Question

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system

• A. Increases by a factor of 4
• B. Decreases by a factor of 2
• C. Remains the same
• D. Increases by a factor of 2

Answer 1

The Correct Option is B

To solve this problem, we first need to understand the scenario presented. Initially, a capacitor is charged by a battery. After the battery is removed, this charged capacitor is connected in parallel with another identical, but uncharged capacitor. We need to calculate the change in the total electrostatic energy of the system due to this new connection.

Let's break it down step by step:

1. When the first capacitor (let's call it C) is charged by a battery with a voltage V, the energy stored in this capacitor (U_1) can be given by the formula:

   U_1 = \frac{1}{2} \cdot C \cdot V^2

2. After charging, the battery is removed, and an identical uncharged capacitor is connected in parallel. When capacitors are connected in parallel, the total capacitance of the system doubles (since the capacitors are identical):

   C_{\text{total}} = C + C = 2C

3. Now, the total charge Q on the initially charged capacitor is: Q = C \cdot V.

4. This charge gets redistributed over the two capacitors equally because they are identical, resulting in a new voltage (V') across each capacitor:

   V' = \frac{Q}{C_{\text{total}}} = \frac{C \cdot V}{2C} = \frac{V}{2}

5. Now, the energy stored in each capacitor in the parallel combination is:

   U' = \frac{1}{2} \cdot C \cdot (V')^2 = \frac{1}{2} \cdot C \cdot \left( \frac{V}{2} \right)^2 = \frac{1}{2} \cdot C \cdot \frac{V^2}{4} = \frac{1}{8} \cdot C \cdot V^2

6. So, the total energy of the two capacitors in parallel is:

   U_{\text{total}} = 2 \times \frac{1}{8} \cdot C \cdot V^2 = \frac{1}{4} \cdot C \cdot V^2

7. Comparing the initial and final total energy:

   \frac{U_{\text{total}}}{U_1} = \frac{\frac{1}{4} \cdot C \cdot V^2}{\frac{1}{2} \cdot C \cdot V^2} = \frac{1}{2}

The final electrostatic energy of the system decreases by a factor of 2. Therefore, the correct answer is:

Decreases by a factor of 2