Interaction energy of two interpenetrating spheres of uniform charge density

[Johe]

Homework Statement

Find the interaction energy of two interpenetrating spheres of uniform charge density $$\rho_{1}$$ and $$\rho_{2}$$ Let the two spheres have equal radii $a$ and let the separation.

Relevant Equations

$$
\Phi(x)=\int \rho\left(x^{\prime}\right) \frac{1}{\left|x-x^{\prime}\right|} d \tau^{\prime}
$$

I am trying to calculate the interaction energy of two interpenetrating spheres of uniform charge density. Here is my work:
First I want to calculate the electric potential of one sphere as following;
$$\Phi(\mathbf{r})=\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho\left(\mathbf{r}^{}\right)}{\bf{r^{\prime}}} d \tau$$
Doing that I got
$$\Phi(r)= \frac{\rho}{2 \epsilon_{0}}\left(R^{2}-\frac{1}{3} r^{2}\right)$$
Now I need to solve the following;
$$\int_{V} \rho_{2}(\vec{r}) \Phi_{1}(\vec{r}) \mathrm{d}^{3} r$$ But I did not know how to do it.

I general,

$$U_{i n t}=\frac{}{} \int_{V} \rho_{2}(\vec{r}) \Phi_{1}(\vec{r})+\rho_{1}(\vec{r}) \Phi_{2}(\vec{r}) \mathrm{d}^{3} r=\int_{V} \rho_{1}(\vec{r}) \Phi_{2}(\vec{r}) \mathrm{d}^{3} r=\int_{V} \rho_{2}(\vec{r}) \Phi_{1}(\vec{r}) \mathrm{d}^{3} r$$
which tells that the total work done by one sphere on the other sphere as you bring the isolated objects from super far away to their current positions. Since the forces were equal and opposite, the work done by one on the other is equal to the work done by the other on the one.
Please help me. I attached the problem statement.

 

[anuttarasammyak]

As for a uniformly charged sphere of radius a, its potential $\phi$ is
for distance r>a
$$\phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{r}$$
and for r<a,
$$\phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{a}$$
,constant, where $Q=4\pi a^2 \rho$ is total charge. This fact would be useful in your calculation.

 

[Johe]

As for a uniformly charged sphere of radius a, its potential $\phi$ is
for distance r>a
$$\phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{r}$$
and for r<a,
$$\phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{a}$$
,constant, where $Q=4\pi a^2 \rho$ is total charge. This fact would be useful in your calculation.

This for a one sphere, not in case of overlapping. Which is not trivial.

 

[anuttarasammyak]

Another sphere interpenetrates it. For inside part constant potential is applied. For outside part potential of a function of the distance from the original sphere center is applied.

 

[Johe]

Another sphere interpenetrates it. For inside part constant potential is applied. For outside part potential of a function of the distance from the original sphere center is applied.

Would you sketch what are you think and placed your vectors? and position vector?

 

[anuttarasammyak]

As for calculation of outside part
distance from the original sphere center to ring part of penetrating sphere of area
$$2\pi a \sin \theta d\theta$$
is
$$r=\sqrt{d^2+a^2+2ad\cos\theta}$$
where $\theta$ is angle measured from the center of intepenetrating spere in polar coordinate whose z axis is shared by the spheres. Integration should be done for the interval $\theta[0,\theta_0]$ where
$\cos \theta_0=-\frac{d}{2a}$

As for inside part, the interval $\theta[\theta_0,\pi]$ is used for integration of interpenetrating sphere charge inside to be multiplied with constant potential.

 

[Johe]

As for calculation of outside part
distance from the original sphere center to ring part of penetrating sphere of area
$$2\pi a \sin \theta d\theta$$
is
$$r=\sqrt{d^2+a^2+2ad\cos\theta}$$
where $\theta$ is angle measured from the center of intepenetrating spere in polar coordinate whose z axis is shared by the spheres. Integration should be done for the interval $\theta[0,\theta_0]$ where
$\cos \theta_0=-\frac{d}{2a}$

As for inside part, the interval $\theta[\theta_0,\pi]$ is used for integration of interpenetrating sphere charge inside to be multiplied with constant potential.

Thanks! If you could sketch that, I really appreciate it? moreover, would you willing to provide half of the solution?

 

[anuttarasammyak]

Drawing a sketch by yourself is a good practice. Have a good try!

 

[haruspex]

For inside part constant potential is applied.

I assumed these are solid spheres, not spherical shells.

 

[anuttarasammyak]

Oh, I took it wrong.

 

[haruspex]

Thanks! If you could sketch that, I really appreciate it? moreover, would you willing to provide half of the solution?

Draw the diagram showing two equal overlapping circles, centres A and B.
(According to the question, each centre lies inside the other circle, though I suspect a mistake here. I think they meant d<2a, not d<a. Correspondingly, the "extreme case" where the value is well known should be at d=2a.)
Now draw a thin spherical cap, thickness dr, radius r<a centred on A and lying inside the other sphere (extending to its surface).

What is the potential at this cap due to the charge on sphere A? This will require integration, but it's easy.
What is the volume of this cap? An easy way to get this is to use the way Archimedes found the surface area of a sphere without quite inventing calculus.
What is the potential energy of the cap in respect of the charges on it and on the sphere A?
Integrate to get the total for r<a, then follow the same procedure for r>a.

 

[minimikanix]

You could use
$$
\begin{equation*}
\epsilon_0 \int\limits_{\text{all space}} \mathbf E_1 \cdot \mathbf E_2 =
\frac{1}{2} \int\limits_{\text{all space}}
\left(
\rho_1 \, V_2 + \rho_2 \, V_1
\right) \, d\tau.
\end{equation*}
$$