A battery does 100J of work, in charging a capacitor, and then final energy in capacitor is:
(A) 200J
(B) 150J
(C) 100J
(D) 50J

Hint: A capacitor charged by a battery. So the final energy in the capacitor is half the work done by the capacitor. Though work and energy are closely related they are not entirely the same. Energy is the capacity to do work on the application of a certain force. In this problem, we need to calculate the final energy when the work done by the battery of the capacitor is given.

Formula used: E = $\dfrac{W}{2}$

Complete step by step answer:
Given, the amount of work done by the battery of a capacitor while charging = 100J
To find: Final energy stored in the capacitor
We know that the charging of a capacitor is due to the battery in the circuit. The work done by the capacitor while charging is given by, W = QV ...................... (1)
Where, Q is the total charge and V is the potential of the capacitor while charging.
The energy stored by a capacitor is given by, E =$\dfrac{QV}{2}$ ..................... (2)
From equation (1) and (2), we have
E = $\dfrac{W}{2}$
Plugging the value of work that is given in the question, we have
$\Rightarrow E=\dfrac{100}{2}$
$\Rightarrow E=50$ J

So, the correct answer is “Option D”.

Additional Information: The energy stored in a capacitor can be expressed in terms of work done by the capacitor while charging as given by the relation. Storing energy on the capacitor involves doing work to transport charge from one plate of the capacitor to the other against the electrical force so the two terms are related.

Note: While solving this question it is important to properly establish the relationship between work done during charging of a capacitor and the energy stored in it during the charging process. The entire problem depends on this relationship.