ELECTROSTATICS BYU 35

1. Energy Analysis

The total electrostatic energy of a point charge $Q$ at the center of a conducting shell (inner radius $R$, thickness $t$) differs from its free-space energy because the electric field is zero within the conductor’s volume. This reduction in energy is:

$$ U_{reduction} = \int_{R}^{R+t} \frac{1}{2}\epsilon_0 E^2 d\tau \approx \frac{kQ^2}{2} \left( \frac{1}{R} – \frac{1}{R+t} \right) \approx \frac{kQ^2 t}{2R^2} $$

(Assuming $t \ll R$).

2. Calculating States

Initial State ($U_i$):

• Sphere 1 ($R=r$): Charge $q$. Reduction $\propto \frac{q^2}{r^2}$.
• Sphere 2 ($R=3r$): Charge $2q$. Reduction $\propto \frac{(2q)^2}{(3r)^2} = \frac{4q^2}{9r^2}$.

Final State ($U_f$):

• Sphere 1 ($R=r$): Charge $2q$. Reduction $\propto \frac{(2q)^2}{r^2} = \frac{4q^2}{r^2}$.
• Sphere 2 ($R=3r$): Charge $q$. Reduction $\propto \frac{q^2}{(3r)^2} = \frac{q^2}{9r^2}$.

3. Work Calculation

The “Base” energy (infinite self-energy of points) is constant. We only look at changes in the reduction term (binding energy). A larger reduction means lower total energy.

$$ W = U_f – U_i = (U_{red, i}) – (U_{red, f}) $$

(Since $U_{sys} = U_{base} – U_{red}$).

$$ \Delta U_{red} = \frac{kt}{2} \left[ \left( \frac{4q^2}{r^2} + \frac{q^2}{9r^2} \right) – \left( \frac{q^2}{r^2} + \frac{4q^2}{9r^2} \right) \right] $$ $$ \Delta U_{red} = \frac{ktq^2}{2r^2} \left[ 4 + \frac{1}{9} – 1 – \frac{4}{9} \right] = \frac{ktq^2}{2r^2} \left[ 3 – \frac{1}{3} \right] = \frac{ktq^2}{2r^2} \left( \frac{8}{3} \right) $$

The final state has a larger reduction (field eliminated in a region where it is stronger, i.e., near the smaller sphere). Thus, the final energy is lower. Work done is negative.

$$ W = – \frac{4 k t q^2}{3 r^2} $$

Substitute $t = 0.05r = r/20$:

$$ W = – \frac{4 k (r/20) q^2}{3 r^2} = – \frac{k q^2}{15 r} $$