A uniformly charged non-conducting sphere having charge Q and radius R is kept in gravity free space. Two identical charge particles are released from rest from ‘A’ and ‘B’ simultaneously. The speed of one of the charge particle when separation between them is very large, is K q n m R ( q 2 + 2 Q ) \sqrt{\frac{Kq}{nmR}\left(\frac{q}{2}+2Q\right)} nm R K q ( 2 q + 2 Q ) (mass of one of the particle is m), Find n. [C → centre of sphere and A, B and C in same line. AC = BC = 2R, K = 1 4 π ϵ ο \frac{1}{4\pi\epsilon_{\omicron}} 4 π ϵ ο 1 ]
Physics · Electrostatic
JEE Mains Advanced

Question

A uniformly charged non-conducting sphere having charge Q and radius R is kept in gravity free space. Two identical charge particles are released from rest from ‘A’ and ‘B’ simultaneously. The speed of one of the charge particle when separation between them is very large, is K q n m R ( q 2 + 2 Q ) \sqrt{\frac{Kq}{nmR}\left(\frac{q}{2}+2Q\right)} nm R K q ( 2 q + 2 Q ) (mass of one of the particle is m), Find n. [C → centre of sphere and A, B and C in same line. AC = BC = 2R, K = 1 4 π ϵ ο \frac{1}{4\pi\epsilon_{\omicron}} 4 π ϵ ο 1 ]

Physics · Electrostatic

G1 Physics · Accepted Answer

Step 1: In electrostatic equilibrium, E = 0 inside a conductor and the conductor is an equipotential. Surface charges adjust so the boundary condition field just outside·𝐧 = σ/ε₀ is satisfied. Step 2: For connected conductors set their potentials equal; for sheets use electrostatic pressure p = ε₀E²/2 when appropriate. Step 3: Apply charge conservation and the given geometry. Final: 2.