JEE Advanced 2025 Paper 2 Q05 Physics Electrostatics & Circuits Gauss+Conductor Hard

JEE Advanced 2025 Paper 2 · Q05 · Gauss+Conductor

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 $1/(4\pi\epsilon_0) = 9 \times 10^9$ N·m²/C² (where $\epsilon_0$ is the permittivity of free space), which of the following statements is/are correct:

  1. A. Before the grounding, the electrostatic potential of the sphere is $450$ V.
  2. B. Charge flowing from the sphere to the ground because of grounding is $5 \times 10^{-9}$ C.
  3. C. After the grounding is removed, the charge on the sphere is $-5 \times 10^{-9}$ C.
  4. D. The final electrostatic potential of the sphere is $300$ V.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C

Solution

Before grounding, the (neutral) sphere's potential equals the potential at its center due to the external point charge: $V = k q/d = (9 \times 10^9)(10^{-8})/0.2 = 450$ V. (A) TRUE. After grounding, $V_{\text{sphere}} = 0$. The induced (image) charge satisfies $q_s = -q R/d = -10^{-8}(0.1/0.2) = -5 \times 10^{-9}$ C. So $+5 \times 10^{-9}$ C of positive charge flows from the sphere to ground (sphere ends with $-5 \times 10^{-9}$ C). (B) TRUE. When grounding is removed and the source charge is moved further (to $d' = 30$ cm), the sphere keeps $q_s = -5 \times 10^{-9}$ C. (C) TRUE. Final potential: $V = k q_s/R + k q/d' = (9 \times 10^9)(-5 \times 10^{-9})/0.1 + (9 \times 10^9)(10^{-8})/0.3 = -450 + 300 = -150$ V (NOT $300$ V). (D) FALSE. Correct: (A), (B), (C).

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