1. Given data

• Capacitor $C_1 = C$, charged to potential difference $V_1 = V$
• Capacitor $C_2 = 2C$, charged to potential difference $V_2 = 2V$

They are connected in parallel such that positive terminal of one is connected to negative terminal of the other.

2. Initial charges on the capacitors

Using $Q = CV$:

$Q_1 = C \cdot V = CV$

$Q_2 = (2C)(2V) = 4CV$

Since the connection is opposite-polarity, let us assign node charges after connection:

• Join the positive plate of capacitor 1 to the negative plate of capacitor 2: call this node $A$
• Join the negative plate of capacitor 1 to the positive plate of capacitor 2: call this node $B$

Then initially, node charges are:

For capacitor 1:

• positive plate: $+CV$
• negative plate: $-CV$

For capacitor 2:

• positive plate: $+4CV$
• negative plate: $-4CV$

So,

$Q_A = (+CV) + (-4CV) = -3CV$

$Q_B = (-CV) + (+4CV) = +3CV$

These node charges are conserved.

3. Equivalent capacitance after connection

Since they are connected in parallel between the same two nodes,

$C_{\text{eq}} = C + 2C = 3C$

4. Final common voltage

Let final voltage across the combination be $V_f$.

Using

$Q = C_{\text{eq}} V_f$

Magnitude of charge on the equivalent capacitor is $3CV$, so

$3CV = (3C)V_f$

$V_f = V$

5. Final energy

Energy stored finally is

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

$U_f = \frac{1}{2}(3C)(V^2)$

$U_f = \frac{3}{2}CV^2$

6. Check options

• A: Zero $\quad \times$
• B: $\dfrac{3}{2}CV^2 \quad \checkmark$
• C: $\dfrac{9}{2}CV^2 \quad \times$
• D: $\dfrac{25}{6}CV^2 \quad \times$

Thus the correct option is B.

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.