1. Initial capacitor energy

A capacitor of capacitance $C=2\,\text{F}$ is charged to potential $V$.

The initial energy stored is

$E_1=\frac{1}{2}CV^2$

Substituting $C=2$ F:

$E_1=\frac{1}{2}\cdot 2\cdot V^2=V^2$

2. After connecting to an identical uncharged capacitor

Now this charged capacitor is connected in parallel to another identical uncharged capacitor.

• First capacitor capacitance $=C$
• Second capacitor capacitance $=C$
• Total capacitance in parallel:

$C_{\text{eq}}=C+C=2C$

Initially, total charge on the system is only on the first capacitor:

$Q=CV$

Since the two capacitors are identical and connected in parallel, charge redistributes equally, and the final common potential becomes

$V_f=\frac{Q}{C_{\text{eq}}}=\frac{CV}{2C}=\frac{V}{2}$

3. Final energy of the combination

Now total energy stored in the two-capacitor combination is

$E_2=\frac{1}{2}C_{\text{eq}}V_f^2$

Substitute $C_{\text{eq}}=2C$ and $V_f=\frac{V}{2}$:

$E_2=\frac{1}{2}(2C)\left(\frac{V}{2}\right)^2$

$E_2=C\cdot \frac{V^2}{4}=\frac{CV^2}{4}$

But

$E_1=\frac{1}{2}CV^2$

So,

$\frac{E_2}{E_1}=\frac{\frac{CV^2}{4}}{\frac{1}{2}CV^2}=\frac{1}{4}\cdot 2=\frac{1}{2}$

Thus,

$E_2:E_1=1:2$

So,

$\boxed{\frac{E_2}{E_1}=\frac{1}{2}}$

4. Option check

• A: $1:2$ ✅ Correct
• B: $2:3$ ❌
• C: $2:1$ ❌
• D: $1:4$ ❌

Hence, the correct answer is Option A.