# 20/11 $\mu C$ 90/11 $\mu C$ 10 $\mu C$ 110 $\mu C$

Let the equivalent capacitance of the thre capacitors connected in series be C

$\\\frac{1}{C}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6} \\C=1\ \mu F$

Since the capacitors are connected on series same charge will be accumulated on all of them.

Charge accumulated on $2\ \mu F$ capacitor$=1\times 10=10\ \mu C$

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