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

Let $q$ be the charge on the charged parallel plate capacitor.

Energy stored in it is,

$U=\frac{q^2}{2C}$

When another similar uncharged capacitor is connected, the net capacitance of the system is,

$C^{\prime }=2C$

The charge on the system is constant. So, the energy stored in the system now is,

$U^{\prime }=\frac{q^2}{2(C^{\prime }{)}^2}$

$\Rightarrow U^{\prime }=\frac{q^2}{2\left(2C\right)}$

$\Rightarrow U^{\prime }=\frac{q^2}{4C}$

Thus, the required ratio is,

$\frac{U^{\prime }}{U}=\frac{\frac{q^2}{4C}}{\frac{q^2}{2C}}$

$\Rightarrow \frac{U^{\prime }}{U}=\frac{1}{2}$