Question

A charge \( Q \) is distributed over two concentric hollow spheres of radii \( r \) and \( R \) (\( R>r \)) such that their surface charge densities are equal. Find the electric potential at their common center.

Hint:

When dealing with concentric spheres, use charge distribution rules and superposition principles to calculate the potential.

Answer 1

The surface charge density for both spheres is equal: \[ \sigma = \frac{Q_1}{4 \pi r^2} = \frac{Q_2}{4 \pi R^2}. \] This implies that the charge on the inner sphere is: \[ Q_1 = \frac{Q r^2}{R^2}, \quad Q_2 = Q - Q_1 = Q \left( 1 - \frac{r^2}{R^2} \right). \] The potential at the common center is: \[ V = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q_1}{r} + \frac{Q_2}{R} \right). \] Substituting values: \[ V = \frac{3Q}{4 \pi \epsilon_0 R}. \]