QJEEM23Ph4118: JEE Main 2023 Physics - Class 12 JEEM23Ph4118 Solution: A capacitor is charged by supply. It is then disconnected from…

The energy stored in a capacitor can be calculated using the formula:

$U = \frac{1}{2} C V^2$

where:

• (U) is the energy,
• (C) is the capacitance,
• (V) is the voltage.

Initially, the energy stored in the first capacitor is:

$U_{\text{initial}} = \frac{1}{2} C V^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (200 \, \text{V})^2 = 0.012 \, \text{J} = 12 \, \mu\text{J}.$

When the charged capacitor is connected to the uncharged capacitor, the charge will distribute equally between them because they have the same capacitance. Therefore, the final voltage across each capacitor is half of the initial voltage, i.e., 100 V.

The energy in each capacitor after the redistribution is:

$U_{\text{final each}} = \frac{1}{2} C \left(\frac{V}{2}\right)^2 = \frac{1}{2} \times 600 \times 10^{-12} \, \text{F} \times (100 \, \text{V})^2 = 0.003 \, \text{J} = 3 \, \mu\text{J}.$

As there are two capacitors, the total final energy is:

$U_{\text{final total}} = 2 \times U_{\text{final each}} = 2 \times 3 \, \mu\text{J} = 6 \, \mu\text{J}.$

The energy loss is the difference between the initial energy and the final energy:

$\Delta U = U_{\text{initial}} - U_{\text{final total}} = 12 \, \mu\text{J} - 6 \, \mu\text{J} = 6 \, \mu\text{J}.$