Potential of a Spherical Shell

Tags:
1. May 16, 2015

Alettix

Hi! I have trouble with solving this problem and would be really thankful for some help. :)

1. The problem statement, all variables and given/known data

Inside a thin, spherical metal-shell with a radius of 50 cm, a smaller homogenous metal-sphere with a radius of 20 cm is placed concentrically. The smal sphere is grounded through a very long cable going through a small hole in the spherical shell. The shell is then given a charge of 2 * 10-8 C. Find the potential of the shell.

2. Relevant equations

Electric potential: V = E/Q

Gauss law: The integral of the electric fiel over a surface equals the elcosed charge divided with the electric constant.

Coloumbs law: F = k * Q1 * Q2 / r2

3. The attempt at a solution

The smal metal-sphere is grounded, thus it has zero potential. In order to calculate the electric potential of the outer shell we need to know how much work it takes to move a test charge from the smal sphere to the shell. My plan was to use Gauss law to find the electric field inbetween the sphere and the shell as a function of r, and then integrate it over dr, from r = 0,2m to r = 0,5. However, as there is no enclosed net charge inside the spherical shell, Gauss law yields zero electric field. If I'm not wrong, this means that no work is needed to move a testcharge from the sphere to the shell, and thus the potential of the shell should be zero as well. But my teacher says that this is not the right answer.

My other idea was to use Coloumbs law to find the net force on a test charge as a function of r, and then integrate it from r = 0,2m t or = 0,5m. However, finding an expression for the net force exerted by all the charges along the shell was harder then I thought, so I have not yet been able to make it.

Do you think I should continue searching for an expression for the net force or is there a more simple way to solve this problem? Is there any special meaning in emphasizing that the grounding-cable is very long (I have not been able to use this information yet)?

2. May 16, 2015

Orodruin

Staff Emeritus
Are you familiar with solving Poisson's equation with different geometries?

3. May 16, 2015

BiGyElLoWhAt

A couple of things,
1) Your potential equation is wrong. $V\neq \frac{\vec{E}}{Q}$ ; $V=\int_a^b \vec{E}\cdot d\vec{l}$
2) The purpose of the information (long cable) is to show you that it's grounded at infinity, or at least that's what you should assume.

Hint: With number 2, since the inner sphere is at the same potential as at infinity, then the amount of work to move a charge from the outer shell to the inner shell should be the same as the amount of work to move a charge from the outer shell to infinity.

4. May 16, 2015

Alettix

No, sorry I'm not familiar with Poisson's equation at all.

5. May 16, 2015

Alettix

With E I didn't mean the electrical field, but the electrical potential energy, in some countries denoted by U. Thus: V = U/Q
Sorry for the confusion.

So does this mean that I can use Gauss law to find the field outside the shell and then integrate the work from r = 0,5m to r = ∞ ? :)

6. May 16, 2015

Orodruin

Staff Emeritus
I then suggest assuming an induced charge on the inner sphere. Compute the resulting potential at the inner sphere and fix the induced charge which gives zero potential. Once that is done, you can easily find the potential on the outer sphere.

7. May 16, 2015

Alettix

Thank you, but I'm unsure if I understand your proposed method. With induced charge, do you mean that:
1. the charges in the inner sphere redistribute so we get more negative ones at the surface and more positive on the inside or
2. we get an actual net charge on the inner sphere (equal in magnitude but opposite in sign to the charge on the outer shell) ?

The second option would, if I'm not wrong, result in an electric field between the sphere and the shell, but in zero field outside them. Is it this field I should use in my integral to find the potential or have I completely missunderstod you?

8. May 16, 2015

Alettix

I just tried calculating this but it gives a wrong (really too big) answer.

9. May 16, 2015

Orodruin

Staff Emeritus
We do get a net charge on the inner sphere (it comes from being held at grounded potential). It is generally not equal to the charge on the outer shell. It is going to be the net charge which gives you the correct potential, ie, zero, on the inner sphere.

Remember that charge can come to or leave the inner sphere by the connection to the ground potential.

10. May 16, 2015

Alettix

Hm, I'm sorry but I just realized I don't know how to calculate the potential of the inner sphere, or how it is affected if it is charged. I just thought it was a convection that everything grounded has zero potential.

11. May 16, 2015

Orodruin

Staff Emeritus
Yes, it has zero potential, but charges must be arranged in such a way that it gets that zero potential.

What would the potential be if it had induced charge Q? (Do not care about setting the potential to zero here, doing so will let you fix Q later)

12. May 17, 2015

Alettix

This is what I don't know how to calculate... and I start to feel like I lack some basic knowledge to solve this problem.
I could use Gauss law to find the electric field caused by the induced charge Q. Then I would need to integrate it to find the potential, but integrate it over what? To ground level? We don't know how high up the spheres are.

13. May 17, 2015

Orodruin

Staff Emeritus
What is the electric field of a spherical charge distribution outside the sphere? What is the corresponding potential?

You are right, you could simply use Gauss's law to find the field. The potential reference point is taken to be at infinity.

14. May 19, 2015

Alettix

Gauss's law gives E = k* Q/r2 for the electric field outside a spherical distribution of charge Q.
The potential is the integral over dr: V = k*Q [-2/r]ab

So, could I maybe divide up the integral in two parts? One of the outer field from (r = 0,5m to r = ∞) and another for the inner one (from r = 0,2m to r = 0,5m). For the potential to be zero for the inner sphere the sum of these two integrals should equal zero (which should be possible if the direction of the two fields aren't the same) . From this, I could find the value of the induced charge Q and then use the inner field to calculate the potential of the outer sphere? :)

Last edited: May 19, 2015
15. May 19, 2015

Orodruin

Staff Emeritus
I am not sure what you mean by this. It does not have anything to do with the direction of the fields and everything to do with the fact that different contributions to the potential add linearly. The reason you only integrate the outer shell to 0.5 m is that the field due to the outer surface charge is zero inside of the surface.

16. May 19, 2015

Alettix

Sorry for the confusion. With different direction of the inner and outer fields, I meant that if we integrate from r = 0,2m to r = ∞, to get the potential of the inner sphere, it will be possible for it to be 0 because they add linearly. I think we are meaning the same thing.

I just tested my proposed method and it gave the right answer. Thank you so much for helping me understand and solve this problem! :)