# Question about spherically symmetric charged objects

1. Sep 9, 2009

### simpleton

Hi,

I would like to ask a question about spherically symmetric charged objects. My teacher told me that you can treat spherically symmetric charged objects at point charges. However, my teacher did not prove it. I guess you have to integrate every small volume on the spherically symmetric charge and find its contribution to the force, but I am not sure how to do that.

Therefore, does anyone know how to prove that, given a spherically symmetric object of radius R and charge density rho, if I place a test charge q x metres away from the centre of the charge, the force experienced by the test charge is k*(4/3*pi*R^2*rho)*q/R^2, where k is 1/(4*pi*epilson-nought).

2. Sep 10, 2009

### gabbagabbahey

Assuming you are only interested in points outside the sphere, then your teacher is correct.

To show it, just use Gauss' Law....can you think of a Gaussian surface that will allow you to pull |E| outside of the integral?

3. Sep 10, 2009

### simpleton

Oh right!

I think you mean integral(E dA) = Q/epilson-nought.

If my imaginary surface is a sphere, then I can use symmetry arguments to say that E is constant, so I can pull E out.

E*integral (dA) = Q/epilson-nought

If I take the imaginary sphere to have a radius x and has its centre at its origin, then the area of the imaginary surface is 4*pi*x^2. Thus:

E*4*pi*x^2 = Q/epilson-nought

E = Q/(4*pi*epilson-nought*x^2) = k*Q/(x^2)

So the force on a test charge q' is k*q*q'/(x^2).

Thanks a lot! :)

4. Sep 10, 2009

### Bob S

Is your spherically symmetric object a dielectric or a conductor? Is the charge density a surface charge density or volume charge density?

5. Sep 10, 2009

### simpleton

If I understand correctly, dielectric is an insulator, and yes, the object I am talking about is an insulator, because if it is a conductor, all the charges will accumulate on the surface.

EDIT: Can I know whether I should post such questions on this forum or on the homework forum? I have many more such questions

Last edited: Sep 10, 2009