Question

Two identical capacitors have the same capacitance $C$. One of them is charged to potential $V_1$ and the other to $V_2$. The negative ends of the capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is

• A. $\frac{1}{4} C (V_1 - V_2)^2$
• B. $\frac{1}{2} C (V_1^2 + V_2^2)$
• C. $\frac{1}{2} C (V_1^2 - V_2^2)$
• D. $\frac{1}{2} C (V_1 + V_2)^2$

Hint:

The expression for energy loss during parallel combination always has the structural format of a reduced mass formula: $\frac{1}{2} C_{reduced} (\Delta V)^2$. Since the reduced capacitance of two identical capacitors in series-like fraction matching is $\frac{C \cdot C}{C+C} = \frac{C}{2}$, the overall multiplier must become $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$.

Answer 1

The Correct Option is A

Step 1: Note the starting charges.
Each capacitor has the same capacitance $C$. One holds charge $Q_1 = CV_1$, the other $Q_2 = CV_2$.
Step 2: Find the common potential.
Connecting like plates together puts them in parallel. Total charge is shared over total capacitance: \[ V = \frac{Q_1 + Q_2}{C + C} = \frac{C(V_1 + V_2)}{2C} = \frac{V_1 + V_2}{2}. \]
Step 3: Write the initial energy.
\[ U_i = \tfrac{1}{2}CV_1^2 + \tfrac{1}{2}CV_2^2. \]
Step 4: Write the final energy.
The combined capacitance $2C$ now sits at potential $V$: \[ U_f = \tfrac{1}{2}(2C)\left(\frac{V_1+V_2}{2}\right)^2 = \frac{C(V_1+V_2)^2}{4}. \]
Step 5: Subtract to get the loss.
\[ \Delta U = U_i - U_f = \frac{C}{2}(V_1^2+V_2^2) - \frac{C}{4}(V_1+V_2)^2. \] Expanding gives $\dfrac{C}{4}\big(2V_1^2 + 2V_2^2 - V_1^2 - 2V_1V_2 - V_2^2\big)$.
Step 6: Simplify the bracket.
The bracket becomes $V_1^2 - 2V_1V_2 + V_2^2 = (V_1 - V_2)^2$, so \[ \Delta U = \frac{1}{4}C(V_1 - V_2)^2. \] That is option (A). \[ \boxed{\Delta U = \tfrac{1}{4}C(V_1 - V_2)^2} \]