Table of Contents:

• Capacitor, Types and Capacitance
• Combination of Capacitors
• Energy Stored in a Capacitor

In this module, we will explore the amount of energy that can be stored in a capacitor, the parameters that affect the amount of energy stored, and their relationships. We already know that capacitors are used to store energy.

How to Calculate the Energy Stored in a Capacitor?

The work done to transfer charges onto a conductor, against the force of repulsion from the existing charges, is stored as potential energy in the electric field of the conductor.

The charge on plate A and B after the transfer of dQ’ will be Q’ + dQ’ and -Q’ - dQ’ respectively. The work done by an external force will be dQ’ * (Q’ + dQ’ + (-Q’ - dQ’)) = 2 * dQ’ * Q'

\begin{array}{l}dW = VdQ’ = \frac{Q’}{C}dQ \end{array}

Total work done = $\frac{Q^2}{2C}$

\begin{align} \therefore \text{Energy Stored in a Capacitor} &= \frac{Q^2}{2C} \ &= \frac{1}{2}CV^2 \ &= \frac{1}{2}QV \end{align}

Energy Density in an Electric Field

Energy density is the amount of energy stored per unit volume. It is represented by:

\begin{array}{l}U = \frac{1}{2}\varepsilon_0 E^2\end{array}

$$U = \frac{1}{2}k\varepsilon {{E}^{2}}$$ where k is the dielectric constant of the dielectric medium and E is the net electric field in the dielectric medium.

Problems on Energy Stored in a Capacitor

Problem 1: Calculate the energy stored in the capacitors in the steady state when a 20 V battery is connected to 3 capacitors in series, with two of them being 20μF and one being 10μF.

Sol:

$\frac{1}{C_{eff}}=\frac{1}{5}$

C_eff = 5μF

The energy stored $$=\frac{1}{2}C{{V}^{2}}=\frac{1}{2}\times 5\times {{10}^{-6}}\times {{20}^{2}}={{10}^{-3}}J$$

Problem 2: Find the capacitance, charge, and energy stored in a parallel plate capacitor with plates of area 4 m², separated by a distance of 0.5 mm, and connected across a cell of emf 100 volts, after a dielectric slab of dielectric constant k = 3 and thickness 0.5 mm has been inserted inside the capacitor.

Sol: When the capacitor is without a dielectric

\begin{array}{l}{C}_{0} = \frac{{\varepsilon }_{0}A}{d} = \frac{8.85 \times {{10}^{-12}} \times 4}{0.5 \times {{10}^{-3}}} = 7.08 \times {{10}^{9}}\end{array}

\begin{array}{l}{C_0}=7.08\times {10^{-2}}\mu F.\end{array}

\begin{array}{l}{Q_0} = {C_0}{V_0}\end{array}

$$(7.08 \times 10^{-2} \times 100)_{\mu C} = 7.08\mu C$$

The charge on the capacitor remains constant as the cell has been disconnected, since $${U_0} = \frac{1}{2}{C_0}V_0^2 = 354 \times {10^{-6}}J$$

\begin{array}{l}C = \frac{k \cdot \varepsilon_0 \cdot A}{d} = k \cdot C_0 = 0.2124 \mu F\end{array}

\begin{array}{l}V = \frac{Q}{C} = \frac{{Q_0}}{kC_0} \times \frac{{V_0}}{k} = \frac{100}{3} \text{ volts}\end{array}

$$U = \frac{1}{2}\frac{Q_0^2}{kC_0} \times \frac{U_0}{k} = 118 \times 10^{-6}J$$

Frequently Asked Questions on Energy Stored in Capacitors

Does energy get stored in a capacitor?

Yes, energy can be stored in a capacitor.

In a capacitor, the energy is stored in the space between the two plates.

The type of energy stored in a capacitor is electrical energy.

The electrostatic potential energy stored in a capacitor is related to the charge and voltage between its plates.

Where is the energy stored in a capacitor located?

The energy of a charged capacitor remains in the field between its plates when it is disconnected from a battery.

What are the Ways to Increase the Energy Stored in a Capacitor?

The energy stored by a capacitor is dependent on its capacitance and voltage. If either of these parameters are increased, the energy stored by the capacitor will also increase. To further increase the capacitance of the capacitor, a dielectric slab can be placed between its plates.