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# A network of four $10 \mu F$ capacitors is connected to a $500\; V$ supply, as shown in Figure. the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.)

Clearly, from the figure, the charge on each of the capacitors, $C_1, C_2$ and $C_3$ is the same, say Q.
Let the charge on $C_4$ be $Q′$.
Now, since the potential difference across AB is $Q/C_1$, across BC is $Q/C_2$, across CD is $Q/C_3$, we have
$\large\frac{Q}{V_1} +\frac{Q}{C_2} +\frac{Q}{C_3}$$=500 V Also , \large\frac{Q'}{C_4}$$=500 V$
$Q= 500 V \times \large\frac{10}{3}$$\mu F =1.7 \times 10^{-3} C$
and $Q'= 500 V \times 10 \mu F$
$\qquad= 5.0 \times 10^{-3}$