A thin spherical insulating shell of radius R carries a uniformly… - Vidyadip

MCQ

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 $\alpha 4 \pi R ^2(\alpha<<1)$ is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?

✓
The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2} R$ from center towards the hole will be $\frac{1-\alpha}{1-2 \alpha}$
B
The magnitude of electric field at the center of the shell is reduced by $\frac{\alpha V_0}{2 R}$
C
The magnitude of electric field at a point, located on a line passing through the hole and shell's center on a distance $2 R$ from the center of the spherical shell will be reduced by $\frac{\alpha V_0}{2 R}$
D
The potential at the center of the shell is reduced by $2 \alpha V _0$

✓

Answer

Correct option: A.

The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2} R$ from center towards the hole will be $\frac{1-\alpha}{1-2 \alpha}$

a
Let charge on the sphere initially be $Q$.

$\therefore \frac{ kQ }{ R }= V _0$

and charge removed $=\alpha Q$

(image)

and $V _{ p }=\frac{ kQ }{ R }-\frac{2 K \alpha Q }{ R }=\frac{ kQ }{ R }(1-2 \alpha)$

$V _{ c } =\frac{ kQ (1-\alpha)}{ R }$

$\therefore \frac{ V _{ c }}{ V _{ p }} =\frac{1-\alpha}{1-2 \alpha}$

$(2)$ $\left( E _{ C }\right)_{\text {inritial }}=$ zero

$\left(E_c\right)_{\text {fimal }}=\frac{k \alpha Q}{R^2}$

$\Rightarrow$ Electric field increases

(image)

$(3)$ $\left( E _{ p }\right)_{\text {iritizl }}=\frac{ kQ }{4 R ^2}$

$\left( E _{ P }\right)_{\text {fimel }}=\frac{ kQ }{4 R ^2}-\frac{ k \alpha Q }{ R ^2}$

$\Delta E _{ p }=\frac{ kQ }{4 R ^2}-\frac{ kQ }{4 R ^2}+\frac{ k \alpha Q }{ R ^2}=\frac{ k \alpha Q }{ R ^2}=\frac{ V _0 \alpha}{ R }$

(image)

(4) $\left(V_c\right)_{\text {minitil }}=\frac{k Q}{R}$

$\left(V_c\right)_{\text {finel }}=\frac{k Q(1-\alpha)}{R}$

$\Delta V_C=\frac{k Q}{R}(\alpha)=\alpha V_0$ (image)

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