Class 11 > Unit # 09: Capacitors > Energy Stored in Capacitors

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Energy Stored in a Capacitor

Video Lecture

Watch the complete video lecture below to understand how energy is stored in a capacitor and follow the step-by-step mathematical derivation.

Energy Stored in a Capacitor

"Charging of a capacitor always involves some expenditure of energy by the charging agency. This energy is stored up in the electrostatic field set up in the dielectric medium. On discharging the capacitor, the field collapses and the stored energy is released."

Many individuals working with electronic equipment have at some time verified that a capacitor can store energy. If the plates of a charged capacitor are connected by a conductor such as a wire, charge moves between each plate and its connecting wire until the capacitor is fully discharged. This discharge can often be observed as a visible spark.

Expression for Energy Stored in a Capacitor

Let us consider a capacitor connected to a source of potential difference $V$. Initially, when the capacitor is uncharged, the potential difference between the plates is zero. As charges $+Q$ and $-Q$ are gradually deposited on the plates by the source, the potential difference between the plates increases linearly from zero to $V$.

The average voltage on the capacitor during the entire charging process is given by the mean of the initial and final voltages:

$$\text{Average Voltage} = \frac{0 + V}{2} = \frac{V}{2}$$

By definition, work done ($W$) or potential energy ($U$) stored in moving a total charge $Q$ through an average potential difference is:

$$U = \text{Work Done} = \text{Charge} \times \text{Average Potential Difference}$$

$$U = Q \left(\frac{V}{2}\right)$$

$$U = \frac{1}{2} Q V \quad \text{--- (i)}$$

Since the fundamental relation for a capacitor is $Q = C V$, we can substitute this into equation (i) to express the stored energy in alternative forms:

1. Substituting $Q = C V$:

$$U = \frac{1}{2} (C V) V$$

$$U = \frac{1}{2} C V^2 \quad \text{--- (ii)}$$

2. Substituting $V = \frac{Q}{C}$:

$$U = \frac{1}{2} Q \left(\frac{Q}{C}\right)$$

$$U = \frac{Q^2}{2C} \quad \text{--- (iii)}$$

Thus, the electrostatic potential energy stored in a capacitor can be calculated using any of the three formulas depending on the known quantities: $U = \frac{1}{2}QV$, $U = \frac{1}{2}CV^2$, or $U = \frac{Q^2}{2C}$.

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