JEE Main 2025 · 8 April · evening · Single correct · +4 / −1
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Work done in moving a test charge between two points inside a uniformly charged spherical shell is zero, no matter which path is chosen.
Reason R : Electrostatic potential inside a uniformly charged spherical shell is constant and is same as that on the surface of the shell.
In the light of the above statements, choose the correct answer from the options given below.
- (a) A is true but R is false
- (b) Both A and R are true and R is the correct explanation of A
- (c) Both A and R are true but R is NOT the correct explanation of A
- (d) A is false but R is true
Answer and solution
Answer (b) Both A and R are true and R is the correct explanation of A
Solution
Physics sir · 4 steps
Inside a charged shell the field is zero, so the potential is the same everywhere and no work is needed.
- Field inside. Draw a Gaussian sphere inside the shell. It encloses no charge, so by Gauss's law E = 0 at every point inside.
- Potential inside. Zero field means the potential does not change as you move inside. So it stays equal to its value at the surface, V = kQ/R. Statement R is true.
W_(A → B) = q(V_B - V_A) = q(kQ/R - kQ/R) = 0
- Work done. The electrostatic force is conservative, so the work depends only on the potential difference between the two points, not on the path. Every point inside is at the same potential, so the work is zero on any path. Statement A is true.
- So. A follows directly from R: equal potential at both points is exactly why the work is zero. The answer is (b).
The idea to keep. Work done by the electrostatic force is always q Δ V, whatever the path. Wherever the potential is constant, as inside a conductor or a charged shell, moving a charge costs zero work.
Answer from the NTA final answer key. Solution by PrepWiser.