Physics - Electrostatics Question with Solution | TestHub

Let$‘A’$be the point at a distance$2R$from center and$‘B’$be the point at a distance of$\frac{R}{2}$from center.
Now, let total charge on spherical shell distributed uniformly$=Q$
$\Rightarrow \mathrm{C}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}=\frac{Q}{4\pi r^2}$
Charge on elemental area$‘dA’$.
Given potential at surface${=V}_0$
$\Rightarrow V_0=\frac{KQ}{R}$
Now, potential at center of the shell${=V}_C$
$V_C=\frac{KQ}{R}-\frac{K\left(dQ\right)}{R}=\frac{KQ}{R}-\frac{K\left(\alpha Q\right)}{R}=\frac{KQ}{R}(1-\alpha )$
$\Rightarrow V_C=V_0\left(1-\alpha \right)\Rightarrow \text{Reduced\hspace{0.17em}}\mathrm{b}\mathrm{y}\alpha V_0$
Now potential$=V_B$
$\Rightarrow V_B=\frac{KQ}{R}-\frac{K\left(aQ\right)}{R/2}=V_0\left(1-2a\right)\Rightarrow \mathrm{R}\mathrm{e}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}\mathrm{b}\mathrm{y}2\alpha V_0$
$\therefore \frac{V_C}{V_B}=\frac{1-a}{1-2a}$
Now, electric field at$A=E_A$
Then,$E_A=\frac{KQ}{{\left(2R\right)}^2}-\frac{KaQ}{R^2}=\frac{KQ}{4R^2}-\frac{a{V}_0}{R}$
So reduced by$\frac{a{V}_0}{R}$
Electric field at center$‘C’$due to small change$‘dQ’$
$E_C=\frac{K\left(dq\right)}{R^2}=\frac{K\left(\alpha Q\right)}{R^2}=\frac{\alpha V_0}{R}$
So increased by$\frac{a{V}_0}{R}$