Interaction energy of two interpenetrating spheres of uniform charge density

In summary, according to the question, the potential energy of a sphere of uniform charge at a distance r from another sphere is given by:$$U_{i n t}=\frac{}{} \int_{V} \rho_{2}(\vec{r}) \Phi_{1}(\vec{r})+\rho_{1}(\vec{r}) \Phi_{2}(\vec{r}) \mathrm{d}^{3} r=\int_{V} \rho_{1}(\vec{r}) \Phi_{2}(\vec{r}) \mathrm{d}^{3} r$$
  • #1
Johe
4
0
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.
 

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  • #2
As for a uniformly charged sphere of radius a, its potential ##\phi## is
for distance r>a
[tex] \phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{r}[/tex]
and for r<a,
[tex] \phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{a}[/tex]
,constant, where ##Q=4\pi a^2 \rho## is total charge. This fact would be useful in your calculation.
 
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  • #3
anuttarasammyak said:
As for a uniformly charged sphere of radius a, its potential ##\phi## is
for distance r>a
[tex] \phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{r}[/tex]
and for r<a,
[tex] \phi(r)=\frac{Q}{4\pi\epsilon}\frac{1}{a}[/tex]
,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.
 
  • #4
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.
 
  • #5
anuttarasammyak said:
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?
 
  • #6
As for calculation of outside part
distance from the original sphere center to ring part of penetrating sphere of area
[tex]2\pi a \sin \theta d\theta[/tex]
is
[tex]r=\sqrt{d^2+a^2+2ad\cos\theta}[/tex]
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.
 
  • #7
anuttarasammyak said:
As for calculation of outside part
distance from the original sphere center to ring part of penetrating sphere of area
[tex]2\pi a \sin \theta d\theta[/tex]
is
[tex]r=\sqrt{d^2+a^2+2ad\cos\theta}[/tex]
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?
 
  • #8
Drawing a sketch by yourself is a good practice. Have a good try!
 
  • #9
anuttarasammyak said:
For inside part constant potential is applied.
I assumed these are solid spheres, not spherical shells.
 
  • #10
Oh, I took it wrong.
 
  • #11
Johe said:
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.
 
  • #12
You could use
$$
\begin{equation*}
\epsilon_0 \int\limits_{\text{all space}} \mathbf E_1 \cdot \mathbf E_2 =
\frac{1}{2} \int\limits_{\text{all space}}
\left(
\rho_1 \, V_2 + \rho_2 \, V_1
\right) \, d\tau.
\end{equation*}
$$
 
Last edited:
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1. What is the concept of interaction energy?

The interaction energy of two interpenetrating spheres of uniform charge density refers to the potential energy associated with the interaction between the two spheres due to their electric charges.

2. How is the interaction energy calculated?

The interaction energy can be calculated using the Coulomb's law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. What is the significance of uniform charge density in this concept?

The uniform charge density means that the charge is evenly distributed throughout the surface of the sphere. This simplifies the calculation of the interaction energy as the charge is considered to be concentrated at the center of the sphere.

4. How does the distance between the two spheres affect the interaction energy?

The interaction energy is inversely proportional to the square of the distance between the two spheres. This means that as the distance increases, the interaction energy decreases and vice versa.

5. What are some real-world applications of this concept?

The concept of interaction energy between two interpenetrating spheres of uniform charge density is used in various fields such as physics, chemistry, and materials science. It is particularly important in understanding the behavior of charged particles in biological systems, electrostatic interactions in polymers, and the formation of ionic bonds in chemical compounds.

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