Question

Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is: ____.

• A. 1.0 J
• B. 0.5 J
• C. 1.0 $\times$ 10⁻⁶ J
• D. 0.5 $\times$ 10⁻⁶ J

Hint:

When two \textbf{identical} capacitors are connected (one charged, one uncharged), exactly half of the initial energy is always lost. Initial energy was $\frac{1}{2}CV^2$, so loss is $\frac{1}{4}CV^2$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When a charged capacitor is connected to an uncharged one, charge is redistributed until both reach a common potential. During this redistribution, some energy is always dissipated as heat or electromagnetic radiation.

Step 2: Key Formula or Approach:
Energy loss ($\Delta E$) is given by: \[ \Delta E = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)} \]

Step 3: Detailed Explanation:
Given: $C_1 = C_2 = 200\text{ pF} = 200 \times 10^{-12}\text{ F}$, $V_1 = 100\text{ V}$, $V_2 = 0\text{ V}$. 1. Simplify the formula for equal capacitances ($C$): \[ \Delta E = \frac{C \cdot C \cdot V_1^2}{2(2C)} = \frac{1}{4}CV_1^2 \] 2. Substitute the values: \[ \Delta E = \frac{1}{4} \times (200 \times 10^{-12}) \times (100)^2 \] \[ \Delta E = 50 \times 10^{-12} \times 10^4 = 50 \times 10^{-8} \] \[ \Delta E = 0.5 \times 10^{-6}\text{ J} \]

Step 4: Final Answer:
The amount of energy lost is 0.5 $\times$ 10⁻⁶ J.