# Energy of homogenously charged sphere

1. Jan 15, 2012

### 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?

2. Jan 15, 2012

### 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.

Last edited: Jan 15, 2012
3. Jan 15, 2012

### 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

4. Jan 15, 2012

### 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.

5. Jan 15, 2012

### 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.

6. Jan 15, 2012

### Gavroy

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