A conducting sphere of radius R is given a charge Q. What is the electric field and the electric potential at the centre of the sphere?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → zero and \( \displaystyle\frac{Q}{4\pi\varepsilon_0R} \)
In electrostatic equilibrium the charge given to a conductor resides on its outer surface, and the electric field inside a conductor is zero. So the field at the centre is zero.
Since \( E = -dV/dr \) and \( E = 0 \) inside, the potential does not change inside the sphere. It is constant and equal to its value at the surface.
\( \displaystyle V_{\text{centre}} = V_{\text{surface}} = \frac{Q}{4\pi\varepsilon_0R} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{Q}{4\pi\varepsilon_0R^{2}} \) and \( \displaystyle\frac{Q}{4\pi\varepsilon_0R} \) — Incorrect. \( Q/(4\pi\varepsilon_0R^{2}) \) is the field just outside the surface; inside the conductor the field is zero.
Option (2) \( \displaystyle\frac{Q}{4\pi\varepsilon_0R^{2}} \) and zero — Incorrect. The field at the centre is zero, not \( Q/(4\pi\varepsilon_0R^{2}) \), and the potential there is not zero.
Option (4) Both electric field and electric potential are zero — Incorrect. A zero field means only that the potential does not change inside; the potential keeps its surface value \( Q/(4\pi\varepsilon_0R) \).
NCERT (2026-27):
Electrostatics of Conductors – “… Since E = 0 inside the conductor and has no tangential component on the surface, no work is done in moving a small test charge within the conductor and on its surface. That is, there is no potential difference between any two points inside or on the surface of the conductor.”