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JEE Advanced Physics Electrostatics 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Physics Question (2025) — Solution

Question

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 _0 =9 10^9 Nm ^2 / C ^2 (where _0 is the permittivity of free space), which of the following statements is/are correct: :

Options

  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 10^ -9 C .
  3. C. After the grounding is removed, the charge on the sphere is -5 10^ -9 C .
  4. D. The final electrostatic potential of the sphere is 300 V .

Answer

C. After the grounding is removed, the charge on the sphere is -5 10^ -9 C .

Step-by-step solution

Before grounding aligned & V _ sphere = ( V _ C )_ net = ( V _ C )_ q + ( V _ C )_ ind \\ & V _ sphere = kq +0= 9 10^9 10^ -8 0.2 = 90 0.2 = 900 2 =450 volt aligned After grounding aligned & kQ + kq _ S R = V _ sphere ^ =0 \\ & q _ s =- R q =- 1 2 10^ -8 =-5 10^ -9 \\ & q _ s =-5 10^ -9 Coulomb aligned Charge flower from sphere to ground =5 10^ -9 Coulomb After grounding is removed aligned & ( V _ sphere )_ final = kq ^ + kq _ s R \\ & = 9 10^9 10^2 10^ -8 30 - 9 10^9 5 10^ -9 10^2 10 \\ & = 9 1000 30 -450=300 volt -450 volt =-150 volt aligned

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