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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.)

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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$
This gives for the given value of the capacitances,
$Q= 500 V \times \large\frac{10}{3} $$\mu F =1.7 \times 10^{-3} C$
and $ Q'= 500 V \times 10 \mu F$
$\qquad= 5.0 \times 10^{-3}$
answered Jun 17, 2014 by meena.p
 

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