Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

A unit positive point charge is taken slowly through an infinitesimally thin tube that is inside a charged dielectric sphere of radius R, having uniform positive charge density ρ\rho, as shown in the figure. The initial and final positions of the charge are marked by A and B at distance 2R and 3R respectively, from the centre of the sphere. In this process, the magnitude of the total work done on the point charge is ρR2nε0\frac{\rho R^2}{n\varepsilon_0}. The value of n is :

(ε0(\varepsilon_0 is the permittivity of vacuum)

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Show answer and solution
Sol. Work done W=q(VB−VA)W=q(V_B-V_A)
=q[KQ3R−KQ2R]=KqR[−Q6]= q\left[\frac{KQ}{3R}-\frac{KQ}{2R}\right]=\frac{Kq}{R}\left[\frac{-Q}{6}\right]
∣W∣=KqQ6R=14πε0×qR×16×ρ×43πR3|W|=\frac{KqQ}{6R}=\frac{1}{4\pi\varepsilon_0}\times\frac{q}{R}\times\frac{1}{6}\times\rho\times\frac{4}{3}\pi R^3
∣W∣=ρR218ε0[∵q=1]|W|=\frac{\rho R^2}{18\varepsilon_0} \quad [\because q=1]
∴n=18\therefore n=18

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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