In the circuit shown in figure (A), the capacitors and batteries are ideal and circuit is at steady state. The capacitors $C_1$, $C_2$ and $C_3$ are disconnected from the circuit shown in figure (A) with their charges intact and reconnected as in figure (B) with polarities as indicated. Initially charges on capacitor $C_1$, $C_2$ & $C_3$ before connecting to the circuit A was zero on each
 

$\begin{array}{l}KVL,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+2-\frac{(x-y)}{2}+\frac{y}{3}=0\text{}\\ and\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10-x-\frac{y}{3}=0\text{}\\ After Solving, x=9\mu C and y=3\mu C\end{array}$

$\begin{array}{l}Charge pas\sin g through 10V battery is 9\mu C\text{}\\ \therefore Work done by this battery=qV=90\mu J\text{}\\ Charge pas\sin g through 2V battery is 6\mu C\\ \text{}\therefore Work done by this battery=qV=12\mu J\end{array}$

$\begin{array}{l}Charge conservation,\\ \text{}+3-6-9=-q_1-q_2+q_3\text{}\\ All the three capacitors are in parallel connection \\ V is same ,\text{}\frac{q_1}{C_1}=\frac{q_2}{C_2}=\frac{-q_3}{C_3}\text{}\\ \Rightarrow \frac{q_1}{1}=\frac{q_2}{2}=\frac{-q_3}{3}\text{}\\ After solving, \\ \Rightarrow q_1=+2\mu C, q_2=+4\mu C, q_3=-6\mu C\end{array}$

$V_d-V_e=\frac{4}{2}=2V$