Question

A parallel plate capacitor having capacity \( C_0 \) is charged to \( V_0 \). With battery disconnected, if the separation between the plates is doubled, then the energy stored in it is \( E_1 \). Instead if the separation between the plates is doubled, with battery in connection, the energy stored in it is \( E_2 \). Then the value of \( \frac{E_2}{E_1} \) is

• A. 0.5
• B. 1.5
• C. 2
• D. 0.25

Hint:

For capacitors, if the battery is disconnected, the energy decreases with an increase in plate separation. If connected, the energy remains constant despite changes in capacitance.

Answer 1

The Correct Option is D

The energy stored in a capacitor when the battery is disconnected is given by: \[ E_1 = \frac{1}{2} C_0 V_0^2 \] When the distance between the plates is doubled with the battery disconnected, the capacitance becomes \( \frac{C_0}{2} \). So, the energy becomes: \[ E_1 = \frac{1}{2} \left(\frac{C_0}{2}\right) V_0^2 = \frac{1}{4} C_0 V_0^2 \] When the battery is connected, the voltage remains constant, and doubling the distance between the plates reduces the capacitance by half, but the energy is given by: \[ E_2 = \frac{1}{2} \left(\frac{C_0}{2}\right) V_0^2 = \frac{1}{4} C_0 V_0^2 \] Thus, the ratio \( \frac{E_2}{E_1} \) is: \[ \boxed{\frac{E_2}{E_1} = 0.25} \]