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

${\text{Before making hole , V}}_0=\frac{KQ}{R}=\frac{K\sigma 4\pi r^2}{R}$

 After making hole, potential at center, 

 ${\text{V}}_1=V_0- \text{(potential due to removed element})$ 
$\Rightarrow V_1=V_0-\frac{Kq}{R}=V_0-\frac{K}{R}\left(\sigma \alpha 4\pi R^2\right)$$=V_0-V_0\alpha =V_0\left(1-\alpha \right)$   
At mid point between center and hole is
$V_2=V_0-\frac{kq}{R/2}=V_0-2V_0\alpha =V_0\left(1-2\alpha \right)$ 

Thus, $\frac{V_1}{V_2}=\frac{V_0\left(1-\alpha \right)}{V_0\left(1-2\alpha \right)}=\frac{1-\alpha }{1-2\alpha }\to \left(2\right)$ is correct 

Option A :

$E_{initial}$ =0
$E_{final}$ =$\frac{Kq}{R^z}=\frac{k\sigma \alpha 4\pi R^2}{R^2}=\frac{V_0\alpha }{R}$

$E_i-E_f=\frac{V_0\alpha }{R}$   i.e E will decrease by $\frac{V_0\alpha }{R}$ i.e. (A) is wrong

Option C : 

$E_i=\frac{kQ}{{\left(2R\right)}^2}=\frac{V_0}{4R}$

$E_f=\frac{k\alpha Q}{R^2}=\frac{\alpha V_0}{R}$
Option D:

 $V_{initial} = V_0 and V_{final}$$=V_0-\frac{K\sigma \alpha 4\pi R^2}{R}=V_0-V_0\alpha$
 $\therefore V_{center}$ decreases by $V_0\alpha$ . i.e. (D) is wrong