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(C) Let the charges on the inner and outer spheres be $Q_{1}$ and $Q_{2}$ respectively.Since the surface charge density $\sigma$ is the same for both

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_{0}} \frac{(R+r)}{(R^{2}+r^{2})}Q$

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(C) Let the charges on the inner and outer spheres be $Q_{1}$ and $Q_{2}$ respectively.Since the surface charge density $\sigma$ is the same for both spheres,we have:$\sigma = \frac{Q_{1}}{4 \pi r^{2}} = \frac{Q_{2}}{4 \pi R^{2}} \Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{r^{2}}{R^{2}}$Given that the total charge is $Q_{1} + Q_{2} = Q$,we substitute $Q_{1} = Q_{2} \frac{r^{2}}{R^{2}}$:$Q_{2} \frac{r^{2}}{R^{2}} + Q_{2} = Q \Rightarrow Q_{2} \left( \frac{r^{2} + R^{2}}{R^{2}} \right) = Q \Rightarrow Q_{2} = \frac{Q R^{2}}{R^{2} + r^{2}}$Similarly,$Q_{1} = Q - Q_{2} = Q - \frac{Q R^{2}}{R^{2} + r^{2}} = \frac{Q r^{2}}{R^{2} + r^{2}}$The electric potential at the common centre $O$ is the sum of potentials due to both shells:$V = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{1}}{r} + \frac{1}{4 \pi \varepsilon_{0}} \frac{Q_{2}}{R}$$V = \frac{1}{4 \pi \varepsilon_{0}} \left[ \frac{Q r^{2}}{r(R^{2} + r^{2})} + \frac{Q R^{2}}{R(R^{2} + r^{2})} \right]$$V = \frac{1}{4 \pi \varepsilon_{0}} \left[ \frac{Q r}{R^{2} + r^{2}} + \frac{Q R}{R^{2} + r^{2}} \right]$$V = \frac{1}{4 \pi \varepsilon_{0}} \frac{(R + r)Q}{R^{2} + r^{2}}$

$A$ charge $Q$ is distributed over two concentric conducting thin spherical shells of radii $r$ and $R$ $(R > r)$. If the surface charge densities on the two shells are equal,the electric potential at the common centre is

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