Additional: Energy Density in Electric Field
Energy Density of Electric Field ($ u_E = \frac{1}{2}\epsilon_0 E^2 $)
We have seen that energy is stored in a charged capacitor, and this energy is associated with the electric field between its plates. This suggests that energy is stored in the electric field itself, distributed throughout the region where the field exists. The amount of energy stored per unit volume of space occupied by the electric field is called the energy density of the electric field.
Derivation (for a Parallel Plate Capacitor in Vacuum)
Consider a parallel-plate capacitor with plate area $A$ and separation $d$, containing vacuum between the plates. Let the capacitor be charged to a potential difference $V$ and store a charge $Q$. The capacitance is $C_0 = \frac{\epsilon_0 A}{d}$. The energy stored is $U = \frac{1}{2} C_0 V^2$.
In a parallel-plate capacitor (assuming uniform field), the volume of space between the plates is $V_{space} = A d$. The electric field in this region is uniform and has magnitude $E = V/d$. The potential difference $V = E d$.
Substitute $V = Ed$ and $C_0 = \frac{\epsilon_0 A}{d}$ into the energy formula $U = \frac{1}{2} C_0 V^2$:
$ U = \frac{1}{2} \left(\frac{\epsilon_0 A}{d}\right) (Ed)^2 = \frac{1}{2} \frac{\epsilon_0 A}{d} E^2 d^2 = \frac{1}{2} \epsilon_0 A d E^2 $
The energy density ($u_E$) is the energy stored ($U$) divided by the volume of space ($V_{space}$) where the field exists:
$ u_E = \frac{U}{V_{space}} = \frac{\frac{1}{2} \epsilon_0 A d E^2}{Ad} $
$ u_E = \frac{1}{2} \epsilon_0 E^2 $
This formula gives the energy density of the electric field in vacuum. It shows that the energy density is proportional to the square of the magnitude of the electric field. The units are Joules per cubic metre (J/m$^3$).
Energy Density in a Dielectric Medium
If the space between the capacitor plates is filled with a dielectric material with permittivity $\epsilon = \epsilon_r \epsilon_0$, the capacitance is $C = \epsilon_r C_0 = \frac{\epsilon A}{d}$. The electric field inside the dielectric is $E = V/d$, and the potential difference is $V = Ed$. The stored energy is $U = \frac{1}{2} C V^2 = \frac{1}{2} \left(\frac{\epsilon A}{d}\right) (Ed)^2 = \frac{1}{2} \epsilon A d E^2$.
The energy density in the dielectric medium is:
$ u_E = \frac{U}{Ad} = \frac{\frac{1}{2} \epsilon A d E^2}{Ad} = \frac{1}{2} \epsilon E^2 = \frac{1}{2} \epsilon_r \epsilon_0 E^2 $
So, the energy density formula can be generalized for a medium with permittivity $\epsilon$:
$ u_E = \frac{1}{2} \epsilon E^2 $
where $E$ is the magnitude of the electric field in the medium.
Significance
The concept of energy density of the electric field is important because it suggests that energy is not stored "on" the charges or "in" the capacitor plates, but rather "in" the electric field itself, which fills the space around the charges. This view is consistent with electromagnetic theory and the idea that electromagnetic waves carry energy through space in the form of oscillating electric and magnetic fields. The total electrostatic energy of a system of charges can be calculated by integrating the energy density over all space where the electric field exists.
$ U_{total} = \int_{all\,space} u_E \, dV = \int_{all\,space} \frac{1}{2} \epsilon E^2 \, dV $
Example 1. A parallel plate capacitor has an electric field of $2 \times 10^5$ N/C between its plates in vacuum. Calculate the energy density of the electric field. (Take $\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)$).
Answer:
Magnitude of electric field, $E = 2 \times 10^5$ N/C.
Medium: Vacuum. Permittivity of free space, $\epsilon_0 = 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)$.
Using the formula for the energy density of the electric field in vacuum, $u_E = \frac{1}{2}\epsilon_0 E^2$:
$ u_E = \frac{1}{2} \times (8.854 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2) \times (2 \times 10^5 \text{ N/C})^2 $
$ u_E = \frac{1}{2} \times 8.854 \times 10^{-12} \times (4 \times 10^{10}) $ J/m$^3$ (Units: $\frac{\text{C}^2}{\text{N} \cdot \text{m}^2} \times \frac{\text{N}^2}{\text{C}^2} = \frac{\text{N}}{\text{m}^2} = \frac{\text{J/m}}{\text{m}^2} = \text{J/m}^3$)
$ u_E = \frac{1}{2} \times 8.854 \times 4 \times 10^{(-12+10)} $ J/m$^3
$ u_E = 8.854 \times 2 \times 10^{-2} $ J/m$^3
$ u_E = 17.708 \times 10^{-2} = 0.17708 $ J/m$^3$.
The energy density of the electric field is approximately 0.177 Joules per cubic metre.