Deriving the energy density of the Electric Field

[KvGroOve]

Homework Statement

Taken from Purcell Problem 1.33
Consider the electric field of two protons a distance b apart. The potential energy of the system ought to be given by

U=∫E²dv.

Let E₁ be the field of one particle alone and E₂ that of the other. Evaluate

ε₀∫E₁⋅E₂dv.

Set one of the protons at the origin and the other on the polar axis. Perform the integration over r before the integration over θ. Show that the integral has the value e²/4πε₀b.

2. Homework Equations

The Attempt at a Solution

I uploaded my attempt. I started by applying Coulomb's Law for Electric Fields to both protons. I broke down r₁ and r₂ into their Cartesian components to perform the dot product between the two. Since the two charges were both on the Polar Axis (I think I chose the right axis), I set φ₁=φ₂. I assumed that r₁=r and θ₁=θ. I eventually got an expression involving r₂² and Cos[θ₂-θ]. I was able to rewrite r₂² using the law of cosines. I have no idea how to address θ₂. I figured that I could use the law of sines to figure it out but the expression turns out to be really messy. I'm not really sure if I'm on the right track or if I've made an error anywhere.

Any advice regarding my mistakes and/or pointing me in the right direction would be much appreciated! Thanks.

 

[TSny]

Notice that your $\theta_2 - \theta_1$ is just the angle $\beta$ between the two position vectors. You could have gotten your final expression for $\mathbf r_2 \cdot \mathbf r_1$ more easily by thinking of the dot product in terms of the angle between the vectors.

Your idea of proceeding with the law of cosines and the law of sines sounds good to me.
$\cos \beta$ will simplify fairly nicely in terms of $\theta$, $r_1$ and $r_2$.

 

[NFuller]

Something seems funny to me. The total electric field is $\mathbf{E_{T}}=\mathbf{E_{1}}+\mathbf{E_{2}}$ so the energy density of the field is
$$u=\frac{\epsilon}{2}|\mathbf{E_{T}}|^{2}=\epsilon\left[\frac{1}{2}E_{1}^{2}+\frac{1}{2}E_{2}^{2}+\mathbf{E_{1}}\cdot\mathbf{E_{2}}\right]$$
not
$$u=\mathbf{E_{1}}\cdot\mathbf{E_{2}}$$
as shown in the problem. Are you actually finding the energy of the electric field or do you just need to solve the integral they gave you?

 

[TSny]

Are you actually finding the energy of the electric field or do you just need to solve the integral they gave you?

The E₁² and E₂² contributions are "self-energy" contributions which are infinite for point charges. A physicist just sweeps these under the rug . The dot product part is the energy of interaction.

 

[KvGroOve]

I finally got it. I don't think I did it elegantly because I had to use Mathematica to help me solve an integral. Regardless, thanks for your help, TSny!

 

[TSny]

OK. Good. To do the r integration, you can do a substitution letting u be the expression inside the ( ... )^3/2 in the denominator.

 

[KvGroOve]

Wow. That makes it such a trivial integral to solve - I can't help but to laugh. Thanks again!