A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential…

1. The magnitude of electric field at the center of the shell is reduced by $\frac{\mathrm{\alpha }{\mathrm{V}}_0}{2\mathrm{R}}$
2. The potential at the center of the shell is reduced by $2\mathrm{\alpha }{\mathrm{V}}_0$
3. The ratio of the potential at the center of the shell to that of the point at $\frac{1}{2}\mathrm{R}$ from center towards the hole will be $\frac{1-\mathrm{\alpha }}{1-2\mathrm{\alpha }}$
4. The magnitude of electric field at a point, located on a line passing through the hole and shell's center on a distance $2\mathrm{R}$ from the center of the spherical shell will be reduced by $\frac{\mathrm{\alpha }{\mathrm{V}}_0}{2\mathrm{R}}$

Solution

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 $\frac{a{V}_0}{R}$

Asked in: JEE Advanced 2019 (Paper 1)

Practice more Electrostatics questions on Aicharya