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(A) Let the potential at point $x$ be $0 \ V$ and at point $y$ be $6 \ V$.For the upper branch,the capacitors $4 \ \mu F$ and $2 \ \mu F$ are in serie

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(A) Let the potential at point $x$ be $0 \ V$ and at point $y$ be $6 \ V$.For the upper branch,the capacitors $4 \ \mu F$ and $2 \ \mu F$ are in series. The potential at point $A$ $(V_A)$ is determined by the voltage divider rule for capacitors: $V_A = V_y + (V_x - V_y) 	imes \frac{C_1}{C_1 + C_2} = 6 + (0 - 6) 	imes \frac{4}{4 + 2} = 6 - 6 	imes \frac{4}{6} = 6 - 4 = 2 \ V$.For the lower branch,the capacitors $2 \ \mu F$ and $4 \ \mu F$ are in series. The potential at point $B$ $(V_B)$ is: $V_B = V_y + (V_x - V_y) 	imes \frac{C_3}{C_3 + C_4} = 6 + (0 - 6) 	imes \frac{2}{2 + 4} = 6 - 6 	imes \frac{2}{6} = 6 - 2 = 4 \ V$.The potential difference between $A$ and $B$ is $|V_A - V_B| = |2 \ V - 4 \ V| = 2 \ V$.

What is the potential difference between points $A$ and $B$ in the circuit shown (in $V$)?

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