Problem Analysis

We start with a thin dielectric spherical shell of radius $R$ carrying a total charge $Q$ distributed uniformly. It is divided into two equal hemispherical parts held together. The work done against electrostatic forces to separate these parts to infinity is $W$. We need to find the self-energy of a charge $q$ uniformly distributed on a hemispherical shell of radius $r$.

Concept: The self-energy of a system is the work done to assemble it. For the complete sphere (charge $Q$, radius $R$), the self-energy $U_{sphere}$ is the energy stored in its electric field. The process of separating the two hemispheres to infinity effectively “disassembles” the interaction between them.

The total self-energy of the sphere can be viewed as: $$ U_{sphere} = U_{hemisphere1} + U_{hemisphere2} + U_{interaction} $$ where $U_{hemisphere}$ is the self-energy of one hemisphere. Since the hemispheres are identical, $U_{hemisphere1} = U_{hemisphere2} = U_{hemi}$.

The work done $W$ to separate them is equal to the negative of the interaction energy: $W = -U_{interaction}$. Wait, let’s be precise. To separate them, we must supply work equal to the binding energy. The initial energy is $U_{sphere}$. The final energy (when far apart) is $2 \times U_{hemi}$. $$ W_{ext} = E_{final} – E_{initial} = 2U_{hemi} – U_{sphere} $$ We are given $W$ as the work done *in making separation infinite*. So $W = 2U_{hemi} – U_{sphere}$.

Step-by-Step Calculation

1. Self-Energy of the Complete Shell

The self-energy of a uniformly charged thin spherical shell of charge $Q$ and radius $R$ is a standard result:

2. Relating Given Work W to Self-Energies

The problem states that $W$ is the work done to separate the two halves to infinity. From energy conservation: $$ \text{Energy}_{\text{initial}} + \text{Work Done} = \text{Energy}_{\text{final}} $$ $$ U_{sphere} + W = 2 U_{hemi}(Q/2, R) $$ Note that each hemisphere carries charge $Q/2$. So, $$ 2 U_{hemi}(Q/2, R) = U_{sphere} + W $$ $$ U_{hemi}(Q/2, R) = \frac{1}{2} \left( \frac{Q^2}{8\pi\epsilon_0 R} + W \right) $$

3. Scaling to the Target Hemisphere

We need to find the self-energy of a hemisphere with charge $q$ and radius $r$. Let’s call this $U_{target}$. Electrostatic self-energy is proportional to $\frac{(\text{Charge})^2}{\text{Radius}}$. $$ U \propto \frac{\text{charge}^2}{\text{length}} $$ So, $$ \frac{U_{target}}{U_{hemi}(Q/2, R)} = \frac{q^2/r}{(Q/2)^2/R} = \frac{q^2}{r} \cdot \frac{R}{(Q/2)^2} = \frac{4q^2 R}{Q^2 r} $$ Therefore, $$ U_{target} = \frac{4q^2 R}{Q^2 r} \times U_{hemi}(Q/2, R) $$ Substituting the expression for $U_{hemi}(Q/2, R)$: $$ U_{target} = \frac{4q^2 R}{Q^2 r} \times \frac{1}{2} \left( \frac{Q^2}{8\pi\epsilon_0 R} + W \right) $$ $$ U_{target} = \frac{2q^2 R}{Q^2 r} \left( \frac{Q^2}{8\pi\epsilon_0 R} + W \right) $$ $$ U_{target} = \frac{2q^2 R}{Q^2 r} \frac{Q^2}{8\pi\epsilon_0 R} + \frac{2q^2 R}{Q^2 r} W $$ $$ U_{target} = \frac{2q^2}{8\pi\epsilon_0 r} + \frac{2q^2 R W}{Q^2 r} $$ $$ U_{target} = \frac{q^2}{4\pi\epsilon_0 r} + \frac{2q^2 R W}{Q^2 r} $$