Question

A capacitor having a capacitance of 100 µF is charged to a potential difference of 24 V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12 V with the positive plate of the capacitor joined with the positive terminal of the battery. (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the 12 V battery. (c) Is work done by the battery or is it done on the battery? Find its magnitude. (d) Find the decrease in electrostatic field energy. (e) Find the heat developed during the flow of charge after reconnection.

Answer 1

(a) Before connecting to the battery of 12 volts,

$$\begin{gathered} C=100\mu \mathrm{F}\mathrm{and}V=24\mathrm{V} \\ \therefore q=CV=2400\mathrm{\mu F} \end{gathered}$$

After connecting to the battery of 12 volts,

$$\begin{gathered} C=100\mu \mathrm{F}\mathrm{and}V=12\mathrm{V} \\ \therefore q=CV=1200\mathrm{\mu C} \end{gathered}$$

(b) Charge flown through the 12 V battery is 1200 $\mathrm{\mu C}$.

(c) We know,
$$\begin{gathered} W=Vq=12\times 1200 \\ =14400\mathrm{J} \\ =14.4\mathrm{mJ} \end{gathered}$$

It is the work done on the battery.

(d) Initial electrostatic field energy:
$U_{\mathrm{i}}=\frac{1}{2}C{V}_1^2$
Final electrostatic field energy:
$U_{\mathrm{f}}=\frac{1}{2}C{V}_2^2$

Decrease in the electrostatic field energy:
$$\begin{gathered} ∆U=\frac{1}{2}\left(\mathrm{C}{V}_1^2-{\mathrm{CV}}_2^2\right) \\ =\frac{1}{2}\mathrm{C}\left(V_1^2-V_2^2\right) \\ =\frac{1}{2}=100\left[{\left(24\right)}^2-{\left(12\right)}^2\right] \\ =\frac{1}{2}\times 100\times \left(576-144\right) \\ =21600\mathrm{J}=21.6\mathrm{mJ} \end{gathered}$$

(e) After reconnection,
$\mathrm{C}=100\mathrm{\mu F}\mathrm{and}V=12\mathrm{V}$

Heat developed during the flow of charge after reconnection is given by
$$\begin{gathered} H=\frac{1}{2}\mathrm{C}{V}^2=100\times 144 \\ =7200\mathrm{J}=7.2\mathrm{mJ} \end{gathered}$$

This amount of energy is developed as heat when the charge flows through the capacitor.