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$R$ त्रिज्या के एक पतले गोलीय अचालक कोश (spherical insulating shell) पर आवेश एकसमान रूप से इस तरह से वितरित है कि इसकी सतह पर विभव $V _0$ है। इसमें एक छोटे क्षेत्रफल $\alpha 4 \pi R ^2(\alpha<<1)$ वाला एक छिद्र बकी कोश को प्रभावित किए क्ति काया जाता है। निम्नलिखित कथनों में से कौनसा सही है?
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 $\frac{1-\alpha}{1-2 \alpha}$
The magnitude of electric field at the center of the shell is reduced by $\frac{\alpha V_0}{2 R}$
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 $\frac{\alpha V_0}{2 R}$
The potential at the center of the shell is reduced by $2 \alpha V _0$
Solution
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)
