The energy stored in a capacitor is given by:
$$E = \frac{1}{2} C V^2$$
where \(E\) is the energy stored in the capacitor, \(C\) is the capacitance, and \(V\) is the voltage across the capacitor.
This expression represents the energy stored in an ideal capacitor without considering dielectric losses or resistance.
The energy stored in a capacitor can be derived using the concept of work done in charging the capacitor.
For a capacitor, the basic relation is:
$$C = \frac{q}{V} \quad \Rightarrow \quad q = C V$$
Consider a small amount of charge \(dq\) being transferred to the capacitor. The small work done is:
$$dW = V \, dq$$
Since \(V = \frac{q}{C}\), substituting into the above equation gives:
$$dW = \frac{q}{C} \, dq$$
The total work done, which is the energy stored in the capacitor, is obtained by integrating from \(0\) to \(Q\):
$$E = \int_0^Q \frac{q}{C} \, dq$$
Performing the integration:
$$E = \frac{1}{C} \int_0^Q q \, dq$$
$$E = \frac{1}{C} \left[ \frac{q^2}{2} \right]_0^Q$$
$$E = \frac{Q^2}{2C}$$
Since \(Q = C V\), substituting this into the expression:
$$E = \frac{(C V)^2}{2C} = \frac{1}{2} C V^2$$
Thus, the final expression for the energy stored in a capacitor is:
$$E = \frac{1}{2} C V^2$$
The energy is stored in the electric field between the capacitor plates and the derivation assumes a gradual (quasi-static) charging process.
The energy can also be expressed in alternative forms:
$$E = \frac{1}{2} QV \quad \text{and} \quad E = \frac{Q^2}{2C}$$