Question: (III) A \({\bf{3}}{\bf{.70}}\;{\bf{\mu F}}\) capacitor is charged by a 12.0-V battery. It is disconnected from the battery and then connected to an uncharged \({\bf{5}}{\bf{.00}}\;{\bf{\mu F}}\) capacitor (Fig.17–43). Determine the total stored energy (a) before the two capacitors are connected, and (b) after they are connected. (c) What is the change in energy?

A capacitor is a charge and energy storage device. The stored energy is proportional to the square of the potential difference between the plates.

The expression for stored energy in the capacitor is given as:

\({\rm{PE}} = \frac{1}{2}C{V^2}\) … (i)

Here, C is the capacitance and V is the potential difference.

The capacitance of the first capacitor is,\({C_1} = 3.70\;\mu {\rm{F}} = 3.70 \times {10^{ - 6}}\;{\rm{F}}\)

The potential across the first capacitor is,\({V_1} = 12\;{\rm{V}}\)

The capacitance of the second capacitor is, \({C_2} = 5.00\;\mu {\rm{F}} = 5.00 \times {10^{ - 6}}\;{\rm{F}}\)

From equation (i), the energy stored in the capacitor is,

\({U_1} = \frac{1}{2}{C_1}V_1^2\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{U_1} &= \frac{1}{2}\left( {3.70 \times {{10}^{ - 6}}\;{\rm{F}}} \right)\left( {12.0\;{\rm{V}}} \right)\\ &= 2.66 \times {10^{ - 4}}\;{\rm{J}}\end{aligned}\)

The energy stored in the second capacitor is zero as it is initially uncharged.

Thus, the total energy stored before the capacitors are connected is \(2.66 \times {10^{ - 4}}\;{\rm{J}}\).

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 is,

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

After joining, the capacitance of the parallel combination is,

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

After joining, the final potential difference is,

\(\begin{aligned}{c}V &= \frac{Q}{C}\\ &= \frac{{{C_1}{V_1}}}{{{C_1} + {C_2}}}\\ &= \frac{{\left( {3.70\;\mu {\rm{F}}} \right)\left( {12\;{\rm{V}}} \right)}}{{\left( {3.70\;\mu {\rm{F}}} \right) + \left( {5.00\;\mu {\rm{F}}} \right)}}\\ &= 5.103\;{\rm{V}}\end{aligned}\)

The final total energy stored in the system is,

\(\begin{aligned}{c}U &= \frac{1}{2}\left( {{C_1} + {C_2}} \right){V^2}\\ &= \frac{1}{2}\left( {\left( {3.70 \times {{10}^{ - 6}}\;{\rm{F}}} \right) + \left( {5.00 \times {{10}^{ - 6}}\;{\rm{F}}} \right)} \right){\left( {5.103\;{\rm{V}}} \right)^2}\\ &= 1.13 \times {10^{ - 4}}\;{\rm{J}}\end{aligned}\)

Thus, the total energy stored after the capacitors are connected is \(1.13 \times {10^{ - 4}}\;{\rm{J}}\).

The change in energy is,

\(\begin{aligned}{c}\Delta U &= U - {U_1}\\ &= \left( {1.13 \times {{10}^{ - 4}}\;{\rm{J}}} \right) - \left( {2.66 \times {{10}^{ - 4}}\;{\rm{J}}} \right)\\ &= - 1.53 \times {10^{ - 4}}\;{\rm{J}}\end{aligned}\)

Thus, the change in energy is \( - 1.53 \times {10^{ - 4}}\;{\rm{J}}\).