A thin spherical conducting shell of radius R | vedclass

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Question

Given, charge on spherical conducting shell $\left(q_{1}\right)=q$

Radius of shell $\left(\mathrm{r}_{1}\right)=\mathrm{R}$

Second charge $\left

Vedclass · Accepted Answer

$\frac{{2Q}}{{4\pi {\varepsilon _0}R}} + \frac{{q}}{{4\pi {\varepsilon _0}R}}$

Vedclass · Answer

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 $\mathrm{P}$

$\mathrm{V}=\mathrm{V}_{1}+\mathrm{V}_{2}=\frac{\mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{R}}+\frac{2 \mathrm{Q}}{4 \pi \varepsilon_{0} \mathrm{R}}$

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

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