Question: (III) A \({\bf{7}}{\bf{.7}}\;{\bf{\mu F}}\) capacitor is charged by a 165-V battery (Fig. 17–43a) and then is disconnected from the battery. When this capacitor \(\left( {{{\bf{C}}_{\bf{1}}}} \right)\) is then connected (Fig. 17–43b) to a second (initially uncharged) capacitor, the final voltage on each capacitor is 15 V. What is the value of \({{\bf{C}}_{\bf{2}}}\)? (Hint: Charge is conserved.)

A capacitor is a charge storage device. When a capacitor is connected to a battery, the charge stored on each plate is equal in magnitude but opposite in sign.

The capacitance of the first capacitor is,\({C_1} = 7.7\;\mu {\rm{F}}\)

The potential across the first capacitor is,\({V_1} = 165\;{\rm{V}}\)

The final voltage on each capacitor is, \(V = 15\;{\rm{V}}\)

The initial charge on the first capacitor is \({Q_1} = {C_1}{V_1}\).

The initial charge on the second capacitor is, \({Q_2} = 0\).

After joining, the total charge on the capacitors is,

\(\begin{aligned}{c}Q &= {Q_1} + {Q_2}\\ &= {C_1}{V_1} + 0\\ &= {C_1}{V_1}\end{aligned}\)

After joining, the capacitance in the parallel combination is,

\(C = {C_1} + {C_2}\)

The final potential difference across each capacitor is given as:

\(\begin{aligned}{c}V &= \frac{Q}{C}\\V &= \frac{{{C_1}{V_1}}}{{{C_1} + {C_2}}}\\{C_1} + {C_2} &= \frac{{{C_1}{V_1}}}{V}\\{C_2} &= \frac{{{C_1}{V_1}}}{V} - {C_1}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{C_2} &= \frac{{\left( {7.7\;\mu {\rm{F}}} \right)\left( {165\;{\rm{V}}} \right)}}{{15\;{\rm{V}}}} - \left( {7.7\;\mu {\rm{F}}} \right)\\ &= 77\;\mu {\rm{F}}\end{aligned}\)

Hence, the value of capacitor \({C_2}\) is \(77\;\mu {\rm{F}}\).