A thin spherical insulating shell of radius R | vedclass

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Question

Let charge on the sphere initially be $Q$.

$	herefore \frac{ kQ }{ R }= V _0$

and charge removed $=\alpha Q$

(image)

and $V _{ p }=\frac{

Vedclass · Accepted Answer

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 $\frac{1-\alpha}{1-2 \alpha}$

Vedclass · Answer

Let charge on the sphere initially be $Q$.

$	herefore \frac{ kQ }{ R }= V _0$

and charge removed $=\alpha Q$

(image)

and $V _{ p }=\frac{ kQ }{ R }-\frac{2 K \alpha Q }{ R }=\frac{ kQ }{ R }(1-2 \alpha)$

$V _{ c }  =\frac{ kQ (1-\alpha)}{ R }$

$	herefore \frac{ V _{ c }}{ V _{ p }}  =\frac{1-\alpha}{1-2 \alpha}$

$(2)$ $\left( E _{ C }ight)_{	ext {inritial }}=$ zero

$\left(E_cight)_{	ext {fimal }}=\frac{k \alpha Q}{R^2}$

$\Rightarrow$ Electric field increases

(image)

$(3)$ $\left( E _{ p }ight)_{	ext {iritizl }}=\frac{ kQ }{4 R ^2}$

$\left( E _{ P }ight)_{	ext {fimel }}=\frac{ kQ }{4 R ^2}-\frac{ k \alpha Q }{ R ^2}$

$\Delta E _{ p }=\frac{ kQ }{4 R ^2}-\frac{ kQ }{4 R ^2}+\frac{ k \alpha Q }{ R ^2}=\frac{ k \alpha Q }{ R ^2}=\frac{ V _0 \alpha}{ R }$

(image)

(4) $\left(V_cight)_{	ext {minitil }}=\frac{k Q}{R}$

$\left(V_cight)_{	ext {finel }}=\frac{k Q(1-\alpha)}{R}$

$\Delta V_C=\frac{k Q}{R}(\alpha)=\alpha V_0$  (image)

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