Homework Help: Electric field calculation using Coulomb's law for uniformly charged conducting spher

1. Aug 27, 2006

bobca117

Hi,

When we calculate electric field due to a charged spherical conductor at a point outside the conductor, by Gauss's law, it is equal to the electric field due to a point charge at the center of sphere, with net charge on the sphere. We can also calculate this electric field strength using coulombs law at the same point and we would have to consider the hemispherical charges which are hidden or not seen directly at that point. That is we have to perform integration over the entire spherical surface with uniform charge density. How if the electric field strength at inside the sphere is zero, then at the point where we calculate E, we can still have the effect of the hidden charges through sphere?

Bob

2. Aug 27, 2006

rbj

the clean way to do this is to use Gauss's law (which works for any inverse-square field such as gravity, not just E&M) and spherical symmetry, but if you were to do this with just Coulomb's law, you would have to integrate the resulting field vector over the entire spherical surface (there is no hidden part of the sphere).

but Gauss's law has already dealt with this. and the result outside the sphere is as you say (indistinguishable from a point charge) and inside the sphere, there is no resulting field. you would get the same result by painfully integrating the entire sphere surface.

3. Aug 28, 2006

Meir Achuz

The integral is straightforward, bujt a bit complicated.
It gives the Gauss result.
It is easiest to integrate over a sequence of uniformly charged rings.

4. Sep 24, 2007

Zakyr

And what is the integral? I need to solve this using Coulomb's law. I don't think I'm getting the right integral.

5. Sep 24, 2007

Staff: Mentor

Show us the integral that you got, and someone can probably tell you where your error is.

6. Sep 24, 2007

Zakyr

Alright,

if you look at the attached file, this is the integral I've got at the end.

Please note that this is for a sphere that is only charged at its surface.

Can someone tell me if I got it right? And if so, how do we solve this integral?

Attached Files:

• Integrale.JPG
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Last edited: Sep 24, 2007
7. Sep 24, 2007

Loren Booda

I believe that one integrates over a vector that is the sum of the vector from observer to the center of the sphere, and the vector that spans in spherical coordinates from that center over the sphere's surface charge.

8. Sep 25, 2007

Zakyr

Hmm? :S

9. Sep 25, 2007

ZapperZ

Staff Emeritus
Please note that this type of question belongs in the HW/Coursework forum, and that is where this thread has been moved to.

Zz.