Potential Energy of two protons

[ChowPuppy]

Homework Statement

Consider the electric field of two protons b meters apart.
The potential energy of the system is equal to:

[tex] U = \frac{\epsilon_0}{2}\int {\bf E}^2dv = \int({\bf E}_1+ {\bf E}_2)^2dv [/tex]
[tex] = \frac{\epsilon_0}{2}\int {\bf E}_1^2dv + \frac{\epsilon_0}{2}\int {\bf E}_2^2dv + \epsilon_0 \int{\bf E}_1\cdot {\bf E}_2dv [/tex]

The third integral is not hard to evaluate if you set it up in spherical coordinates with
one proton at the origin and the other along the polar axis (z axis) and perform the integration
over r first. Show that it integrates to
[tex] e^2/4\pi\epsilon_0b [/tex]
2 The attempt at a solution

I have been trying to solve this problem for hours, but cannot find an expression for E1 dot E2 in spherical coordinates in a way that would make this integral easy. Any guidance would be appreciated.

 

[tiny-tim]

Hi ChowPuppy!

If the protons are at O and P, and a typical point is at X,

then you have cosOPX/PX²,

which you can write as PX*cosOPX/PX³ …

then find a simple expression for PX*cosOPX, write it all in terms of r (= OX), and you should get a perfect integral

show us what you get ​

 

[ChowPuppy]

Hey tiny-tim,

Thanks so much for the help!
I was finally able to get it, I think you meant cosPXO instead of cosOPX possibly, but
basically I was able to write PX*cosPXO = [itex] r - bcos\phi [/itex] where [itex] \phi [/itex] is
the usual spherical coordinate phi, and then using the pythagorean theorem I rewrote PX and came up with
this integral:
[tex] \int \frac{r - bcos\phi}{((bsin\phi)^2 + (r-bcos\phi)^2)^{\frac{3}{2}}}dr d\phi d\theta [/tex]

And from there I was only a few substitutions away from my answer!

 

[tiny-tim]

Woohoo! ​

good puppy!

 

[paranormal]

I would first start by understanding the problem and the equations involved. The potential energy of a system is defined as the energy that is stored in the system due to its configuration or position. In this case, we are considering the potential energy of two protons, which are positively charged particles. The electric field is a vector field that describes the electric force experienced by a charged particle at any given point in space.

The equation provided to calculate the potential energy of the two protons involves integrating the electric field squared over the volume of the system. This equation can be broken down into three integrals, the first two represent the potential energy of each proton individually, and the third represents the interaction between the two protons. The interaction integral involves the dot product of the electric fields of the two protons.

To make the integration easier, we are asked to set up the integral in spherical coordinates with one proton at the origin and the other along the polar axis. This means that the electric field of the second proton will have only a radial component, while the electric field of the first proton will have both radial and angular components. By substituting these values into the dot product, we can simplify the integral and solve it in terms of the distance between the two protons, b.

In spherical coordinates, the electric field of a point charge can be expressed as:
E = (q/4πε₀r²) * (r̂ + r̂θ + r̂φ)

Where r is the distance from the origin, and r̂, r̂θ, and r̂φ are unit vectors in the radial, angular, and azimuthal directions, respectively. By substituting these values into the dot product, we get:
E₁ ⋅ E₂ = (q₁q₂/4πε₀b²) * (1 + cosθ)

Where θ is the angle between the two protons. By integrating this expression over the volume of the system, we get:
∫E₁ ⋅ E₂ dv = (q₁q₂/4πε₀) * ∫∫∫ (1 + cosθ) r²sinθ dr dθ dφ

Integrating over r first, we get:
∫E₁ ⋅ E₂ dv = (q₁q₂/4πε₀) * ∫∫ (1 + cosθ) sin