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.
Figure

Answer 1

(a) The charge on each capacitor is equal to 16 µC

(b) electrostatic energy stored in each Capacitor is 32 µJ

(c) Heat produced during the charge transfer from one capacitor to other is 96 µJ

Explanation :

Charge on the 2 $\mu F$ capacitor when it is not connected to the 4 $\mu F$ capacitor is given by

$\boldsymbol{q}=\boldsymbol{C} \times \boldsymbol{V}$

$=12 \times 2 \times 10^{-6}$

$q=24 \times 10^{-6} \mathrm{C}$

(a) On connecting the capacitors, let the charge on each capacitor is $q_1$, $q_2$

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

Therefore Potential difference (V) = Q/C

From the above statement V = $q_1$/$c_1$ =$q_2$ /$c_2$

$\therefore V=\frac{q_{1}}{C_{1}}=\frac{q_{2}}{C_{2}}$

$V=\frac{q_{1}}{2}=\frac{q_{2}}{4}$

$q_{2}=2 q_{1}$

Total charge on the capacitors = $q_{1}+q_{2}=24 \times 10^{-6}$

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

 $q_{1}=8 \times 10^{-6} C=8 \mu C$

 $q_{1}=2 q_{1}=16 \mu C$

(b) Energies stored in the capacitor are given by

$E_{1}=\frac{1}{2} \times \frac{q_{1}^{2}}{C_{1}}=16 \mu J$

$E_{2}=\frac{1}{2} \times \frac{q_{2}^{2}}{C_{2}}=32 \mu J$

(c) Initial energy stored in the 2 $\mu F$capacitor is given by

$E_{i}=\frac{1}{2} C V^{2}$

    $=\frac{1}{2} \times\left(2 \times 10^{-6}\right)(12)^{2}$

$E_{i}=144 \mu J$

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

$\mathrm{E}_{\mathrm{f}}=\mathrm{E}_{1}+\mathrm{E}_{2}$

$\mathrm{Ef}=16+32$

     $=48 \mu \mathrm{J}$

Heat produced during the charge transfer is given by

$E=E_{f}-E_{i}$

$E=144-48$

   $=96 \mu J$