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A small conducting sphere of radius is lying concentrically inside a bigger hollow conducting sphere of radius The bigger and smaller spheres are charged with and $(Q>q)$ and are insulated from each other. The potential difference between the spheres will be
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Verified Answer
The correct answer is:
$\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}}{\mathrm{r}}-\frac{\mathrm{q}}{\mathrm{R}}\right)$
The potential $V_{1}$ of smaller sphere is given by
$$
V_{1}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}
$$
The potential $\mathrm{V}_{2}$ of bigger sphere is given by
So, the potential difference between the plates $\mathrm{V}=\mathrm{V}_{1}-\mathrm{V}_{2}$
or $\mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Q}}{\mathrm{R}}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Q}}{\mathrm{R}}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{R}}$ $=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{R}}$
$$
=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{r}-\frac{q}{R}\right)
$$
$$
V_{1}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}
$$
The potential $\mathrm{V}_{2}$ of bigger sphere is given by
So, the potential difference between the plates $\mathrm{V}=\mathrm{V}_{1}-\mathrm{V}_{2}$
or $\mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{r}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Q}}{\mathrm{R}}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Q}}{\mathrm{R}}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{R}}$ $=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{r}}-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{q}}{\mathrm{R}}$
$$
=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q}{r}-\frac{q}{R}\right)
$$
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