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