1. Initial State

• Top Plate: Charge $+q$.
• Bottom Plate: Charge $-q$.

2. Final State Analysis

Switches close, grounding the top and bottom plates ($V=0$). The middle metal slab (fixed charge $+q$) induces charges on them.
Let slab potential be $V_s$.
Top capacitance (gap $2d$): $C_1 = \frac{\epsilon_0 A}{2d}$.
Bottom capacitance (gap $d$): $C_2 = \frac{\epsilon_0 A}{d}$.
Total charge on slab divides between top and bottom faces: $q_{slab} = q_{top\_face} + q_{bot\_face}$. $$ q_{top\_face} = C_1 V_s, \quad q_{bot\_face} = C_2 V_s $$ Since $C_2 = 2C_1$, $q_{bot\_face} = 2 q_{top\_face}$.
Given $q_{slab} = +q$, we get $3 q_{top\_face} = q \implies q_{top\_face} = q/3$ and $q_{bot\_face} = 2q/3$.

Induced charges on grounded plates:
Top plate (facing slab): $-q/3$.
Bottom plate (facing slab): $-2q/3$.

3. Calculating Flow

Switch A (Top Plate):
Initial: $+q$. Final: $-q/3$.
Change: $\Delta Q = -q/3 – q = -4q/3$.
Result: $4q/3$ flows out (to ground).

Switch B (Bottom Plate):
Initial: $-q$. Final: $-2q/3$.
Change: $\Delta Q = -2q/3 – (-q) = +q/3$.
Result: $q/3$ flows in (from ground).