Question

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and work done by the battery will be (A) \((1 / 2)\) (B) \((2 / 1)\) (C) 1 (D) \((1 / 4)\)

Short answer

The ratio of the energy stored in the capacitor and the work done by the battery is \(\frac{1}{2}\).

Step-by-step solution

Write the formula for energy stored in a capacitor

The formula for the energy stored in a capacitor is given by: \[U_{cap} = \frac{1}{2}CV^2\] where \(U_{cap}\) is the energy stored in the capacitor, \(C\) is the capacitance, and \(V\) is the potential difference across the capacitor.

Write the formula for work done by the battery

The formula for the work done by the battery to charge the capacitor is given by: \[W_{battery} = QV\] where \(W_{battery}\) is the work done by the battery, \(Q\) is the charge on the capacitor, and \(V\) is the potential difference (EMF of the battery in this case) across the capacitor.

Express charge in terms of capacitance and potential difference

Recall that the charge on a capacitor can be expressed as: \[Q = CV\] Substitute this expression for charge into the work done by the battery formula: \[W_{battery} = (CV)V\] \[W_{battery} = CV^2\]

Calculate the ratio of energy stored and work done

Now we can find the ratio of the energy stored in the capacitor to the work done by the battery: \[\frac{U_{cap}}{W_{battery}} = \frac{\frac{1}{2}CV^2}{CV^2}\]

Simplify the ratio

Simplify the expression for the ratio: \[\frac{U_{cap}}{W_{battery}} = \frac{\frac{1}{2}CV^2}{CV^2} = \frac{1}{2}\] According to our calculations, the ratio of the energy stored in the capacitor and the work done by the battery is \(\frac{1}{2}\). Therefore, the correct answer is: (A) \(\frac{1}{2}\)