Question

A capacitor of capacitance 2⋅0 µF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4⋅0 µF as shown in figure (31-E22). Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.

Answer 1

Charge on the 2 µF capacitor when it is not connected to the 4 µF capacitor is given by

$$\begin{gathered} q\mathit{=}C\mathit{\times }V=12\times 2\times {10}^{-6} \\ \Rightarrow q=24\times {10}^{-6}\mathrm{C} \end{gathered}$$

(a) On connecting the capacitors, the charge flows from the 2 µF capacitor to the 4 µF capacitor.
Now, let the charges on the 2 µF and 4 µF capacitors be q₁ and q₂, respectively.

As they are connected in parallel, the potential difference across them is the same.

$$\begin{gathered} \therefore V=\frac{q_1}{C_1}=\frac{q_2}{C_2} \\ \Rightarrow V=\frac{q_1}{2}=\frac{q_2}{4} \\ \Rightarrow q_2=2q_1 \end{gathered}$$

The total charge on the capacitors will be the same as the initial charge stored on the 2 µF capacitor.

$$\begin{gathered} \therefore q_1+q_2=24\times {10}^{-6} \\ \Rightarrow 3q_1=24\times {10}^{-6}\mathrm{C} \\ \Rightarrow q_1=8\times {10}^{-6}\mathrm{C}=8\mathrm{\mu C} \\ \Rightarrow q_1=2q_1=16\mathrm{\mu C} \end{gathered}$$


(b) Energies stored in the capacitors are given by
$$\begin{gathered} E_1=\frac{1}{2}\times \frac{{q_1}^2}{C_1}=16\mathrm{\mu J} \\ \mathrm{and} \\ {E}_2=\frac{1}{2}\times \frac{{q_2}^2}{C_2}=32\mathrm{\mu J} \end{gathered}$$

(c)
Initial energy stored in the 2 µF capacitor is given by
$$\begin{gathered} E_{\mathrm{i}}=\frac{1}{2}C{V}^2=\frac{1}{2}\times \left(2\times {10}^{-6}\right){\left(12\right)}^2 \\ \Rightarrow E_{\mathrm{i}}=144\mathrm{\mu J} \end{gathered}$$

Total energy of the capacitors after they are connected in parallel is given by
E_f = E₁ + E₂
$\Rightarrow$E_f = 16 + 32 = 48 μJ

Heat produced during the charge transfer is given by
E = E_f $-$ E_i
$\Rightarrow$E = 144 $-$ 48 = 96 μJ