# Spherical Ball with uniform Charge Density

1. Nov 12, 2013

### Philosophaie

I want to find $\Phi$ and $\vec{E}$ for the general case of a Spherical Ball with uniform Charge Density centered at the origin radius d.

$\Phi = \frac{\rho}{4*\pi*\epsilon_0}\int\int\int\frac{r^2*sin\theta}{|r-r'|}dr d\theta d\phi$
$E = \frac{\rho}{4*\pi*\epsilon_0}\int\int\int\frac{r^2*sin\theta}{|r-r'|^2}*e_{r-r'}*dr d\theta d\phi$

For the General Case of Point {a, b, c}:

|r-r' | ^ 2 = (x-a) ^2 + (y-b) ^2 + (z-c) ^2
= r ^2 + a ^2 + b ^2 + c^2 -2r*$(a*cos\theta*cos\phi+b*sin\theta*cos\phi+c*sin\phi)$

Could someone check my work?

$\Phi = \frac{\rho}{4*\pi*\epsilon_0}$ $\int\int\int \frac{r^2*sin\theta}{\sqrt{r^2 + a^2 + b^2 + c^2 -2r(a*cos\theta*cos\phi+b*sin\theta*cos\phi+c*sin\phi)}}dr d\theta d\phi$

$E = \frac{\rho}{4*\pi*\epsilon_0}$ $\int\int\int\frac{r^2*sin\theta*\vec{r}}{( r^2 + a^2 + b^2 + c^2 -2r(a*cos\theta*cos\phi+b*sin\theta*cos\phi+c*sin\phi))^{3/2}} dr d\theta d\phi$

Those who wish to appear wise among fools, among the wise seem foolish.
- Quintilian

Last edited: Nov 12, 2013
2. Nov 12, 2013

### Astrum

I'm not really sure what you're doing, your work is very convoluted.

just use Gauss's Law.

$$\oint _S \mathbf E \cdot d\mathbf S = \frac{Q}{\epsilon _0}$$

You don't need to use the more complicated equations to solve this because we have symmetry and the charge density is uniform.

And it's not clear by what you mean with $\Phi$. I've seen that referring to either the electric potential and also the electric flux.

For electric flux, $$\int _S \mathbf E \cdot d \mathbf S = \Phi$$ and if you want the electric potnetial, it can be found by $\mathbf E = - \nabla \Phi$

3. Nov 12, 2013

### Philosophaie

Gauss Law because of symmetry becomes:

$E= \frac{\int_ V \rho dV}{\epsilon_0\int_S dS}$

$E = \frac{\rho*4/3*\pi*d^3}{4*\pi*r^2}$

How does this apply to the general case above?

Wisdom is not a product of schooling but of the lifelong attempt to acquire it.
- Albert Einstein

Last edited: Nov 12, 2013
4. Nov 12, 2013

### Astrum

$$E = \frac{Q}{4\pi r^2 \epsilon _0}$$

That's pretty general. It applies because that's the electric field for a sphere or point charge.