Energy of an electric field

When the electric field between clouds and the ground grows strong enough, the air becomes conductive, and electrons travel from the cloud to the ground.

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.

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 $L?$

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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

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

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?

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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,

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

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.

Dielectric material between the plates of a parallel-plate capacitor is polarized by an electric field and reduces the potential difference between the plates [4].

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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. $_\square$

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 $B = 2.00 \text{ mT}?$

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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.

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