1. Initial State

We have four plates. Plates 2 and 3 are given charges $+q$ and $-q$. Plates 1 and 4 are initially uncharged and isolated.
Charge distribution (Outer faces have zero charge because total system charge is 0):

• Plate 1: 0
• Plate 2: Inner face $+q$
• Plate 3: Inner face $-q$
• Plate 4: 0

Electric field exists only between plates 2 and 3. Initial Energy $U_i$ stored in this capacitor ($C = \frac{\epsilon_0 A}{d}$):

$$ U_i = \frac{q^2}{2C} = \frac{q^2 d}{2 \epsilon_0 A} $$

2. Final State (1 and 4 Connected)

When plates 1 and 4 are connected, $V_1 = V_4$. Let charge $x$ flow from 4 to 1.
New distribution on facing surfaces (sum of potential drops = 0):

$$ E_1 d + E_2 d + E_3 d = 0 $$

Using Gauss’s law, fields are proportional to charges on the left of the gap:

• Gap 1 (1-2): Charge on face $1R$ is $x$. Field $E_1 \propto x$.
• Gap 2 (2-3): Charge on face $2R$ is $q+x$. Field $E_2 \propto q+x$.
• Gap 3 (3-4): Charge on face $3R$ is $x$. Field $E_3 \propto x$.

$$ x + (q+x) + x = 0 \implies 3x = -q \implies x = -q/3 $$

So, charges on the gaps are: $-q/3, +2q/3, -q/3$.

3. Calculation of Heat

Heat dissipated = Loss in electrostatic potential energy ($U_i – U_f$).
Final Energy is the sum of energies in the three gaps:

$$ U_f = \frac{d}{2\epsilon_0 A} \left[ (-q/3)^2 + (2q/3)^2 + (-q/3)^2 \right] $$ $$ U_f = \frac{d}{2\epsilon_0 A} \frac{q^2}{9} (1 + 4 + 1) = \frac{6 q^2 d}{18 \epsilon_0 A} = \frac{q^2 d}{3 \epsilon_0 A} $$

Heat $H = U_i – U_f$:

$$ H = \frac{q^2 d}{2 \epsilon_0 A} – \frac{q^2 d}{3 \epsilon_0 A} = \frac{q^2 d}{\epsilon_0 A} \left( \frac{1}{2} – \frac{1}{3} \right) $$