Electric Field due to infinite charge distribution seems a paradox

1. Apr 21, 2012

sazmat

Suppose everywhere in space charge is distributed with uniform and constant volume charge density. What will be Electric field at any point in space??
1>..Symmetry demands it to be zero,
2>..if I consider the space to be a sphere of infinite radius with constant charge density on its volume then using the formula of field inside a uniformly charged sphere of finite radius I get E=(p*r)/(3*Eo)
where p=charge density
r= distance from center of sphere
Eo=8.82*10^-12 (permittivity of free space)
3>..if I consider the space to be a cylinder of infinite radius and infinite length with constant charge density on its volume then using the formula of field inside a uniformly charged cylinder of finite radius and infinite length I get E=(p*r)/(2*Eo)
where p=charge density
r= distance from the axis of cylinder
Eo=8.82*10^-12 (permittivity of free space)

Different approaches give different answers. Why is that so? and Whats the correct answer?

2. Apr 21, 2012

shriomtiwari

i don't have the answer to your question but i do have a thought
what is the shape of the universe?

3. Apr 21, 2012

Dickfore

Gauss's Law cannot be applied because you identified a preferred point (the center of the Gaussian spheres).

The answer is that field diverges everywhere in space.

4. Apr 21, 2012

sazmat

what do you mean by diverging of field? Would you explain further?

5. Apr 21, 2012

Staff: Mentor

This is not a charge configuration you can have in real life.
If you have some finite universe, it can work, and if it is symmetric (enough) the field is 0.
If you have an infinite universe, where do the charges come from?

The problem with your calculation is that you sum over something which is not absolute convergent, that means that the order of your summation (an integral is like a sum) does matter. In other words: Maxwell's equations are not applicable here.

6. Apr 21, 2012

Dickfore

I was on my mobile device, so I didn't have time to type it.

If you don't have boundary conditions, then the Green's function for the problem is:
$$-\nabla^2_x \, G(\mathbf{x}, \mathbf{x}') = \delta(\mathbf{x} - \mathbf{x}')$$
$$G(\mathbf{x} - \mathbf{x}') = \frac{1}{4 \pi} \, \frac{1}{\vert \mathbf{x} - \mathbf{x}' \vert}$$

Then, the potential due to a charge distribution $\rho(\mathbf{x})$ is given as:
$$\Phi(\mathbf{x}) = \frac{1}{4 \pi \epsilon_0} \, \int{ \frac{\rho(\mathbf{x}')}{\vert \mathbf{x} - \mathbf{x}' \vert} \, d\mathbf{x}'}$$

For a uniform charge distribution $\rho(\mathbf{x}') = \rho$, you may take the charge density out of the volume integral:
$$\Phi(\mathbf{x}) = \frac{\rho}{4 \pi \epsilon_0} \, \int{ \frac{1}{\vert \mathbf{x} - \mathbf{x}' \vert} \, d\mathbf{x}'}$$

Then, you may be tempted to make the substitution $\mathbf{x}' \rightarrow \mathbf{x}' + \mathbf{x}$, so that the integrand no longer depends parametrically on $\mathbf{x}$! A constant electric potential would give a zero gradient, i.e. no electric field.

However, the integral in spherical coordinates is:
$$\int{ \frac{1}{\vert \mathbf{x}' \vert} \, d\mathbf{x}'} = \int_{0}^{2\pi}{\int_{0}^{\pi}{\int_{0}^{\infty}{d\phi \, d\theta \, dr \, r \, \sin \theta}}}$$