Electric Field due to infinite charge distribution seems a paradox

In summary: Thus, in summary, the electric field at any point in space is zero due to the symmetry and boundary conditions of the charge distribution. Different approaches and calculations may give different answers, but the correct answer is that the field diverges everywhere in space, making it impossible to accurately calculate or define.
  • #1
sazmat
4
0
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?
 
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  • #2
i don't have the answer to your question but i do have a thought
what is the shape of the universe?
 
  • #3
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
what do you mean by diverging of field? Would you explain further?
 
  • #5
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
sazmat said:
what do you mean by diverging of field? Would you explain further?

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:
[tex]
-\nabla^2_x \, G(\mathbf{x}, \mathbf{x}') = \delta(\mathbf{x} - \mathbf{x}')
[/tex]
[tex]
G(\mathbf{x} - \mathbf{x}') = \frac{1}{4 \pi} \, \frac{1}{\vert \mathbf{x} - \mathbf{x}' \vert}
[/tex]

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

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

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

However, the integral in spherical coordinates is:
[tex]
\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}}}
[/tex]
As you can see, the radial integral diverges quadratically.

Thus, the above substitution is illegitimate, and the electric potential is undefined.

If you wanted to impose an upper cutoff in the radial integral, then that would delimit a ball of charge with a large radius R. Then, the field rises linearly with distance and is radially distributed.
 
  • #7
sazmat said:
Suppose everywhere in space charge is distributed with uniform and constant volume charge density. What will be Electric field at any point in space??
Consider a closed conducting sphere enclosing a volume with mobile space charge (like in a gas or plasma). There can not be an electric field within the sphere conductor, so the volume charge density inside the sphere is matched by an equal and opposite surface charge density on the inside wall of the sphere. Motion of charges in this volume will quickly neutralize on the sphere wall. So the "everywhere" space charge would be neutralized by enclosed conducting surfaces.
 

1. What is an infinite charge distribution?

An infinite charge distribution is a theoretical concept used in physics to model a system of charges that extends infinitely in all directions. This allows for simpler mathematical calculations and analysis.

2. How does an infinite charge distribution result in a paradox for electric fields?

The paradox arises because in an infinite charge distribution, the electric field strength at any given point would be infinite due to the infinite number of charges. This contradicts the laws of physics, which state that the strength of an electric field should decrease with distance.

3. Why is the electric field strength infinite in an infinite charge distribution?

This is because the electric field is directly proportional to the number of charges and inversely proportional to the square of the distance. In an infinite charge distribution, the distance between any two charges is effectively zero, resulting in an infinite electric field strength.

4. How is the paradox of infinite electric field strength resolved in real-life situations?

In real-life situations, an infinite charge distribution does not exist. Instead, there is a finite number of charges with a finite amount of space between them, allowing for the electric field strength to decrease as distance increases, in accordance with the laws of physics.

5. Can the paradox of infinite electric field strength be applied to other physical phenomena?

Yes, similar paradoxes can arise in other areas of physics, such as the concept of black holes in astronomy. However, these paradoxes are usually resolved through further research and the development of new theories and models.

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