Question

A 2.0 µF capacitor is charged to 12 V. The voltage supply is removed and then a 4.0 µF capacitor is fitted in parallel with the 2.0 µF one. Calculate the charge stored in the 2.0 µF capacitor finally.

Answer 1

Voltage across $2\mu F$ capacitor =$12V$

Charge stored in it $(Q) =C\times V=2\times12=24\mu C$

When the source is removed and a $4\mu F$ is connected across, the charge $Q$ is conserved.

Hence $24\mu C$ has to be shared between $4\mu F$ and $2\mu F$

As both the capacitors are connected across, let the voltage be $V$ across both.

Let charge on $2\mu F$ be $Q1$

Let charge on $4\mu F$ be $Q2$

We have, $Q=Q1+Q2=2V+4V=24.$

$6V=24 \implies V=4V$

Charge stored in $2\mu F$ finally $=2\times 4=8\mu C$ .