Question

Two independent electrostatic configurations are shown in the figure. Configuration (I) consists of an isolated point charge \(q = 1\ \text{C}\), and configuration (II) consists of another identical charge surrounded by a thick conducting shell of inner radius \(R_1 = 1\ \text{m}\) and outer radius \(R_2 = 2\ \text{m}\), with the charge being at the center of the shell. \[ W_I = \frac{\epsilon_0}{2} \int E_I^2 dV \text{and} W_{II} = \frac{\epsilon_0}{2} \int E_{II}^2 dV, \] where \(E_I\) and \(E_{II}\) are the magnitudes of the electric fields for configurations (I) and (II) respectively, \(\epsilon_0\) is the permittivity of vacuum, and the volume integrations are carried out over all space. If \[ \frac{8\pi}{\epsilon_0} |W_I - W_{II}| = \frac{1}{n}, \] what is the value of the integer \(n\)? 

Hint:

For spherical charge distributions, the energy calculations often rely on the integration of the square of the electric field. A conducting shell's influence inside and outside must be carefully considered.

Answer 1

- The electric field \(E_I\) for the point charge (I) is radial and given by: \[ E_I = \frac{q}{4\pi \epsilon_0 r^2}. \] - For the configuration (II), the electric field inside the shell is zero and outside the shell, the electric field is similar to that of a point charge at the center of the shell: \[ E_{II} = \frac{q}{4\pi \epsilon_0 r^2}. \] - The energy \(W_I\) for configuration (I) is: \[ W_I = \frac{\epsilon_0}{2} \int E_I^2 dV = \frac{\epsilon_0}{2} \int_0^\infty \left(\frac{q}{4\pi \epsilon_0 r^2}\right)^2 4\pi r^2 dr = \frac{q^2}{8\pi \epsilon_0}. \] - The energy \(W_{II}\) for configuration (II) is: \[ W_{II} = \frac{\epsilon_0}{2} \int E_{II}^2 dV = \frac{\epsilon_0}{2} \int_0^{R_1} \left(0\right)^2 + \int_{R_1}^{R_2} \left(\frac{q}{4\pi \epsilon_0 r^2}\right)^2 4\pi r^2 dr = \frac{q^2}{8\pi \epsilon_0}. \] - Hence, \(W_I = W_{II}\), and the difference is zero. Therefore, \[ \frac{8\pi}{\epsilon_0} |W_I - W_{II}| = 1/n ⇒ n = 2. \]