1. Conceptual Analysis

We are analyzing a spherical capacitor formed by two concentric conducting shells of radii \( a \) and \( b \). The space between them is divided into two hemispheres filled with different dielectric materials having permittivities \( \epsilon_1 \) and \( \epsilon_2 \).

Key Physics Principles:

• Symmetry & Potential: Since the inner and outer shells are conductors, they are equipotential surfaces. The potential depends only on the radial distance \( r \). Consequently, the electric field is purely radial: \( \vec{E} = E(r) \hat{r} \).
• Continuity: Because the potential difference between the shells is the same for both the top and bottom hemispheres, the electric field magnitude \( E(r) \) must be the same in both dielectrics at any specific radius \( r \).
• Parallel Combination: Physically, this arrangement behaves like two spherical capacitors connected in parallel.

2. Calculating the Electric Field

To find the electric field, we apply Gauss’s Law for dielectrics. We construct a Gaussian spherical surface of radius \( r \) (where \( a < r < b \)).

The total free charge enclosed by this Gaussian surface is \( q \).

$$ \oint \vec{D} \cdot d\vec{A} = q_{\text{free}} $$

The displacement vector \( \vec{D} \) is related to the electric field by \( \vec{D} = \epsilon \vec{E} \). Since the surface area of the sphere is split into two hemispheres:

• Flux through Hemisphere 1 (permittivity \( \epsilon_1 \)): \( \Phi_1 = (\epsilon_1 E) \cdot (2\pi r^2) \)
• Flux through Hemisphere 2 (permittivity \( \epsilon_2 \)): \( \Phi_2 = (\epsilon_2 E) \cdot (2\pi r^2) \)

Summing these fluxes and equating to the enclosed charge:

$$ 2\pi r^2 \epsilon_1 E + 2\pi r^2 \epsilon_2 E = q $$ $$ 2\pi r^2 E (\epsilon_1 + \epsilon_2) = q $$

Solving for \( E \):

$$ E(r) = \frac{q}{2\pi (\epsilon_1 + \epsilon_2) r^2} $$

3. Calculating the Potential Difference

The potential difference \( V \) between the spheres is found by integrating the electric field from the inner radius \( a \) to the outer radius \( b \).

$$ V = \int_a^b E(r) \, dr $$

Substitute the expression for \( E(r) \):

$$ V = \int_a^b \frac{q}{2\pi (\epsilon_1 + \epsilon_2) r^2} \, dr $$

Since \( q, \pi, \epsilon_1, \) and \( \epsilon_2 \) are constants, we pull them out of the integral:

$$ V = \frac{q}{2\pi (\epsilon_1 + \epsilon_2)} \int_a^b \frac{1}{r^2} \, dr $$ $$ V = \frac{q}{2\pi (\epsilon_1 + \epsilon_2)} \left[ -\frac{1}{r} \right]_a^b $$ $$ V = \frac{q}{2\pi (\epsilon_1 + \epsilon_2)} \left( \frac{1}{a} – \frac{1}{b} \right) $$ $$ V = \frac{q}{2\pi (\epsilon_1 + \epsilon_2)} \left( \frac{b – a}{ab} \right) $$ $$ V = \frac{q(b – a)}{2\pi (\epsilon_1 + \epsilon_2) ab} $$