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RE-NEET2026Physics-Electrostatics

RE-NEET 2026 Physics Electric Potential MCQ Question

Type: MCQ-numerical-Hard-Class 12

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, as shown in the figure. The initial and final positions of the charge are marked by A and B at distances 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).

Question diagram
A

6

B

9

C

18

D

2

Correct Answer

Option C

Detailed Explanation

To find the work done on a unit positive charge moving from point A to point B, we use the concept of electric potential due to a uniformly charged sphere. The electric field outside the sphere behaves as if all the charge were concentrated at the center.

The potential VV at a distance rr from the center of the sphere is given by:

V=Q4πε0rV = \frac{Q}{4\pi \varepsilon_0 r}

where Q=43πR3ρQ = \frac{4}{3} \pi R^3 \rho.

Calculating the potential at points A (2R) and B (3R):

VA=Q8πε0RV_A = \frac{Q}{8\pi \varepsilon_0 R} VB=Q12πε0RV_B = \frac{Q}{12\pi \varepsilon_0 R}

The work done WW is:

W=VA−VB=Q8πε0R−Q12πε0RW = V_A - V_B = \frac{Q}{8\pi \varepsilon_0 R} - \frac{Q}{12\pi \varepsilon_0 R}

After simplifying, we find:

W=ρR218ε0W = \frac{\rho R^2}{18 \varepsilon_0}

Thus, n=18n = 18. The other options fail because they do not match the derived expression for work done.

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