MHT-CET Physics Electrostatics PYQ - Battery & Capacitor Energy

14) A battery is used to charge a parallel plate capacitor until the potential difference between the plates becomes equal to the emf of the battery. The ratio of the energy stored in the capacitor to the work done by the battery is:

A. \(1\)

B. \(\frac{1}{2}\)

C. \(2\)

D. \(\frac{1}{4}\)

Solution:

When a battery charges a capacitor, the final potential difference across the capacitor becomes equal to the battery’s emf, say \(V\). The charge on the capacitor is then: \[ Q = CV. \]

The energy stored in the capacitor is given by: \[ U = \frac{1}{2} C V^2. \]

However, the work done by the battery is not equal to the energy stored in the capacitor. As the capacitor charges, the battery does work to move the charge against the increasing potential. The total work done by the battery is: \[ W_{\text{battery}} = QV = CV^2. \]

Therefore, the ratio of the energy stored in the capacitor to the work done by the battery is: \[ \frac{U}{W_{\text{battery}}} = \frac{\frac{1}{2} C V^2}{C V^2} = \frac{1}{2}. \]

Hence, the correct answer is option B.