Question

Consider three metal spherical shells A, B, and C, each of radius \( R \). Each shell has a concentric metal ball of radius \( R/10 \). The spherical shells A, B, and C are given charges \( +6q, -4q, \) and \( 14q \) respectively. Their inner metal balls are also given charges \( -2q, +8q, \) and \( -10q \) respectively. Compare the magnitude of the electric fields due to shells A, B, and C at a distance \( 3R \) from their centers.

Hint:

According to Gauss’s law, the electric field outside a spherical shell behaves as if all the charge were concentrated at its center.

Answer 1

The electric field at a distance \( 3R \) from the center of a spherical shell depends only on the net charge enclosed and is given by Gauss’s law: \[ E = \frac{1}{4\pi \epsilon_0} \frac{Q_{\text{net}}}{r^2} \] where \( Q_{\text{net}} \) is the total charge enclosed by each shell. #### Step 1: Calculate Net Charge on Each Shell - For Shell A: \[ Q_A = 6q + (-2q) = 4q \] - For Shell B: \[ Q_B = -4q + 8q = 4q \] - For Shell C: \[ Q_C = 14q + (-10q) = 4q \] Since the total charge enclosed for all three shells is the same (\( 4q \)), the magnitude of the electric field at a distance \( 3R \) is identical for all: \[ E_A = E_B = E_C = \frac{1}{4\pi \epsilon_0} \frac{4q}{(3R)^2} \] Thus, the electric fields due to shells A, B, and C at a distance \( 3R \) are equal.