 10⁻⁸ m apart. Calculate the electric dipole moment.
Answer: Electric dipole moment: p = q × 2l
where 2l = distance between charges = 10⁻⁸ m
p = 0.2 × 10⁻⁶ × 10⁻⁸
p = 2 × 10⁻¹⁵ C·m
Q23. The electric force between two electrons is same as the electric force between two protons when placed at the same distance. Is this statement true?
Answer: YES, the statement is TRUE.
Explanation:
- Charge on electron = -1.6 × 10⁻¹⁹ C
- Charge on proton = +1.6 × 10⁻¹⁹ C
F = kq₁q₂/r²
For electrons: F = k(e)(e)/r² = ke²/r² For protons: F = k(e)(e)/r² = ke²/r²
Magnitude of force is same (both are repulsive)
Q24. Three charges of 1 C, 2 C and 3 C are placed at the corners of an equilateral triangle of side 1 m. Find the force on 2 C charge.
Answer: Force from 1 C: F₁ = k(1)(2)/(1)² = 2k = 2 × 9×10⁹ = 1.8×10¹⁰ N
Force from 3 C: F₂ = k(3)(2)/(1)² = 6k = 6 × 9×10⁹ = 5.4×10¹⁰ N
Angle between F₁ and F₂ = 60°
Resultant: F = √(F₁² + F₂² + 2F₁F₂cos60°)
F = √((1.8×10¹⁰)² + (5.4×10¹⁰)² + 2(1.8×10¹⁰)(5.4×10¹⁰)(0.5))
F = √(3.24×10²⁰ + 29.16×10²⁰ + 9.72×10²⁰)
F = √(42.12×10²⁰)
F = 6.49 × 10¹⁰ N
Q25. What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?
Answer: Given: q₁ = 2×10⁻⁷ C, q₂ = 3×10⁻⁷ C, r = 30 cm = 0.3 m
F = kq₁q₂/r²
F = (9×10⁹ × 2×10⁻⁷ × 3×10⁻⁷)/(0.3)²
F = (9×10⁹ × 6×10⁻¹⁴)/(0.09)
F = (54×10⁻⁵)/(0.09)
F = 6 × 10⁻³ N = 6 mN
CUET Level Questions
Q26. Two identical conducting spheres with charges +20 μC and -8 μC are placed at certain distance. They are brought in contact and then separated to the same distance. What is the ratio of initial to final force?
Answer: Initial force: F₁ = k(20)(-8)/r² = -160k/r² (attractive) |F₁| = 160k/r²
After contact: Total charge = 20 + (-8) = 12 μC Each gets = 6 μC
Final force: F₂ = k(6)(6)/r² = 36k/r² (repulsive)
Ratio = |F₁|/F₂ = 160/36 = 40/9
F₁:F₂ = 40:9
Q27. If the force between two charges in air is F, what will be the force when they are immersed in water? (Dielectric constant of water K = 80)
Answer: In air: F = kq₁q₂/r²
In water: F' = kq₁q₂/(Kr²) = F/K
F' = F/80
Force becomes F/80
Q28. Two point charges +8q and -2q are located at x = 0 and x = L respectively. The point on the x-axis at which the net electric field is zero is:
Answer: For opposite charges, zero field point lies outside, near smaller charge.
Let point be at distance x from -2q (beyond -2q)
k(8q)/(L+x)² = k(2q)/x²
8/((L+x)²) = 2/x²
8x² = 2(L+x)²
4x² = (L+x)²
2x = L+x
x = L
Point is at distance L from -2q, or 2L from +8q
Q29. Two charges placed in air repel each other with a force of 10⁻⁴ N. When oil is introduced between the charges, the force becomes 2.5 × 10⁻⁵ N. The dielectric constant of oil is:
Answer: In air: F = 10⁻⁴ N
In oil: F' = 2.5 × 10⁻⁵ N
F'/F = 1/K
2.5 × 10⁻⁵/10⁻⁴ = 1/K
0.25 = 1/K
K = 4
Q30. Two point charges of +3 μC each are kept at a distance of 1 m. What is the magnitude of force acting on each?
Answer: F = kq₁q₂/r²
F = (9×10⁹ × 3×10⁻⁶ × 3×10⁻⁶)/(1)²
F = 9×10⁹ × 9×10⁻¹²
F = 8.1 × 10⁻² N = 0.081 N
Q31. The force of attraction between two charges separated by certain distance in air is 10 N. What will be the force if the distance is reduced to half?
Answer: F ∝ 1/r²
F₁/F₂ = r₂²/r₁²
10/F₂ = (r/2)²/r²
10/F₂ = 1/4
F₂ = 40 N
Q32. Two spheres carrying charges +6 μC and -8 μC attract each other with a force of 18 N. If each sphere is given an additional charge of +2 μC, what will be the new force?
Answer: Let distance = r
Initial: k(6)(8)/r² = 18 So, k(48)/r² = 18
After adding +2 μC: q₁' = 6 + 2 = 8 μC q₂' = -8 + 2 = -6 μC
New force: F' = k(8)(6)/r² = k(48)/r² = 18 N
F' = 18 N (attractive)
Q33. A charge Q is divided into two parts q and Q-q. For what value of q will the force between them be maximum when placed at a fixed distance?
Answer: F = kq(Q-q)/r²
For maximum F, dF/dq = 0
k(Q-2q)/r² = 0
Q - 2q = 0
q = Q/2
Q34. Two pith balls each weigh 0.1 g and have equal charges. When suspended by silk threads of length 50 cm each from a common point, they repel to a distance of 6 cm. Find the charge.
Answer: From geometry: sinθ = 3/50 = 0.06 tanθ ≈ 0.06
For equilibrium: tanθ = F/(mg) = kq²/(mgr²)
q² = (mgr²tanθ)/k
q² = (10⁻⁴ × 10 × (0.06)² × 0.06)/(9×10⁹)
q² = (2.16 × 10⁻⁷)/(9×10⁹)
q = 4.9 × 10⁻⁹ C = 4.9 nC
Q35. State the vector form of Coulomb's law.
Answer: Vector form:
F₁₂ = k(q₁q₂/r²)r̂₁₂
or
F₁₂ = (1/4πε₀)(q₁q₂/r₁₂³)r₁₂
where:
- F₁₂ = force on charge q₁ due to q₂
- r̂₁₂ = unit vector from q₂ to q₁
- r₁₂ = position vector from q₂ to q₁
Basic Level Questions
Q36. What is principle of superposition of charges? State it.
Answer: Principle of Superposition:
The resultant force on a charge due to a number of other charges is the vector sum of the individual forces acting on that charge due to all other charges.
Mathematically: F = F₁ + F₂ + F₃ + ... + Fₙ
Each force is calculated independently, ignoring the presence of other charges.
Q37. Two charges 1 μC and 2 μC are kept at a distance of 30 cm. Find the electric field at a point midway between them.
Answer: Distance of midpoint from each charge = 15 cm = 0.15 m
E₁ = k(1×10⁻⁶)/(0.15)² = (9×10⁹ × 10⁻⁶)/0.0225 = 4 × 10⁵ N/C (towards left)
E₂ = k(2×10⁻⁶)/(0.15)² = (9×10⁹ × 2×10⁻⁶)/0.0225 = 8 × 10⁵ N/C (towards right)
Both fields are in same direction (towards 1 μC)
Net E = E₁ + E₂ = 4×10⁵ + 8×10⁵
E = 12 × 10⁵ N/C = 1.2 × 10⁶ N/C
Q38. Define electric field intensity. Write its SI unit.
Answer: Electric field intensity (E) at a point is defined as the force experienced by a unit positive charge placed at that point.
E = F/q₀
where q₀ is a test charge (very small positive charge)
SI Unit: N/C (Newton per Coulomb) or V/m (Volt per meter)
Dimension: [MLT⁻³A⁻¹]
Q39. Two charges of 10 μC and -10 μC are placed at points A and B separated by a distance of 10 cm. Find the electric field at point C on the perpendicular bisector at a distance of 6 cm from the midpoint of AB.
Answer: Distance from each charge to C: r = √(5² + 6²) = √61 cm
E₁ = E₂ = k(10×10⁻⁶)/(61×10⁻⁴)
Components perpendicular to AB cancel. Components along AB add up.
E_net = 2E₁cosθ where cosθ = 5/√61
E = 2 × (9×10⁹ × 10×10⁻⁶)/(61×10⁻⁴) × 5/√61
E = 2 × 9×10⁷/61 × 5/√61
E = 1.19 × 10⁷ N/C
Q40. What happens to the force between two charges if the distance between them is tripled and each charge is doubled?
Answer: Initial: F = kq₁q₂/r²
Final: F' = k(2q₁)(2q₂)/(3r)²
F' = k(4q₁q₂)/(9r²)
F' = (4/9) × kq₁q₂/r²
F' = (4/9)F
Force becomes 4/9 times the original
Q41. Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them is the electric field zero?
Answer: Zero field point lies between charges, closer to smaller charge.
Let distance from +q = x
k(4q)/(0.3-x)² = kq/x²
4/((0.3-x)²) = 1/x²
4x² = (0.3-x)²
2x = 0.3-x
3x = 0.3
x = 0.1 m = 10 cm from +q charge
Q42. A charge q is placed at the center of the line joining two equal charges Q. Show that the system of three charges will be in equilibrium if q = -Q/4.
Answer: Let distance between two Q charges = 2a Charge q is at center (distance a from each Q)
For Q to be in equilibrium: Force from other Q = Force from q
kQ²/(2a)² = kQq/a²
Q²/4a² = Qq/a²
Q/4 = q
q = -Q/4 (negative for equilibrium)
Q43. Two identical positive charges are placed at opposite corners of a square. Another identical positive charge is placed at one of the remaining corners. Where should the fourth identical positive charge be placed so that all four charges are in equilibrium?
Answer: For symmetry and equilibrium, the fourth charge must be placed at the remaining corner of the square.
All four charges at the four corners create a symmetric configuration where:
- Diagonal forces cancel
- Adjacent forces balance
Fourth charge at the remaining corner
Q44. A charge of 10 μC is distributed uniformly over a circular ring of radius 20 cm. What is the electric field at the center of the ring?
Answer: For a uniformly charged ring, the electric field at the center is ZERO.
This is because:
- Every element dq on the ring produces a field dE at center
- For every element, there is a diametrically opposite element
- Their fields cancel each other due to symmetry
E = 0 at center
Q45. Can two like charges attract each other? Explain.
Answer: Generally NO, but in special cases YES.
Normal case: Two like charges always repel by Coulomb's law.
Special case: If charges are induced or in presence of conducting medium:
- Two like charges on a conductor can attract due to induced charges
- In a dielectric medium with complex geometry
- When charges are on flexible strings and system energy is considered
Under standard conditions, like charges REPEL.
Q46. Three charges +q, -q and +q are placed at the corners of an equilateral triangle. What is the resultant force on each charge?
Answer: On +q at vertex A:
- Force from -q: F₁ = kq²/a² (attractive)
- Force from +q: F₂ = kq²/a² (repulsive)
- Angle between = 60°
Resultant = √(F₁² + F₂² + 2F₁F₂cos60°) = √(3) × kq²/a²
F = √3 kq²/a² on each charge
Q47. Calculate the electric field due to a charge of 4 × 10⁻⁸ C at a distance of 2 m from it.
Answer: E = kq/r²
E = (9×10⁹ × 4×10⁻⁸)/(2)²
E = (36×10)/4
E = 90 N/C
Q48. Two point charges of +1 μC and -1 μC are separated by 1 cm. Calculate the electric field at a point on the perpendicular bisector at 10 cm from the center.
Answer: r = √(0.5² + 10²) = √100.25 cm ≈ 10.01 cm
For dipole on perpendicular bisector: E = 2E₁sinθ where sinθ = 0.5/10.01
E = 2 × k(1×10⁻⁶)/(10.01×10⁻²)² × 0.5/10.01
E ≈ k × 10⁻⁶/(10⁻²)³
E ≈ 9 × 10³ N/C
Q49. Write two important properties of electric field lines.
Answer: Two Important Properties:
- Electric field lines never intersect each other.
- If they intersect, there would be two directions of electric field at that point, which is impossible.
- Electric field lines start from positive charges and end on negative charges.
- For isolated positive charge, lines go to infinity
- For isolated negative charge, lines come from infinity
Q50. The electric field in a region is given by E = 3i + 4j N/C. Find the magnitude of electric field.
Answer: E = 3i + 4j N/C
Magnitude: |E| = √(E_x² + E_y²)
|E| = √(3² + 4²)
|E| = √(9 + 16)
|E| = √25
|E| = 5 N/C
Quick Formula Reference
- Coulomb's Law: F = kq₁q₂/r² where k = 9×10⁹ Nm²/C²
- Electric Field: E = F/q = kq/r²
- Force in medium: F = kq₁q₂/(Kr²)
- Dipole moment: p = q × 2l
- Superposition: F_net = ΣF_i
Coulomb's Law - 2 Marks Questions (Q51-Q100)
NEET Level Questions
Q51. Two charges of magnitude +5 μC and -5 μC are placed 20 cm apart. Calculate the electric field intensity at a point on the axial line at a distance of 20 cm from the midpoint on the side of positive charge.
Answer: Distance from +5 μC = 20 - 10 = 10 cm = 0.1 m Distance from -5 μC = 20 + 10 = 30 cm = 0.3 m
E₁ = k(5×10⁻⁶)/(0.1)² = (9×10⁹ × 5×10⁻⁶)/0.01 = 4.5×10⁶ N/C (→)
E₂ = k(5×10⁻⁶)/(0.3)² = (9×10⁹ × 5×10⁻⁶)/0.09 = 5×10⁵ N/C (→)
Both in same direction: E_net = E₁ + E₂ = 4.5×10⁶ + 5×10⁵
E = 5 × 10⁶ N/C
Q52. Two equal charges q are placed at a distance 2a and a third charge -2q is placed at the midpoint. What is the potential energy of the system?
Answer: U = U₁₂ + U₁₃ + U₂₃
U₁₂ = k(q)(q)/(2a) = kq²/2a
U₁₃ = k(q)(-2q)/a = -2kq²/a
U₂₃ = k(q)(-2q)/a = -2kq²/a
U_total = kq²/2a - 2kq²/a - 2kq²/a
U = kq²/2a - 4kq²/a
U = -7kq²/2a
Q53. An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius 0.53 Å. Calculate the electrostatic force of attraction between them. (e = 1.6×10⁻¹⁹ C)
Answer: r = 0.53 Å = 0.53 × 10⁻¹⁰ m
F = ke²/r²
F = (9×10⁹ × (1.6×10⁻¹⁹)²)/(0.53×10⁻¹⁰)²
F = (9×10⁹ × 2.56×10⁻³⁸)/(0.2809×10⁻²⁰)
F = (23.04×10⁻²⁹)/(0.2809×10⁻²⁰)
F = 8.2 × 10⁻⁸ N
Q54. A point charge causes an electric flux of -1.0 × 10³ Nm²/C to pass through a spherical Gaussian surface of 10 cm radius centered on the charge. What is the value of the point charge?
Answer: By Gauss's law: Φ = q/ε₀
q = Φ × ε₀
q = -1.0×10³ × 8.85×10⁻¹²
q = -8.85 × 10⁻⁹ C = -8.85 nC
Q55. Two small charged spheres repel each other with a force 2×10⁻³ N. The charge on one sphere is twice that on the other. What is the sum of charges if the distance between them is 30 cm?
Answer: Let charges be q and 2q
F = k(q)(2q)/r²
2×10⁻³ = (9×10⁹ × 2q²)/(0.3)²
2×10⁻³ = (18×10⁹ × q²)/0.09
q² = (2×10⁻³ × 0.09)/(18×10⁹)
q² = 10⁻¹⁴
q = 10⁻⁷ C
Sum = q + 2q = 3q = 3×10⁻⁷ C
Sum = 3 × 10⁻⁷ C = 0.3 μC
Q56. Three charges -q, +Q and -q are placed at equal distances on a straight line. For the total force on +Q to be zero, what should be the ratio Q/q?
Answer: Let distance between adjacent charges = d
Force on +Q from left -q: F₁ = kqQ/d² (←) Force on +Q from right -q: F₂ = kqQ/d² (→)
Since both are equal and opposite: Net force is ZERO for any value of Q/q
The system is always in equilibrium.
Q/q can be any value
Q57. Two identical spheres having unequal opposite charges are placed at a distance of 0.9 m apart. After touching each other, they are again placed at the same distance apart. Now the force of repulsion between them is 0.025 N. Calculate the final charge on each sphere.
Answer: After touching, charges become equal = q each
F = kq²/r²
0.025 = (9×10⁹ × q²)/(0.9)²
0.025 = (9×10⁹ × q²)/0.81
q² = (0.025 × 0.81)/(9×10⁹)
q² = 2.25×10⁻¹²
q = 1.5 × 10⁻⁶ C = 1.5 μC on each sphere
Q58. A cube of side 'a' has charges at all its 8 vertices. If a charge +q is at one vertex, then what is the force on it due to the other 7 charges?
Answer: 7 charges at other vertices create forces:
- 3 adjacent vertices at distance a: Force = 3kq²/a²
- 3 face diagonal vertices at distance a√2: Force = 3kq²/2a²
- 1 body diagonal vertex at distance a√3: Force = kq²/3a²
Vector sum along body diagonal: F = kq²/a² [3/√3 + 3/(2√2) + 1/3√3]
After calculation: F = kq²/a² × [√3 + 3/(2√2) + 1/(3√3)]
Simplified: F ≈ 2.77 kq²/a²
Q59. The ratio of the forces between two small spheres with constant charge (a) in air (b) in a medium of dielectric constant K, at the same distance is:
Answer: Force in air: F_air = kq₁q₂/r²
Force in medium: F_medium = kq₁q₂/(Kr²)
Ratio = F_air/F_medium = K/1
Ratio = K : 1
Q60. An electric dipole of moment p is placed in a uniform electric field E with p parallel to E. Find the work done in rotating the dipole by 180°.
Answer: Work done: W = -pE(cosθ₂ - cosθ₁)
Initial angle θ₁ = 0° Final angle θ₂ = 180°
W = -pE(cos180° - cos0°)
W = -pE(-1 - 1)
W = -pE(-2)
W = 2pE
CUET Level Questions
Q61. Two point charges of +16 μC and -9 μC are placed 8 cm apart. At what distance from +16 μC will the electric potential be zero on the line joining them?
Answer: Let distance from +16 μC = x cm
V = V₁ + V₂ = 0
k(16)/x + k(-9)/(8-x) = 0
16/x = 9/(8-x)
16(8-x) = 9x
128 - 16x = 9x
128 = 25x
x = 5.12 cm from +16 μC charge
Q62. A charge Q is divided into two parts q and (Q-q). What is the ratio of q to Q so that the force between them is maximum when separated by a fixed distance?
Answer: F = kq(Q-q)/r²
For maximum F: dF/dq = 0
k(Q-2q)/r² = 0
Q - 2q = 0
q = Q/2
Ratio: q/Q = 1/2 or 1:2
Q63. Two charges of +25 μC and -25 μC are placed 6 mm apart. Find the electric field at a point 4 mm away from the midpoint on the equatorial line.
Answer: For dipole on equatorial line:
E = kp/r³ where p = q × 2l
p = 25×10⁻⁶ × 6×10⁻³ = 1.5×10⁻⁷ C·m
r = 4 mm = 4×10⁻³ m
E = (9×10⁹ × 1.5×10⁻⁷)/(4×10⁻³)³
E = 1.35×10³/64×10⁻⁹
E = 2.11 × 10¹⁰ N/C
Q64. The force between two electrons when placed in air is F. What will be the force when they are placed at the same distance in a medium of dielectric constant 4?
Answer: F_medium = F_air/K
F_medium = F/4
Force = F/4
Q65. Four charges +q, +q, -q and -q are placed at the corners of a square taken in order. What is the electric field and potential at the center?
Answer: At center:
- Distance from each charge = a/√2 (a = side)
Electric field: All four fields cancel due to symmetry E = 0
Potential: V = k[q/(a/√2) + q/(a/√2) - q/(a/√2) - q/(a/√2)]
V = 0
Both E = 0 and V = 0
Q66. Two insulating small spheres are rubbed against each other and placed 1 cm apart. If they attract each other with a force of 0.1 N, how many electrons were transferred from one sphere to the other?
Answer: F = ke²n²/r² where n = number of electrons
0.1 = (9×10⁹ × (1.6×10⁻¹⁹)² × n²)/(0.01)²
0.1 = (9×10⁹ × 2.56×10⁻³⁸ × n²)/10⁻⁴
n² = (0.1 × 10⁻⁴)/(9×10⁹ × 2.56×10⁻³⁸)
n² = 4.34×10¹⁸
n = 6.6 × 10⁹ electrons
Q67. What is the amount of work done in moving a point charge Q around a circular arc of radius r at the center of which another point charge q is located?
Answer: Work done by electrostatic force in a closed path = 0
Since electrostatic force is conservative:
W = 0
Q68. The distance between charges +5 μC and +10 μC is 20 cm. What is the distance of the point from +5 μC where the net electric field is zero?
Answer: Let distance from +5 μC = x
k(5)/x² = k(10)/(0.2-x)²
5(0.2-x)² = 10x²
(0.2-x)² = 2x²
0.2-x = √2 x
0.2 = x(1+√2)
x = 0.2/(1+√2) = 0.2/2.414
x = 0.0828 m = 8.28 cm from +5 μC
Q69. Two point charges 4 μC and 9 μC are separated by 12 cm in air. At what distance from 4 μC charge should a third charge be placed so that it experiences no net force?
Answer: Let distance from 4 μC = x
k(4)/x² = k(9)/(12-x)²
4(12-x)² = 9x²
2(12-x) = 3x
24 - 2x = 3x
24 = 5x
x = 4.8 cm from 4 μC charge
Q70. A hollow metal sphere of radius 10 cm is charged such that the potential on its surface is 80 V. What is the potential at the center of the sphere?
Answer: For a hollow conducting sphere: Potential inside = Potential on surface
V_center = V_surface = 80 V
Q71. Two charges +4 μC and +16 μC are separated by a distance of 0.6 m. Where should a third charge be placed so that it does not experience any force?
Answer: Point must be between charges, closer to smaller charge.
Let distance from +4 μC = x
k(4)/x² = k(16)/(0.6-x)²
4(0.6-x)² = 16x²
(0.6-x)² = 4x²
0.6-x = 2x
0.6 = 3x
x = 0.2 m = 20 cm from +4 μC
Q72. A proton is placed in a uniform electric field of 2.5 × 10³ N/C. Calculate the force experienced by the proton.
Answer: F = qE
F = 1.6×10⁻¹⁹ × 2.5×10³
F = 4 × 10⁻¹⁶ N
Q73. Two point charges 2 μC and -2 μC are placed at points A and B, 6 cm apart. Calculate electric dipole moment.
Answer: p = q × 2l
where 2l = 6 cm = 6×10⁻² m
p = 2×10⁻⁶ × 6×10⁻²
p = 1.2 × 10⁻⁷ C·m
Q74. The electric field at a point on equatorial line of a dipole is E. What will be the field at the same distance on the axial line?
Answer: E_axial = 2kp/r³
E_equatorial = kp/r³
Ratio: E_axial/E_equatorial = 2
E_axial = 2E_equatorial
E_axial = 2E
Q75. Can the potential at a point be zero while the electric field is not zero? Give example.
Answer: YES, it is possible.
Example: At the midpoint between two equal and opposite charges (+q and -q):
- V = kq/r + k(-q)/r = 0
- E ≠ 0 (fields add up)
Electric potential is scalar, electric field is vector.
Potential = 0, but E ≠ 0
Basic Level Questions
Q76. What is the SI unit of electric dipole moment?
Answer: Electric dipole moment: p = q × 2l
SI Unit: Coulomb-meter (C·m)
Dimension: [AT·L]
Q77. Two charges +3 μC and -3 μC are placed 10 cm apart. Calculate the dipole moment.
Answer: p = q × 2l
p = 3×10⁻⁶ × 10×10⁻²
p = 3 × 10⁻⁷ C·m
Q78. State the difference between electrostatic force and gravitational force.
Answer:
| Electrostatic Force | Gravitational Force |
|---|---|
| Can be attractive or repulsive | Always attractive |
| Depends on medium (K) | Independent of medium |
| Much stronger (10⁴² times) | Much weaker |
| Acts between charges | Acts between masses |
| F = kq₁q₂/r² | F = Gm₁m₂/r² |
Q79. Define linear charge density. Give its unit.
Answer: Linear charge density (λ): Charge per unit length of a linear conductor.
λ = Q/L
SI Unit: C/m (Coulomb per meter)
Dimension: [AT L⁻¹]
Q80. Four charges each equal to +q are placed at the four corners of a square. What charge should be placed at the center so that all charges are in equilibrium?
Answer: Let charge at center = Q
Force on corner charge due to:
- Two adjacent charges
- One opposite charge
- Center charge
For equilibrium, force from center must balance others.
By symmetry and calculation:
Q = -q(2+√2)/2 = -q(1+√2/2)
Or approximately: Q = -1.707q
Q81. An electric dipole is placed in a uniform electric field. In which position will it be in stable equilibrium?
Answer: Torque on dipole: τ = pE sinθ
Potential energy: U = -pE cosθ
For stable equilibrium: U should be minimum
U is minimum when cosθ = maximum = 1 θ = 0°
When dipole moment is parallel to electric field (θ = 0°)
Q82. Two charges of 1 C each are placed 1 km apart. Find the force between them.
Answer: F = kq₁q₂/r²
F = (9×10⁹ × 1 × 1)/(1000)²
F = 9×10⁹/10⁶
F = 9000 N = 9 kN
Q83. What is the nature of electrostatic force - conservative or non-conservative?
Answer: Electrostatic force is CONSERVATIVE.
Properties:
- Work done in closed loop = 0
- Work done is path-independent
- Potential energy can be defined
- Curl of E = 0
Q84. An electron falls through a distance of 1.5 cm in a uniform electric field of 2×10⁴ N/C. Calculate the work done by the field.
Answer: W = qEd
W = 1.6×10⁻¹⁹ × 2×10⁴ × 1.5×10⁻²
W = 1.6×10⁻¹⁹ × 2×10⁴ × 1.5×10⁻²
W = 4.8×10⁻¹⁷ J
W = 4.8 × 10⁻¹⁷ J
Q85. Why do two field lines never cross each other?
Answer: If two field lines cross:
- At the point of intersection, there would be two tangents
- This means two directions of electric field at same point
- This is physically impossible
Electric field has unique direction at every point.
Therefore, field lines never intersect.
Q86. An electric dipole is placed at an angle of 30° with an electric field of intensity 2×10⁵ N/C. The dipole moment is 4×10⁻²⁹ C·m. Calculate the torque acting on the dipole.
Answer: τ = pE sinθ
τ = 4×10⁻²⁹ × 2×10⁵ × sin30°
τ = 8×10⁻²⁴ × 0.5
τ = 4 × 10⁻²⁴ N·m
Q87. What is the electric flux through a cube of side 1 cm if a point charge of 1 μC is at its center?
Answer: By Gauss's law: Φ = q/ε₀
Φ = (1×10⁻⁶)/(8.85×10⁻¹²)
Φ = 1.13 × 10⁵ Nm²/C
(Note: Size of cube doesn't matter)
Q88. Two point charges 4q and q are separated by a distance r. At what point on the line joining them is electric field zero?
Answer: Let point be at distance x from q
k(q)/x² = k(4q)/(r-x)²
(r-x)² = 4x²
r-x = 2x
r = 3x
x = r/3 from charge q
Q89. State Gauss's law in electrostatics.
Answer: Gauss's Law:
The total electric flux through a closed surface is equal to 1/ε₀ times the total charge enclosed by the surface.
Φ = q_enclosed/ε₀
or
∮E·dA = q/ε₀
Q90. An α-particle (He nucleus) is accelerated through a potential difference of 1000 V. Calculate the kinetic energy gained.
Answer: Charge on α-particle = 2e = 2 × 1.6×10⁻¹⁹ C
K.E. = qV
K.E. = 2 × 1.6×10⁻¹⁹ × 1000
K.E. = 3.2×10⁻¹⁶ J
Converting to eV: K.E. = 1000 × 2 = 2000 eV
K.E. = 3.2 × 10⁻¹⁶ J = 2 keV
Q91. Two charges 2 μC and 8 μC are placed 10 cm apart. Where will the electric potential be zero on the line joining them?
Answer: Let point be at distance x from 2 μC
k(2)/x = k(8)/(0.1-x)
2(0.1-x) = 8x
0.2 - 2x = 8x
0.2 = 10x
x = 0.02 m = 2 cm from 2 μC charge
Q92. What is the dimensional formula of electric field intensity?
Answer: E = F/q
[E] = [Force]/[Charge]
[E] = [MLT⁻²]/[AT]
[E] = [MLT⁻³A⁻¹]
Q93. Two point charges of +1 μC and -1 μC are kept at a distance of 2 cm. Calculate electric potential at the midpoint.
Answer: Distance from midpoint to each charge = 1 cm = 0.01 m
V = V₁ + V₂
V = k(1×10⁻⁶)/0.01 + k(-1×10⁻⁶)/0.01
V = k×10⁻⁴ - k×10⁻⁴
V = 0
Q94. Define surface charge density. Give its SI unit.
Answer: Surface charge density (σ): Charge per unit area of a surface.
σ = Q/A
SI Unit: C/m² (Coulomb per square meter)
Dimension: [AT L⁻²]
Q95. A charge of 5 μC experiences a force of 2000 N in an electric field. Find the electric field intensity.
Answer: E = F/q
E = 2000/(5×10⁻⁶)
E = 4×10⁸ N/C
E = 4 × 10⁸ N/C
Q96. Why is electrostatic potential constant throughout the volume of a conductor?
Answer: Inside a conductor at electrostatic equilibrium:
- Electric field E = 0
- Since E = -dV/dr = 0
- Therefore dV = 0
- V = constant
Potential is constant because E = 0 inside conductor.
Q97. Two charges 3 μC and 5 μC are separated by 8 cm. Calculate force between them in (a) air (b) water (K = 81).
Answer: (a) In air: F = kq₁q₂/r²
F = (9×10⁹ × 3×10⁻⁶ × 5×10⁻⁶)/(0.08)²
F = (135×10⁻³)/6.4×10⁻³
F_air = 21.09 N
(b) In water: F_water = F_air/K = 21.09/81
F_water = 0.26 N
Q98. What is the direction of electric field at a point on the perpendicular bisector of an electric dipole?
Answer: On the perpendicular bisector (equatorial line):
Electric field is directed opposite to the dipole moment (from +q to -q side).
Direction: Parallel to dipole axis, opposite to p
Q99. An electron experiences a force of 6.4×10⁻¹⁵ N in an electric field. Find the intensity of electric field.
Answer: E = F/q
E = (6.4×10⁻¹⁵)/(1.6×10⁻¹⁹)
E = 4 × 10⁴ N/C
Q100. Define volume charge density. Write its SI unit.
Answer: Volume charge density (ρ): Charge per unit volume of a body.
ρ = Q/V
SI Unit: C/m³ (Coulomb per cubic meter)
Dimension: [AT L⁻³]
Important Constants to Remember
- k = 9 × 10⁹ Nm²/C²
- ε₀ = 8.85 × 10⁻¹² C²/Nm²
- e = 1.6 × 10⁻¹⁹ C
- G = 6.67 × 10⁻¹¹ Nm²/kg²
- m_e = 9.1 × 10⁻³¹ kg
- m_p = 1.67 × 10⁻²⁷ kg
