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JEE Advanced Physics Electrostatics 2024 JEE Advanced 2024 (Paper 2)

JEE Advanced Physics Question (2024) — Solution

Question

An infinitely long thin wire, having a uniform charge density per unit length of 5 nC / m , is passing through a spherical shell of radius 1 ~m , as shown in the figure. A 10 nC charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points P and R , in Volt, is _______ . [Given: In SI units 1 4 _0 =9 10^9, 2=0.7. Ignore the area pierced by the wire.]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

due to wire aligned & dV =- E dx \\ & _ v _ P ^ v _ R dV =- _ 0.5 ^2 2 k x dx \\ & v _ R - v _ P =-2 k 2 0.5 \\ & =-2 9 10^9 3 10^ -9 2 0.7=-126 ~V aligned due to sphere aligned & v _ R - v _P= kQ 2 - kQ 1 =- kQ 2 = -9 10^9 10 10^ -9 2 \\ &=-45 ~V \\ & v _ R - v _ P =-126-45=-171 ~V \\ & v _ P - v _ R =171 ~V aligned

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