#### 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 . 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.

#### Solution

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

`q = C xx V = 12 xx 2 xx 10^-6`

`⇒ q = 24 xx 10^-6 "C"`

(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 q_{1} and q_{2}, respectively.

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

`therefore V = q_1/C_1 = q_2/C_2`

`⇒ V = q_1/2 = q_2/4`

`⇒ 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.

`therefore q_1 + q_2 = 24 xx 10^-6`

`⇒ 3 q_1 = 24 xx 10^-6 C`

`⇒ q_1 = 8 xx 10^-6 C = 8 muC`

`⇒ q_1 = 2 q_1 = 16 muC`

(b) Energies stored in the capacitors are given by

`E_1 = 1/2 xx q_1^2/C_1 = 16 muJ`

and

`E_2 = 1/2 xx q_2^2/C_2 = 32 muJ`

(c) Initial energy stored in the 2 µF capacitor is given by

`E_i = 1/2 CV^2 = 1/2 xx (2 xx 10^-6) (12)^2`

`⇒ E_i = 144 muJ`

Total energy of the capacitors after they are connected in parallel is given by

E_{f} = E_{1} + E_{2}

⇒ Ef = 16 + 32 = 48 μJ

Heat produced during the charge transfer is given by

`E = E_f - E_i`

`⇒ E = 144 - 48 = 96 muJ`