1. Physical Setup

A capacitor has two sections with areas \( A_1, A_2 \) and separations \( d_1, d_2 \). It is placed in a uniform external electric field \( E \) directed perpendicular to the plates. The plates are isolated conductors. We need to find the fields \( E_1 \) and \( E_2 \) inside the gaps.

Key Conditions:

1. Equipotential Conductor: The plates are conductors, so the potential difference across the gap must be the same for both sections (since the plates are connected/continuous). $$ E_1 d_1 = E_2 d_2 \quad \text{(Equation 1)} $$
2. Charge Conservation / Gauss’s Law: Since the plates are isolated and neutral initially, the net charge induced on them is zero. Applying Gauss’s Law to a box enclosing one of the plates:
   Flux In = \( E(A_1 + A_2) \) (from external field)
   Flux Out = \( E_1 A_1 + E_2 A_2 \) (into the gaps)
   Net Flux = 0. $$ E(A_1 + A_2) = E_1 A_1 + E_2 A_2 \quad \text{(Equation 2)} $$

2. Solving the Equations

From Equation 1, substitute \( E_2 = E_1 \frac{d_1}{d_2} \) into Equation 2:

$$ E(A_1 + A_2) = E_1 A_1 + \left( E_1 \frac{d_1}{d_2} \right) A_2 $$ $$ E(A_1 + A_2) = E_1 \left( A_1 + A_2 \frac{d_1}{d_2} \right) $$ $$ E(A_1 + A_2) = E_1 \left( \frac{A_1 d_2 + A_2 d_1}{d_2} \right) $$

Solving for \( E_1 \):

$$ E_1 = E \frac{d_2(A_1 + A_2)}{A_1 d_2 + A_2 d_1} $$

Similarly, solving for \( E_2 \):

$$ E_2 = E \frac{d_1(A_1 + A_2)}{A_1 d_2 + A_2 d_1} $$