Energy Density and Electric Field

claimed by Samir Nileshwar

Mathematical Model

This section takes an alternative view and treats electric fields as if they have energy stored in them. To learn how this works, consider moving one plate of a capacitor. For a very small gap, the force of one capacitor plate on another capacitor plate is charge Q times the field made by the other plate: [math]\displaystyle{ E_{one plate}=\frac{(Q/A)}{2\epsilon_0 } }[/math]

Then the force on one plate is: [math]\displaystyle{ F=Q\frac{(Q/A)}{2\epsilon_0} }[/math]

Now, suppose that you pull the positive plate of the capacitor slowly away from the negative plate, exerting a force that is just barely larger than the one exerted by the other plate. This work that you do, moving the plate a distance Δs, increases the potential energy: [math]\displaystyle{ ΔU=W=Q\frac{(Q/A)}{2\epsilon_0}Δs }[/math]

For the sake of simplification, working out the math, we get this representation: [math]\displaystyle{ ΔU=\frac{1}{2}\epsilon_0{(\frac{Q/A}{\epsilon_0})^2}AΔs }[/math]

The expression in parenthesis that we are squaring is the same as the electric field inside the capacitor. Substituting, we get: Field Energy Density= [math]\displaystyle{ \frac{ΔU}{Δ(volume)}=\frac{1}{2}\epsilon_0E^2 }[/math]

The units of Field Energy Density are [math]\displaystyle{ J/m^3 }[/math].

Keep in mind the above equation is solved for the electric field from a capacitor. You can actually use anything with an electric field to derive this above equation.

Examples

Simple

Problem: What is the energy density of an electric field of magnitude 600V/m?

Solution: Simply, plug in the electric field into the equation: [math]\displaystyle{ \frac{1}{2}\epsilon_0E^2 }[/math] and you will get [math]\displaystyle{ 4.07*10^{16} J/m^3 }[/math]

Middling

Problem: What is the energy density a distance of 1 cm away on the axis of the positive end a dipole? (s=.1mm, q=4nC)

Solution:

Step 1: First, you must solve for the magnitude of the electric field 1 cm away from the dipole. Use the equation for the E field of a dipole on the axis: [math]\displaystyle{ E=\frac{1}{4\pi\epsilon_0}\frac{2qs}{r^3} }[/math] Plugging in, we get E=72 V/m.

Step 2: Then we plug the value for E into the energy density equation: [math]\displaystyle{ \frac{1}{2}\epsilon_0E^2 }[/math]. The answer is <math> 2.26\e-8