JEE Advanced 2025 · Paper 2 · Physics Q5

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PhysicsMultiple correct+4 / -2Question 21 of 48

A positive point charge of $10^{-8}$ 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 $\dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \mathrm{Nm^2/C^2}$ (where $\epsilon_0$ is the permittivity of free space), which of the following statements is/are correct:

Before the grounding, the electrostatic potential of the sphere is 450 V.

Charge flowing from the sphere to the ground because of grounding is

5\times10^{-9}

After the grounding is removed, the charge on the sphere is

-5\times10^{-9}

The final electrostatic potential of the sphere is 300 V.

Answer and worked solution

Official answer: A, B, C

Here $k=\frac1{4\pi\epsilon_0}$ , $q=10^{-8}$ C and $R=0.1$ m.

(A) True. The sphere is neutral, so its potential equals that at its centre: $\frac{kq}{0.2}=450$ V.
(B) True. When grounded, the potential is zero: $\frac{kQ'}{R}+\frac{kq}{0.2}=0$ , so $Q'=-\frac{qR}{0.2}=-5\times10^{-9}$ C. A charge of $+5\times10^{-9}$ C therefore flows to the ground.
(C) True. After the ground is removed, the sphere keeps $-5\times10^{-9}$ C.
(D) False. With $q$ at 0.3 m, $V=9\times10^9\left(\frac{-5\times10^{-9}}{0.1}+\frac{10^{-8}}{0.3}\right)=-150$ V, not 300 V.

Topics:Conductors and charge distribution Electric potential

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JEE Advanced 2025 · Paper 2, Physics question 5