JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Electrostatics

A positive point charge of

1 0^{- 8}

C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking

\frac{1}{4 π ϵ _{0}} = 9 \times 1 0^{9}

Nm²/C² (where

ϵ_{0}

is the permittivity of free space), which of the following statements is/are correct:

(A)Before the grounding, the electrostatic potential of the sphere is 450 V.✓ Correct
(B)Charge flowing from the sphere to the ground because of grounding is $5 \times 1 0^{- 9}$ C.✓ Correct
(C)After the grounding is removed, the charge on the sphere is $- 5 \times 1 0^{- 9}$ C.✓ Correct
(D)The final electrostatic potential of the sphere is 300 V.

Answer: (A), (B), (C)

Solution

Before grounding: The potential at the sphere's center due to the point charge is:

V = \frac{k q}{ℓ} = \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 8}}{0.2} = 450 V

After grounding: The sphere acquires charge

q_{s}

such that its potential becomes zero:

\frac{k q}{ℓ} + \frac{k q _{s}}{R} = 0 ⟹ q_{s} = - \frac{R}{ℓ} q = - \frac{0.1}{0.2} \times 1 0^{- 8} = - 5 \times 1 0^{- 9} C

Charge flowing to ground =

5 \times 1 0^{- 9}

C.

After moving the charge: New distance

ℓ^{'} = 0.3

m. Final potential:

V_{final} = \frac{k q}{ℓ ^{'}} + \frac{k q _{s}}{R} = \frac{9 \times 1 0 ^{9} \times 1 0 ^{- 8}}{0.3} - \frac{9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 9}}{0.1} = 300 - 450 = - 150 V

Options (A), (B), (C) are correct; (D) is incorrect.

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · One or more correct options

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JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Electrostatics

Solution

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