A parallel plate capacitor of capacitance 2 F is charged to a potential $$V$$. The energy stored in the capacitor is $$E_1$$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $$E_2$$. The ratio $$\frac{E_2}{E_1}$$ is

We have a capacitor of capacitance $$C = 2$$ F initially charged to potential $$V$$.

The initial energy is:

$$E_1 = \frac{1}{2}CV^2 = \frac{1}{2}(2)V^2 = V^2$$

Now, when this capacitor is connected to an identical uncharged capacitor in parallel, the total capacitance becomes $$C_{total} = 2 + 2 = 4$$ F. By conservation of charge, $$Q = CV = 2V$$, so the new voltage across the combination is:

$$V' = \frac{Q}{C_{total}} = \frac{2V}{4} = \frac{V}{2}$$

The energy of the combination is:

$$E_2 = \frac{1}{2}C_{total}(V')^2 = \frac{1}{2}(4)\left(\frac{V}{2}\right)^2 = \frac{1}{2}(4)\frac{V^2}{4} = \frac{V^2}{2}$$

So the ratio is:

$$\frac{E_2}{E_1} = \frac{V^2/2}{V^2} = \frac{1}{2}$$

Hence, $$E_2 : E_1 = 1 : 2$$.