Question

A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.

Answer 1

$\text{Step 1: Initial energy stored in capacitor}$

$U_i=\frac{1}{2}C{V}^2$ $....\left(1\right)$

$\text{Step 2: Charge conservation[Ref. Fig.]}$

Let the Initial charge be $Q$

After connecting with another capacitor of the equal capacitance, the charge will flow until the potential becomes the same.

$V_1=V_2$

$\Rightarrow \frac{Q_1}{C_1}=\frac{Q_2}{C_2}$

$\Rightarrow Q_1=Q_2$ $[\because C_1=C_2=C]$

Applying charge conservation: $Q_1+Q_2=Q$

$\therefore Q_1=Q_2=\frac{Q}{2}$

$\text{Step 3: Potential on each capacitor}$

$V_1=\frac{Q_1}{C}=\frac{Q}{2C},V_2=\frac{Q_2}{C}=\frac{Q}{2C}$

$\Rightarrow V_1=\frac{V}{2},V_2=\frac{V}{2}$

$\text{Step 4: Final energy stored}$

$U_f=\frac{1}{2}{C}_1{V}_1^2+\frac{1}{2}C{V}_2^2$

$=\frac{1}{2}C{\left(\frac{V}{2}\right)}^2+\frac{1}{2}C{\left(\frac{V}{2}\right)}^2$

$\Rightarrow U_f=\frac{C{V}^2}{4}$ $....\left(2\right)$

$\text{Step 5: Ratio of energy}$

From equation $\left(1\right)$ and $\left(2\right)$

$\frac{U_f}{U_i}=\frac{C{V}^2/4}{C{V}^2/2}$

$\Rightarrow \frac{V_f}{V_i}=\frac{1}{2}$