A charge Q is distributed over the two concentric hollow spheres of...

Let $q_1 and$q_2$be the charges on the inner and outer spheres respectively.<br><br>Given$q_1 + q_2 = Q$<br><br>Since the surface charge densities are equal,$\sigma_1 = \sigma_2 = \sigma$<br><br>$\dfrac{q_1}{4\pi r^2} = \dfrac{q_2}{4\pi R^2} = \sigma$<br><br>$q_1 = \sigma 4\pi r^2$and$q_2 = \sigma 4\pi R^2$<br><br>Total charge$Q = \sigma 4\pi (r^2 + R^2) \Rightarrow \sigma = \dfrac{Q}{4\pi (r^2 + R^2)}$<br><br>The potential at the common centre is the sum of the potentials due to both spheres:<br><br>$V = \dfrac{1}{4\pi\epsilon_0} \dfrac{q_1}{r} + \dfrac{1}{4\pi\epsilon_0} \dfrac{q_2}{R}$<br><br>$V = \dfrac{1}{4\pi\epsilon_0} $\left( \dfrac{\sigma 4\pi r^2}{r} + \dfrac{\sigma 4\pi R^2}{R} \right)<br><br>$V = $\dfrac{\sigma}{\epsilon_0} ($ r + R)$<br><br>Substituting the value of \sigma:<br><br>$ V = $\dfrac{Q}{4\pi\epsilon_0 (r^2 + R^2)} ($ r + R)$<br><br>Given$K = $\dfrac{Q}{4\pi\epsilon_0}, we get:<br><br>$ V = $\dfrac{K(r+R)}{R^2+r²}<br><br>Answer: \dfrac{K(r+R)}{(R²+r²)}$