Two conducting spheres of radii $R_1$ and $R_2$ (where $R_1 > R_2$) are kept at a very large distance from each other. They are charged to the same electrostatic potential and then connected by a fine conducting wire. The ratio of their final surface charge densities ($\sigma_1 / \sigma_2$) will be:

Two conducting spheres of radii R₁ and R₂ (R₁ > R₂) are kept at a very large distance from each other. They are charged to the same electrostatic potential and then connected by a fine conducting wire. The ratio of their final surface charge densities +

Correct Answer

σ₁ / σ₂ = R₂ / R₁

Step-by-Step Solution

For an isolated conducting sphere, the potential is:

V = kQ/R

Initially, both spheres are charged to the same potential. Therefore:

kQ₁/R₁ = kQ₂/R₂

Hence:

Q₁/Q₂ = R₁/R₂

After connecting them with a conducting wire, they must again reach the same final potential. Therefore:

Q₁/R₁ = Q₂/R₂

So the final charges are also proportional to their radii:

Q₁ : Q₂ = R₁ : R₂

Surface Charge Density

Surface charge density is:

σ = Q/(4πR²)

Therefore:

σ₁/σ₂ = [Q₁/(4πR₁²)] / [Q₂/(4πR₂²)]

Using:

Q₁/Q₂ = R₁/R₂

we get:

σ₁/σ₂ = (R₁/R₂) × (R₂²/R₁²)

Therefore:

σ₁/σ₂ = R₂/R₁

Important Concept

Connected conducting spheres → Same potential.

For isolated spheres: V ∝ Q/R, so Q ∝ R when potentials are equal.

But surface area is proportional to R². Therefore:

σ = Q/(4πR²) ∝ 1/R

JEE Quick Trick

For two connected spheres:

V₁ = V₂  →  Q₁/R₁ = Q₂/R₂

σ ∝ 1/R

So the smaller sphere has the greater surface charge density.

30 Related MCQs with Solutions

1. Two conducting spheres are connected by a wire. Their final potentials are:

A. Different

B. Equal

C. Zero

D. Infinite

Answer: B. Equal

2. The potential of an isolated conducting sphere is:

A. kQR

B. kQ/R

C. kR/Q

D. Q/R²

Answer: B. kQ/R

3. For equal potential, charge on a conducting sphere varies as:

A. 1/R

B. R

C. R²

D. R³

Answer: B. R

4. Surface charge density of a sphere is:

A. Q/(4πR)

B. Q/(4πR²)

C. QR²

D. Q/R³

Answer: B. Q/(4πR²)

5. If R₁ > R₂, which sphere has greater final surface charge density?

A. Sphere 1

B. Sphere 2

C. Both equal

D. Cannot be determined

Answer: B. Sphere 2

6. The ratio σ₁/σ₂ is:

A. R₁/R₂

B. R₂/R₁

C. (R₁/R₂)²

D. 1

Answer: B. R₂/R₁

7. If R₁ = 2R₂, then σ₁/σ₂ equals:

A. 2

B. 1/2

C. 4

D. 1/4

Answer: B. 1/2

8. If R₁ = 3R₂, then σ₂/σ₁ is:

A. 1/3

B. 3

C. 9

D. 1

Answer: B. 3

9. When two conducting spheres are connected, charge flows until:

A. Charges become equal

B. Potentials become equal

C. Radii become equal

D. Surface areas become equal

Answer: B. Potentials become equal

10. Which sphere has more total charge when both have equal potential?

A. Smaller sphere

B. Larger sphere

C. Both equal

D. Neither

Answer: B. Larger sphere

11. For equal potential, Q₁/Q₂ is:

A. R₂/R₁

B. R₁/R₂

C. 1

D. R₁²/R₂²

Answer: B. R₁/R₂

12. Surface area of a sphere is proportional to:

A. R

B. R²

C. 1/R

D. R³

Answer: B. R²

13. For equal potential, surface charge density varies as:

A. R

B. R²

C. 1/R

D. R³

Answer: C. 1/R

14. The electric field just outside a conductor is related to surface charge density by:

A. E = σ/ε₀

B. E = ε₀σ

C. E = σR

D. E = R/σ

Answer: A. E = σ/ε₀

15. Therefore, at the surfaces of the two spheres:

A. E₁/E₂ = R₁/R₂

B. E₁/E₂ = R₂/R₁

C. E₁ = E₂ always

D. E = 0

Answer: B. R₂/R₁

16. Which sphere has the stronger electric field immediately outside its surface?

A. Larger sphere

B. Smaller sphere

C. Both equal

D. Neither

Answer: B. Smaller sphere

17. If the two spheres have equal radii, their final surface charge densities are:

A. Equal

B. Unequal

C. Opposite

D. Zero

Answer: A. Equal

18. The conductor wire ensures that both spheres have:

A. Equal charge

B. Equal potential

C. Equal radius

D. Equal surface area

Answer: B. Equal potential

19. If R₁ = 4R₂, then Q₁/Q₂ is:

A. 1/4

B. 4

C. 16

D. 1

Answer: B. 4

20. If R₁ = 4R₂, then σ₁/σ₂ is:

A. 4

B. 1/4

C. 16

D. 1/16

Answer: B. 1/4

21. Capacitance of an isolated spherical conductor is proportional to:

A. R

B. R²

C. 1/R

D. R³

Answer: A. R

22. A larger conducting sphere at the same potential carries:

A. Less charge

B. More charge

C. No charge

D. Same charge always

Answer: B. More charge

23. The relation between charge and radius at equal potential is:

A. Q ∝ R

B. Q ∝ R²

C. Q ∝ 1/R

D. Q ∝ R³

Answer: A. Q ∝ R

24. Which quantity is inversely proportional to radius for equal potential spheres?

A. Total charge

B. Surface charge density

C. Capacitance

D. Surface area

Answer: B. Surface charge density

25. If σ₁ < σ₂, then:

A. R₁ < R₂

B. R₁ > R₂

C. R₁ = R₂

D. No relation

Answer: B. R₁ > R₂

26. At electrostatic equilibrium, the potential throughout a conductor is:

A. Variable

B. Constant

C. Zero always

D. Infinite

Answer: B. Constant

27. When connected spheres are far apart, their mutual influence can be:

A. Neglected approximately

B. Infinite

C. Exactly zero in every case

D. Dominant

Answer: A. Neglected approximately

28. For two connected spheres, which statement is correct?

A. Equal charge density

B. Equal potential

C. Equal charge

D. Equal electric field everywhere

Answer: B. Equal potential

29. The smaller sphere has greater surface charge density because:

A. It carries no charge

B. Its surface area is smaller for its charge

C. Its potential is different

D. Its capacitance is zero

Answer: B

30. The final ratio σ₁/σ₂ for R₁ > R₂ is:

A. R₁/R₂

B. R₂/R₁

C. 1

D. R₁²/R₂²

Answer: B. R₂/R₁

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A parallel-plate capacitor is charged by a battery of voltage $V_0$. The battery is then completely disconnected, and a dielectric slab of dielectric constant $K$ is slowly inserted to completely fill the space between the plates. Which of the following statements correctly describes the changes in the system?

A parallel-plate capacitor is charged by a battery of voltage V₀. The battery is then completely disconnected, and a dielectric slab of dielectric constant K is slowly inserted to completely fill the space between the plates. Which of the following statements correctly describes the changes in the system? +

Correct Answer

Charge remains constant, capacitance becomes K times, voltage decreases by K times, electric field decreases by K times, and stored energy decreases by K times.

Step-by-Step Explanation

The most important point is that the battery is disconnected before inserting the dielectric. Therefore, the capacitor is isolated and no charge can flow onto or away from the plates.

Q = constant

Initially, without dielectric:

C₀ = ε₀A/d

Since the dielectric completely fills the space, the new capacitance becomes:

C = K C₀

Because Q remains constant:

V = Q/C

Therefore:

V = V₀/K

What Happens to Electric Field?

For a parallel-plate capacitor:

E = V/d

Since the plate separation d is unchanged and voltage becomes V₀/K:

E = E₀/K

Thus, the electric field decreases by a factor of K.

What Happens to Stored Energy?

For an isolated capacitor, charge is constant. Hence we use:

U = Q²/(2C)

Since C becomes KC₀:

U = U₀/K

So the stored electrostatic energy decreases by a factor of K. The decrease in electrostatic energy is associated with the work done by the electric field while the dielectric is pulled into the capacitor.

Final Changes at a Glance

Quantity	Initially	After dielectric	Change
Charge Q	Q0	Q0	Constant
Capacitance C	C0	KC0	Increases K times
Voltage V	V0	V0/K	Decreases K times
Electric field E	E0	E0/K	Decreases K times
Energy U	U0	U0/K	Decreases K times

JEE Quick Trick

Battery disconnected → Q constant.
Then remember:

C ↑ K times → V ↓ K times → E ↓ K times → U ↓ K times.

30 Related MCQs with Solutions

1. A charged capacitor is disconnected from the battery. A dielectric of constant K is inserted completely. The charge becomes:

A. KQ

B. Q/K

C. Q

D. Zero

Answer: C. Q

2. The capacitance after completely inserting dielectric is:

A. C/K

B. C

C. KC

D. K²C

Answer: C. KC

3. The voltage across the isolated capacitor becomes:

A. KV₀

B. V₀/K

C. V₀

D. K²V₀

Answer: B. V₀/K

4. The electric field between the plates becomes:

A. KE₀

B. E₀/K

C. E₀

D. K²E₀

Answer: B. E₀/K

5. The energy stored in the isolated capacitor becomes:

A. KU₀

B. U₀/K

C. U₀

D. K²U₀

Answer: B. U₀/K

6. Which quantity remains constant after battery disconnection?

A. Voltage

B. Capacitance

C. Charge

D. Electric field

Answer: C. Charge

7. If K = 4, the final capacitance is:

A. C/4

B. 2C

C. 4C

D. C

Answer: C. 4C

8. If K = 5, the final voltage is:

A. 5V₀

B. V₀/5

C. V₀

D. 25V₀

Answer: B. V₀/5

9. If K = 3, the electric field becomes:

A. 3E₀

B. E₀/3

C. E₀

D. 9E₀

Answer: B. E₀/3

10. For an isolated capacitor, the appropriate energy formula is:

A. U = ½CV²

B. U = Q²/(2C)

C. U = CV

D. U = Q/C

Answer: B. U = Q²/(2C)

11. Why does charge remain constant?

A. Battery continues charging

B. Plates are isolated

C. Dielectric creates charge

D. Voltage source increases

Answer: B. Plates are isolated

12. If capacitance increases while Q is constant, voltage:

A. Increases

B. Decreases

C. Remains constant

D. Becomes infinite

Answer: B. Decreases

13. Dielectric insertion changes capacitance because:

A. Permittivity increases

B. Plate area becomes zero

C. Charge disappears

D. Plate separation doubles

Answer: A

14. The permittivity of the dielectric is:

A. ε₀/K

B. Kε₀

C. K²ε₀

D. ε₀

Answer: B. Kε₀

15. For K > 1, dielectric insertion in an isolated capacitor causes capacitance to:

A. Increase

B. Decrease

C. Remain unchanged

D. Become zero

Answer: A

16. The relation Q = CV gives, for constant Q:

A. V ∝ C

B. V ∝ 1/C

C. V ∝ C²

D. V is independent of C

Answer: B. V ∝ 1/C

17. If K = 2, energy stored becomes:

A. 2U₀

B. U₀/2

C. 4U₀

D. U₀

Answer: B. U₀/2

18. The force between plates of an isolated capacitor after dielectric insertion generally:

A. Increases

B. Decreases

C. Remains exactly unchanged

D. Becomes infinite

Answer: B. Decreases

19. Which condition is essential for Q to remain constant?

A. Battery remains connected

B. Battery is disconnected

C. Voltage is increased

D. Plate area is changed

Answer: B

20. If the battery were still connected, the voltage would:

A. Remain V₀

B. Become V₀/K

C. Become zero

D. Become KV₀

Answer: A. V₀

21. If the battery remains connected, charge on the capacitor becomes:

A. Q/K

B. KQ

C. Q

D. Zero

Answer: B. KQ

22. The key difference between connected and disconnected cases is:

A. Connected → V constant; disconnected → Q constant

B. Both have Q constant

C. Both have V constant

D. Neither changes

Answer: A

23. When dielectric completely fills the capacitor, capacitance is:

A. ε₀A/d

B. Kε₀A/d

C. ε₀d/A

D. Kd/A

Answer: B

24. The electric field in a completely filled isolated capacitor is:

A. E₀

B. KE₀

C. E₀/K

D. K²E₀

Answer: C

25. The electrostatic energy density inside a dielectric-filled capacitor is:

A. Always increased by K

B. Reduced in this isolated case

C. Zero

D. Infinite

Answer: B

26. For the isolated capacitor, inserting dielectric causes V to change because:

A. Q changes

B. C changes while Q remains constant

C. Plate area becomes zero

D. Battery supplies more charge

Answer: B

27. Which relation is always valid for a capacitor?

A. Q = CV

B. Q = C/V

C. V = QC

D. C = QV

Answer: A. Q = CV

28. The dielectric reduces the electric field for an isolated capacitor because:

A. Polarization opposes the original field

B. Charge becomes infinite

C. Battery increases voltage

D. Plate separation becomes zero

Answer: A

29. Which statement is correct?

A. Disconnected capacitor → V constant

B. Disconnected capacitor → Q constant

C. Disconnected capacitor → C constant

D. Disconnected capacitor → U constant

Answer: B

30. If the dielectric constant is K, the final energy compared with initial energy is:

A. KU₀

B. U₀/K

C. K²U₀

D. U₀

Answer: B. U₀/K

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