Question

A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor?

Answer 1

Given:
$$\begin{gathered} C_1=100\mathrm{pF} \\ V=24\mathrm{V} \end{gathered}$$
Charge on the given capacitor, $q=C_{1}V=24\times 100\mathrm{pC}$

Capacitance of the uncharged capacitor,$C_2=20\mathrm{pF}$
When the charged capacitor is connected with the uncharged capacitor, the net charge on the system of the capacitors becomes
$q_1+q_2=24\times 100\mathrm{qC}...\left(\mathrm{i}\right)$

The potential difference across the plates of the capacitors will be the same.
Thus,
$$\begin{gathered} \frac{q_1}{C_1}=\frac{q_2}{C_2} \\ \Rightarrow \frac{q_1}{100}=\frac{q_2}{20} \\ \Rightarrow q_1=5q_2...\left(\mathrm{ii}\right) \end{gathered}$$

From eqs. (i) and (ii), we get
$$\begin{gathered} q_1+\frac{q_1}{5}=24\times 100\mathrm{pC} \\ \Rightarrow 6q_1=5\times 24\times 100\mathrm{pC} \\ \Rightarrow q_1=\frac{5\times 24\times 100}{6}\mathrm{pC} \\ \mathrm{Now}, \\ {V}_1=\frac{q_1}{C_1} \\ =\frac{5\times 24\times 100\mathrm{pC}}{6\times 100\mathrm{pF}}=20\mathrm{V} \end{gathered}$$