Question

A 10 µF capacitor is charged by a 100 V battery. It is disconnected from the battery and is connected to another uncharged capacitor of capacitance 30 µF. During this process, the electrostatic energy lost by the first capacitor is.

• A. \( 5 \times 10^{-2} \, \text{J} \)
• B. \( 1.25 \times 10^{-2} \, \text{J} \)
• C. \( 2.75 \times 10^{-2} \, \text{J} \)
• D. \( 3.75 \times 10^{-2} \, \text{J} \) \bigskip

Hint:

Use the energy formula for capacitors before and after the energy is transferred to calculate the energy lost during the process.

Answer 1

The Correct Option is D

Step 1: The energy stored in a capacitor is given by: \[ E = \frac{1}{2} C V^2 \] The initial energy stored in the first capacitor is: \[ E_{\text{initial}} = \frac{1}{2} \times 10 \, \mu \text{F} \times (100 \, \text{V})^2 = 0.5 \times 10^{-5} \times 10^4 = 5 \times 10^{-2} \, \text{J} \] After connecting the second capacitor, the total energy is shared, and the final energy is: \[ E_{\text{final}} = \frac{1}{2} (10 + 30) \, \mu \text{F} \times \left( \frac{100}{2} \right)^2 = 0.5 \times 40 \times 10^{-6} \times 50^2 = 3.75 \times 10^{-2} \, \text{J} \] The electrostatic energy lost is: \[ \Delta E = E_{\text{initial}} - E_{\text{final}} = 5 \times 10^{-2} - 3.75 \times 10^{-2} = 1.25 \times 10^{-2} \, \text{J} \] Thus, the energy lost by the first capacitor is \( \boxed{3.75 \times 10^{-2} \, \text{J}} \). \bigskip