JEE Advanced 2019 · Paper 1 · Physics Q2

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A thin spherical insulating shell of radius $R$ carries a uniformly distributed charge such that the potential at its surface is $V_0$ . A hole with a small area $\alpha4\pi R^2\ (\alpha \ll 1)$ is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

The potential at the center of the shell is reduced by

2\alpha V_0

The magnitude of electric field at the center of the shell is reduced by

\dfrac{\alpha V_0}{2R}

The ratio of the potential at the center of the shell to that of the point at

\frac12R

from center towards the hole will be

\dfrac{1-\alpha}{1-2\alpha}

The magnitude of electric field at a point, located on a line passing through the hole and shell's center, on a distance

2R

from the center of the spherical shell will be reduced by

\dfrac{\alpha V_0}{2R}

Answer and worked solution

Official answer: C

Treat the shell with a hole as an intact shell plus a small patch carrying charge $-\alpha Q$ at the hole, where $V_0=kQ/R$ .

(A) At the centre the patch adds $-k\alpha Q/R=-\alpha V_0$ . The potential falls by $\alpha V_0$ , not $2\alpha V_0$ . False.
(B) The field at the centre was zero and becomes $k\alpha Q/R^2=\alpha V_0/R$ . It increases. False.
(C) Halfway to the hole the patch is $R/2$ away and adds $-2\alpha V_0$ . So
$\frac{V_{centre}}{V_{R/2}}=\frac{V_0(1-\alpha)}{V_0(1-2\alpha)}=\frac{1-\alpha}{1-2\alpha}.$
True.
(D) At $2R$ the field drops by $\dfrac{k\alpha Q}{4R^2}=\dfrac{\alpha V_0}{4R}$ , not $\dfrac{\alpha V_0}{2R}$ . False.

(C)

Topics:Electric potential Electric field Coulomb's law and superposition

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