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: Understanding the Question:
Two identical independent capacitors of capacitance $C$ are initially charged to different electrical potentials, $V_1$ and $V_2$.
They are then wired in parallel by connecting their matching negative terminals together and their matching positive terminals together. We need to calculate the resulting total electrostatic potential energy loss ($\Delta U$) of the combined system due to charge redistribution.

Step 2: Key Formula or Approach:
When two pre-charged capacitors are linked together in parallel, the net decrease or loss in electrostatic energy during charge sharing is given by the standard formula:
$$\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2$$ This formula applies directly because like-polarity plates are connected to like-polarity plates.

Step 3: Detailed Explanation:
The two capacitors are completely identical, which means their capacitance values are equal:
$$C_1 = C \quad \text{and} \quad C_2 = C$$ Substitute these identical parameters into the energy loss formula:
$$\Delta U = \frac{1}{2} \frac{C \cdot C}{C + C} (V_1 - V_2)^2$$ Simplify the capacitive fraction part of the relation:
$$\frac{C \cdot C}{C + C} = \frac{C^2}{2C} = \frac{C}{2}$$ Now substitute this simplified equivalent capacity term back into our energy loss expression:
$$\Delta U = \frac{1}{2} \left(\frac{C}{2}\right) (V_1 - V_2)^2 = \frac{1}{4} C (V_1 - V_2)^2$$

Step 4: Final Answer:
The total decrease in energy of the combined system is $\frac{1}{4} C (V_1 - V_2)^2$, which matches option (A).