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NEET 2024 · Electric Potential & Equipotentials

Electric Potential & Equipotentials question from NEET 2024

Q43NEET 2024Easy

Electric Potential & Equipotentials

A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in V) shown in the figure is:

(Take $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^{9}$ SI units)

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Answer: (d)

Inside a charged spherical shell $E = 0$ , so the potential is constant and equal to the surface value.
$C$ (centre) and $P$ (on the surface) are therefore at the same potential: $V_C = V_P = \frac{kq}{R}$
$V_C - V_P = 0$
So the answer is (d) Zero.

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