# Energy of an electric field

The **energy of an electric field** results from the excitation of the space permeated by the electric field. It can be thought of as the potential energy that would be imparted on a point charge placed in the field.

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## Energy of a point charge distribution

The energy stored in a pair of point charges $q_1$ and $q_2$ a distance $r$ from one another can be thought of in terms of the work-energy theorem and the work required to bring one of the charges from far away.

$\Delta U = -W = -\int_\infty^r F dr' = \int_r^\infty \frac{kq_1 q_2}{r'^2} dr' = -\frac{kq_1 q_2}{r'} \Big|_r^\infty =-(0 - \frac{kq_1 q_2}{r}) = \frac{kq_1 q_2}{r}$

$U_f - U_0 = U_f-0 = U_f = \frac{kq_1 q_2}{r}$

The energy of each pair of charges in an arrangement is

$U=\frac{k q_1 q_2}{r}.$

What is the energy stored in an arrangement of three charges of magnitude $Q$ arranged in an equilateral triangle with side length $L?$

Since there are three charges, there are three pairs. Each pair has energy

$U = \frac{kq_1 q_2}{r} = \frac{kQ^2}{L}.$

So for three identical pairs, the energy is

$U = 3*\frac{kQ^2}{L} = \frac{3kQ^2}{L}.$

## Energy stored in a capacitor

Storing charge on the isolated conductors of a capacitor requires work to move the charge onto the conductors. By definition of the potential difference, if charge $dQ$ is added to one of the conductors, causing a potential difference $dV$, then a work of $dW=VdQ = \frac{Q}{C} dQ$ is required. So the total work required to charge one of the conductors from neutrality up to charge $Q$ is

$W = \int dW = \int_0^Q \frac{Q'}{C} dQ' = \frac{Q^2}{2C}.$

By the law of conservation of energy, the work done in charging the capacitor is stored as potential energy $U$ in the electric field of the capacitor. Using $Q=CV$ this can be rewritten several ways:

$U = \frac{Q^2}{2C} = \frac12 CV^2 = \frac12 QV.$

## What is the energy stored on a parallel-plate capacitor?

As computed above, the capacitance of the parallel-plate capacitor (area $A$, plate separation $d$, charge $Q$) is

$C = \frac{A\epsilon_0}{d}.$

Plugging into the formula for the potential energy stored in a capacitor,

$U = \frac{Q^2}{2C} = \frac{Q^2 d}{2 A \epsilon_0}.\ _\square$

If a capacitor is composed of two isolated conductors, after charging the oppositely charged plates will experience a Coulombic attraction. Given a spherical capacitor of inner radius $a$ and outer radius $b$, find the attractive force exerted on the outer conductor assuming that each conductor holds charge $\pm Q$.

Assume the conductors are mechanically held fixed, so the force is constant in time, and let negative forces correspond to attraction and vice versa.

The capacitance of a capacitor and thus the energy stored in a capacitor at fixed voltage can be increased by use of a **dielectric**. A dielectric is an insulating material that is polarized in an electric field, which can be inserted between the isolated conductors in a capacitor. That is, when an electric field is applied to a dielectric, the positive and negative charges in the insulating material shift slightly from their neutral equilibrium, creating a small electric field opposing the applied field.

Mathematically, the effect of a dielectric is to modify the permittivity of free space $\epsilon_0$ by some constant:

$\epsilon_0 \to \epsilon = \kappa \epsilon_0.$

The constant $\kappa$ is often called the **dielectric constant**, and takes into account how the presence of the dielectric modifies the strength of the electric field in the insulating material. In vacuum, $\kappa = 1$; otherwise, $\kappa > 1$ since any atoms present may be slightly polarized.

## Find the capacitance of a parallel-plate capacitor with a dielectric of constant $\kappa$ inserted between the plates.

The capacitance of the parallel-plate capacitor was derived to be

$C = \frac{A\epsilon_0}{d},$

with $A$ the area of each plate and $d$ the plate separation. With the dielectric inserted, $\epsilon_0 \to \kappa \epsilon_0$. The resulting capacitance is

$C = \frac{A \kappa \epsilon_0}{d}.$

Since $\kappa > 1$, inserting the dielectric increases the capacitance and therefore increases the energy stored at a fixed voltage. $_\square$

## Energy density of an electric field

The electric field component of an electromagnetic wave carries an electric energy density $u_E$ given by

$u_E =\frac12 \varepsilon E$

where $E$ is the amplitude of the electric field and $\varepsilon =8.85 \times 10^{-12} \frac{\text{s}^4 \text{A}^2}{\text{m}^3 \text{kg}}$ is the permittivity of free space.

What is the electric energy density of an electromagnetic wave with magnetic field amplitude $B = 2.00 \text{ mT}?$

First, express the magnetic field amplitude in terms of the electric field amplitude.

$c=\frac{E}{B}$

$\rightarrow E=cB$

Then, calculate the energy density.

$\begin{aligned} u_E &= \frac12 \varepsilon E \\ &=\frac12 \varepsilon cB \\ &=\frac12 (8.85\times 10^{-12})(3.00\times 10^8) (2.00\times 10^{-3}) \\ &= 2.66 \frac{\text{T}}{\text{m}^3} \text{ }_\square \\ \end{aligned}$

**Cite as:**Energy of an electric field.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/energy-of-an-electric-field/