# Electromagnetism II - Potential and Electrical Field of a Spherical Shell

1. Jan 21, 2007

### Asrai

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

Consider a thin spherical shell having surface charge density σ, and radius a as shown in the diagram (see attachment).

Find, by integration over the sphere, the potential at a point P, a distance r fromthe centre of the sphere, for r > a. Using this result also find the electric filed at the same point. Hence verify that for points outside the sphere the charge acts as if it were all concentrated at the centre of the shell.

NB: Do not use Gauss' Law in answering this question!

(Hint: consider the spherical shell to be built up from infinitesimal rings of charge, centred on their axis of symmetry through OP):

2. Relevant equations

n/a

3. The attempt at a solution

I thought that calculating the electric field first might be easier, as we've done roughly the same problem with only one ring of charge; this gave the result E = (Q*r)/((4*Pi*Epsilon)*(a^2+r^2)^1.5).

What I've done so far for the above problem is to set the surface charge density as Q/4*Pi*a^2.

The electric field as I've written it down so far is:

dE = (2*σ*dS*cos(theta))/(4*Pi*Epsilon*(a^2+r^2)).

I've simplified this equation quite a bit, writing it as a surface integral, pulling the constants out of the integral, etc. until I ended up with this:

E = Q/(8*Pi^2*Epsilon)*integral((r*dS)/(a^2*(a^2+r^2)^0.5)).

Now, what I want to know is whether I'm on the right path at all, or if I've missed out something; also, if this is correct, how do I set up the surface integral? There's two parts to it, and one is obviously the rings and their radius going from 0 to a, but what's the other bit?

Any help on this would be very much appreciated.

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2. Jan 21, 2007

### Kurdt

Staff Emeritus
Why are you starting with electric field? For a general surface charge distribution the potential is given by:

$$V = \frac{1}{4\pi \epsilon_0} \int_S \frac{\sigma dS}{R}$$

The electric field is then the negative gradient of the potential.

3. Jan 22, 2007

### Asrai

Okay, I suppose I can do it this way too. However, this does not solve my problem in setting up dS properly; I'm a bit stuck there. One bit should be the changing radius of all the circles going from 0 to a, but what's the other one?

4. Jan 22, 2007

### Kurdt

Staff Emeritus
Well its fairly easy to work out. The radius of the rings changes with $$d\phi$$ by $$a sin(\phi) d\phi$$ and the thickness of the rings is $$ad\theta$$

Then you need to work out the limits.

5. Jan 22, 2007

### Asrai

For $$a sin(\phi) d\phi$$ I set the limits as 0 to Pi; for $$ad\theta$$ I set the limits as 0 to 2*Pi. Integrating this and calculating the potential gives me $$V = Q/(8 \pi \epsilon r)$$.

Is this correct?

Last edited: Jan 22, 2007
6. Jan 22, 2007

### Kurdt

Staff Emeritus
I think you made a slight error it should be 4 pi on the bottom.

7. Jan 22, 2007

### Asrai

Ah, I caught the error, thank you. The electric field should then be E = Q/(4*Pi*Epsilon*r^2), right?

8. Jan 22, 2007

### Kurdt

Staff Emeritus
Yes which is exactly the same as for a point charge at the centre of the sphere with the same total charge as the sphere.

EDIT: Just to add, make sure you understand how dS is composed the way it is.

Last edited: Jan 22, 2007
9. Jan 22, 2007

### Asrai

Thank you ever so much, that was really helpful!