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A thin spherical conducting shell of radius $R$ has charge $q$. Another charge $Q$ is placed at the centre of the shell. The electrostatic potential at a point $P$ at a distance $R/2$ from the centre of the shell is
$\frac{{2Q}}{{4\pi {\varepsilon _0}R}}$
$\frac{{{{(q + Q)}^2}}}{{4\pi {\varepsilon _0}R}}$
$\frac{{2Q}}{{4\pi {\varepsilon _0}R}} - \frac{{2q}}{{4\pi {\varepsilon _0}R}}$
$\frac{{2Q}}{{4\pi {\varepsilon _0}R}} + \frac{{q}}{{4\pi {\varepsilon _0}R}}$
Solution
Given, charge on spherical conducting shell $\left(q_{1}\right)=q$
Radius of shell $\left(\mathrm{r}_{1}\right)=\mathrm{R}$
Second charge $\left(\mathrm{q}_{2}\right)=\mathrm{Q}$
Distance of point $\mathrm{P}$ from the centre $\left(r_{2}\right)=\frac{R}{2}$
We known that electrostatic pootential at $\mathrm{P}$ due to the spherical conductor.
$\left(\mathrm{V}_{1}\right)=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{1}}{\mathrm{r}_{1}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{R}}$
Similarly, electrostatic potential at $\mathrm{P}$ due to second charge,
$\left(\mathrm{V}_{2}\right) =\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}_{2}}{\mathrm{r}_{2}} $
$=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{\mathrm{R} / 2}=\frac{2 \mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}}$
There, net electrostatic potential at $\mathrm{P}$
$\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2}=\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{R}}+\frac{2 \mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}}$
