Energy of homogenously charged sphere

[Gavroy]

hi

i know that the energy of a conducting charged sphere is E=Q²/(8epsilon_0*pi*r) and i know how to calculate this.

now i found that a homogenously charged sphere has E=3/5*Q^2/(epsilon_0*pi*r).

does anybody have an idea how to calculate this? i don't know where this 3/5 come from?

 

[vanhees71]

What you have there is the energy in the electrostatic field of the sphere. This one you need first. You get it from the corresponding Maxwell equations for static fields,

$$\vec{\nabla} \cdot \vec{E}=\frac{\rho}{\epsilon_0}, \quad \vec{\nabla} \times \vec{E}=0.$$

[The above is corrected for typos according to #3; thanks for pointing them out!]

From the second equation you see that the electrostatic field is a potential field, i.e.,

$$\vec{E}=-\vec \nabla \Phi.$$

Substituting this into the 2nd equation yields\

$$\Delta \Phi=-\frac{\rho}{\epsilon_0}.$$

Now you have

$$\rho(\vec{x})=\begin{cases} \rho_0=\frac{3Q}{4 \pi R^3} & \text{for} \quad r=|\vec{x}<R|\\ 0 & \text{for} \quad r \geq R. \end{cases}.$$

Due to symmetry, $\Phi$ depends only on $r$ and thus writing the Laplacian in spherical coordinates for $r<R$ leads to

$$\frac{1}{r} \frac{\mathrm{d}}{\mathrm{d} r^2} [r \Phi(r)]=-\rho_0.$$

From this you get

$$\epsilon_0 [r \Phi(r)]''=-\rho_0 r \; \Rightarrow \; \epsilon_0 [r \Phi(r)]'=-\frac{\rho_0}{2} r^2+C_1 \; \Rightarrow \; \Phi(r)=-\frac{\rho_0}{6 \epsilon_0} r^2+C_1+\frac{C_2}{r}.$$

Since for $r=0$ the potential should be regular, we must have $C_2=0$.

For $r>R$ the same calculation leads to

$$\Phi(r)=\frac{C_3}{r},$$

where we have determined one constant such that $\Phi(r) \rightarrow 0$ for $r \rightarrow \infty$. The constant, $C_3$, is determined by Gauss's Law to be $C_3=Q/(4 \pi \epsilon_0)$.

Continuity at $r=R$ leads to

$$-\frac{Q}{8 \pi \epsilon_0 R}+C_2=\frac{Q}{4 \pi \epsilon_0 R}$$

i.e.

$$C_2=\frac{3 Q}{8 \pi \epsilon_0 R}.$$

For the total energy of the electric field you find

$$\mathcal{E}=\frac{\epsilon_0}{2} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \vec{E}^2(\vec{x})=\frac{1}{2} \int_{\mathbb{R^3}} \mathrm{d}^3 \vec{x} \rho(\vec{x}) \Phi(\vec{x}).$$

This gives from our above result

$$\mathcal{E}=2 \pi \rho_0 \int_0^R \mathrm{d} r \Phi(r) \; r^2 =\frac{3Q^2}{20 \pi \epsilon_0 r}.$$

In your solution, obviously a factor of $4 \pi$ is missing.

 

[Simon Bridge]

Um ... I always thought:
$$\vec{\nabla}\cdot\vec{E} = \frac{\rho}{\epsilon_0}$$$$\nabla^2 \varphi = -\frac{\rho}{\epsilon_0}$$... for the potential formulation.

The approach is basically correct though.
Compare:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesph.html

 

[vanhees71]

Sure, that were typos in the 1st and 3rd equation (I hate the SI because of this damn $\epsilon_0$...). The rest is fine. I'll correct it also in the original posting.

 

[Simon Bridge]

No worries - there is a thread in PF someplace about the why's and why-nots of that $\epsilon_0$ but if you trained in SI the pain is the other way around.

At first I thought maybe the typo would affect the calculation but I was the epsilon vanished at the spherical coordinates but was recovered in the correct place on the next line. I was concerned this may puzzle someone.

I think we need to hear from Gavroy about the source of his equation.

 

[Gavroy]

you are totally correct, i accidentally lost 4 in the denominator. thank you for your help!