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

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)