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 $4.0 μF .$

Find:

$(a)$ The charge flow through connecting wire till steady-state on each of the two capacitors after the connection in $μC$ .

$(b)$ The total electrostatic energy stored in both capacitors in $μJ$ .

$(c)$ The heat produced during the charge transfer from one capacitor to the other in $μJ .$

Accepted Answer

Initial charge stored in the $2 µ F$ capacitor is $12 V$ .

$(a)$ On connecting the capacitors, the charge flows from the $μJ$ capacitor to the $4 µ F$ capacitor.

Let $q_{1}$ and $q_{2}$ be the charge on $2 μF$ and $4 μF$ capacitor.

The capacitors are connected in parallel, thus, the potential difference across them is the same.

$∴ V = \frac{q_{1}}{C_{1}} = \frac{q_{2}}{C_{2}}$ $\Rightarrow \frac{q_{1}}{2} = \frac{q_{2}}{4}$ $\Rightarrow q_{2} = 2 q_{1}$ .

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

$∴ q_{1} + q_{2} = 24 \times 10^{- 6}$

$\Rightarrow 3 q_{1} = 24 \times 10^{- 6}$

$\Rightarrow q_{1} = 8 \times 10^{- 6} C=8 μC$ and $q_{2} = 16 \times 10^{- 6} C = 16 μC$ .

$(b)$ Energies stored in the capacitors are given by $E_{1} = \frac{1}{2} \frac{q_{1}^{2}}{C_{1}} = \frac{1}{2} \times \frac{8^{2}}{2} = 16 μJ$ and $E_{2} = \frac{1}{2} \frac{q_{2}^{2}}{C_{2}} = \frac{1}{2} \times \frac{16^{2}}{4} = 32 μJ$ .

Hence, the total electrostatic energy stored in both capacitors is $16 + 32 = 48 μJ$ .

$(c)$ Initial energy stored in the $2 µ F$ capacitor is given by $E_{i} = \frac{1}{2} C V^{2} = \frac{1}{2} \times 2 \times 10^{- 6} \times 12^{2} = 144 μJ$ .

Total energy of the capacitors after they are connected in parallel is $48 μJ$ .

Thus, the heat produced during the charge transfer from one capacitor to the other is $= 144 - 48 = 96 μJ$ .

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