E and V respectively represent magnitudes of electrostatic field and potential inside a charged metallic shell. We have :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( E = 0, V \ne 0 \)
The charge on a conductor resides on its outer surface. By Gauss's law, a Gaussian surface drawn inside the shell encloses no charge, so the field inside is zero.
Since \( E = -dV/dr = 0 \) inside, the potential does not change from point to point there: it is constant and equal to its value at the surface.
\( \displaystyle V_{\mathrm{inside}} = V_{\mathrm{surface}} = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R} \ne 0 \)
Option (1) \( E = 0, V = 0 \) — Incorrect. The field is zero, but the potential equals the non-zero surface value.
Option (2) \( E = 0, V \ne 0 \) — Correct. Zero field, and a constant non-zero potential equal to that at the surface.
Option (3) \( E \ne 0, V = 0 \) — Incorrect. There is no field inside a charged conducting shell.
Option (4) \( E \ne 0, V \ne 0 \) — Incorrect. The potential is non-zero, but the field inside is zero.
NCERT (2026-27):
Electrostatics of Conductors – “… 4. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface … That is, there is no potential difference between any two points inside or on the surface of the conductor.”
Electrostatic Shielding – “… We have proved a simple case of this result already: the electric field inside a charged spherical shell is zero.”