The potential of a uniformly charged spherical shell at a point r/2 from its centre is 5 V. The potential at its surface is: (Given radius of shell = r)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5 V
By Gauss's law the electric field inside a uniformly charged shell is zero. Since \( E = -dV/dr \), a zero field means the potential does not change anywhere inside, so it equals the potential on the surface.
\( \displaystyle V_{\text{inside}} = V_{\text{surface}} = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r} \)
The point at \( r/2 \) is inside the shell, so the potential at the surface is also 5 V.
NCERT (2026-27):
Potential Due to a System of Charges – “… The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\,\frac{q}{R} \qquad [2.19(b)] \)”