Question

A capacitor of capacitance 5⋅00 µF is charged to 24⋅0 V and another capacitor of capacitance 6⋅0 µF is charged to 12⋅0 V. (a) Find the energy stored in each capacitor. (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the new charges on the capacitors. (c) Find the loss of electrostatic energy during the process. (d) Where does this energy go?

Answer 1

Given:

$$\begin{gathered} C_1=5\mathrm{\mu F}\mathrm{and}{V}_1=24\mathrm{V} \\ \therefore q_1=C_1{V}_1=5\times 24=120\mathrm{\mu C} \\ \mathrm{and} \\ {C}_2=6\mathrm{\mu F}\mathrm{and}{V}_2=12\mathrm{V} \\ \therefore q_2=C_2{V}_2=6\times 12=72\mathrm{\mu C} \end{gathered}$$

(a)
Energy stored in the first capacitor:
$$\begin{gathered} U_1=\frac{1}{2}{C}_1{V}_1^2 \\ =1440\mathrm{J}=1.44\mathrm{mJ} \end{gathered}$$

Energy stored in the second capacitor:
$$\begin{gathered} U_{\mathit{2}}=\frac{1}{2}{\mathrm{C}}_1{\mathrm{V}}_2^2 \\ =432\mathrm{J}=0.432\mathrm{mJ} \end{gathered}$$

(b) The capacitors are connected to each other in such a way that the positive plate of the first capacitor is connected to the negative plate of the second capacitor and vice versa.

∴ Net change in the system, Q_net = 120 $-$ 72 = 48

Now, let V be the common potential of the two capacitors.
From the conservation of charge before and after connecting, we get
$$\begin{gathered} V=\frac{Q_{\mathrm{net}}}{(C_1+C_2)} \\ =\frac{48}{(5+6)} \\ =4.36\mathrm{V} \end{gathered}$$

New charges:

$$\begin{gathered} q_1\mathit{\text{'}}\mathit{=}{C}_{1}V=5\times 4.36=21.8\mathrm{\mu C} \\ \mathrm{and} \\ {q}_2\text{'}=C_{2}V=6\times 4.36=26.2\mathrm{\mu C} \end{gathered}$$

(c)

$$\begin{gathered} \mathrm{Given}: \\ {U}_1=\frac{1}{2}{C}_1{V}^2 \\ \mathrm{and} \\ {U}_2=\frac{1}{2}{C}_2{V}^2 \\ \therefore U_f=\frac{1}{2}{\mathrm{V}}^2\left({\mathrm{C}}_1+{\mathrm{C}}_2\right) \\ =\frac{1}{2}{\left(4.36\right)}^2\left(5+6\right) \\ =\frac{1}{2}\times 19\times 11 \\ =104.5\times {10}^{-6}\mathrm{J} \\ =0.1045\mathrm{mJ} \end{gathered}$$

$$\begin{gathered} \mathrm{But}{U}_i=1.44+0.433=1.873 \\ \mathrm{Loss}\mathrm{of}\mathrm{energy}: \\ ∆U=1.873-0.1045 \\ =1.7678 \\ =1.77\mathrm{mJ} \end{gathered}$$

(d) The energy is dissipated as heat.