# Electric Potential with a hollow insulated sphere

1. Oct 8, 2009

### Brian-san

1. The problem statement, all variables and given/known data
Consider a hollow metallic sphere of finite thickness, with the inner radius a and the outer radius b. A point charge q is placed inside the sphere at a distance a/2 from the center of the sphere (the sphere is insulated).
a) What is the potential at a point r outside the sphere (r > b)?
b) What are the potentials at the inner (r = a) as well as the outer (r = b) surfaces of the sphere?
c) What is the potential at the center of the sphere? (Use the method of images to calculate this result.)

2. Relevant equations
$$\Phi=\frac{q}{r}$$
Law of cosines.

3. The attempt at a solution
For part a I was thinking that $$\Phi=\frac{q}{r}$$, behaving as if the charge were a point charge concentrated at the origin. I'm not sure if this is right as I can't justify to myself why it would be the case. We've only been discussing properties of conductors in class and said very little about insulators.

In b, I assume the center of the sphere is at the origin, and d is the distance from q to a point on the inner surface at radius a. By the law of cosines,
$$d^2=a^2+\frac{a^2}{4}-a^2cos\theta, d=a\sqrt{\frac{5}{4}-cos\theta}$$

so the potential on the surface at radius a is given by
$$\Phi=\frac{q}{a\sqrt{\frac{5}{4}-cos\theta}}$$

Logically, I used the same process for the outer surface b and got
$$d^2=b^2+\frac{a^2}{4}-abcos\theta$$

so the potential is just
$$\Phi=\frac{q}{\sqrt{b^2+\frac{a^2}{4}-abcos\theta}}$$

I then generalized the expression for any r inside the sphere to
$$\Phi=\frac{q}{\sqrt{r^2+\frac{a^2}{4}-arcos\theta}}$$

Using this, the potential at the origin should be simply $$\Phi=\frac{2q}{a}$$ and thus there is no need for the method of images.

I just have this feeling that at least some of the above answers are wrong, most likely because the sphere is insulated and I have not properly utilized that fact in the calculations. That's what has been throwing me off about this problem.

2. Oct 8, 2009

### kuruman

Part (a), you forgot to multiply by 1/4πε0.
Part (b), you forgot that conductors are equipotentials - "metallic" means "conducting".

** Edit **
I interpret that "the sphere is insulated" to mean that the sphere is not grounded or otherwise connected to a battery.

Last edited: Oct 8, 2009
3. Oct 8, 2009

### Brian-san

We're using the CGS units so those constants are being dropped (Coulomb's constant =1 in CGS), which was the complete opposite of what I was used to.

So, the point charge causes the charge on the sphere to be rearranged such that the electric field in the region a<r<b is zero (-q is the induced charge on the inner surface and q is induced on the outer surface). Thus the potential is constant here, which leads to Φ(a)=Φ(b), so only one of the expressions I gave was needed, my best guess would be the one involving q/a. By this same reasoning, the rearranging of charges, does the net charge of the sphere appear to be q when outside the sphere? This is the only way my answer for part a makes sense.

Knowing the potential at a and b (and consequently for the entire region between), that is where the method of images would seem to come into play.

4. Oct 9, 2009

### gabbagabbahey

If the induced charge on the outer surface of the was uniformly distributed, then sure, the potential outside would be $\Phi=\frac{q}{r}$.

However, I don't see any reason why that would be the case. I'd either use Separation of Variables or the method of images to find what the potential really is.

5. Oct 9, 2009

### kuruman

I believe the method of images should be used to find the potential inside the shell (r<a). I think it is correct to say that outside (r > b) the potential is q/r. This is my reasoning
1. The function q/r is a solution to Laplace's equation.
2. This potential reduces on the outer surface to constant = q/b.
3. This potential is zero at infinity.

So q/r is a solution to Laplace's equation in the region r > b and satisfies the boundary conditions. Therefore, by the uniqueness theorem, it is the solution. Moving the charge around inside the cavity affects the charge distribution on the inner surface only.

6. Oct 9, 2009

### gabbagabbahey

You're correct as usual kuruman!

7. Oct 9, 2009

### Brian-san

I set up some relations involving the image charge based on the boundary condition that Φ(a)=Φ(b)=q/b.
$$\frac{2q}{a}+\frac{q'}{d'-a}=\frac{q}{b}, \frac{2q}{2b-a}+\frac{q'}{d'-b}=\frac{q}{b}$$

Solving them gave me
$$q'=\frac{qa(2b-a)}{4b}, d'=\frac{a(4b-a)}{4b^2}$$

which do give back the correct boundary conditions when put into the potential expression. Then the potential inside the sphere should be given by
$$\Phi(r,\theta)=q\left(\frac{1}{\sqrt{r^2+\frac{a^2}{4}-arcos\theta}}+\frac{a(2b-a)}{4b\sqrt{r^2+\frac{a^2(4b-a)^2}{16b^4}-\frac{ra(4b-a)}{4b^2}cos\theta}}\right)$$

It's not pretty and can probably be simplified a bit, but all I need is the potential at the origin, so,
$$\Phi(0)=q\left(\frac{8b-2a+2ab^2-a^2b}{4ab-a^2}\right)$$

I'm faurly sure that expression is as simplified as possible, I tried factoring a few different ways, but didn't get any more cancellation.

8. Oct 9, 2009

### kuruman

That's quite a bit of algebra you've done, however you must have slipped somewhere. If you do dimensional analysis on your expressions, you will see that q' has dimensions of charge*length and d' is dimensionless. The error is propagated to the expressions that follow.

9. Oct 10, 2009

### Brian-san

I took the most obvious fix and went with the following image charge (I checked and these still work for the original boundary equations)
$$q'=\frac{qa(2b-a)}{4b^2}, d'=\frac{a(4b-a)}{4b}$$

Now the potential should be
$$\Phi(r,\theta)=q\left(\frac{1}{\sqrt{r^2+\frac{a^2}{4}-arcos\theta}}+\frac{a(2b-a)}{4b^2\sqrt{r^2+\frac{a^2(4b-a)^2}{16b^2}-\frac{ra(4b-a)}{4b}cos\theta}}\right)$$

Then the potential at the origin is just
$$\Phi(0)=q\left(\frac{8b^2-a^2}{ab(4b-a)}\right)$$

The image charge and potential at the origin now have correct units, I now believe this to be the correct answer. The only thing I see that might be a problem is I looked along the line of theta=0 at r=a and r=b to generate my equations to find q', d', instead of taking some general angle.

I also figured out my justification for the potential being q/r outside the sphere. We know the electric field is zero at all points inside the conductor, so if we draw a spherical Gaussian surface of radius r, then for a<r<b 4πQenclosed=0, by Gauss' Law. So the total enclosed charge is zero, and this implies the total induced charge on the surface at radius a is -q. Since the conductor began as a neutrally charged object and no charge was transferred to/from it, then the total induced charge on the surface at radius b is q. So outside the sphere, the enclosed charge is just q for Gaussian surfaces in this region.

Also, we can say the potential at r=b is q/b, since the potential must be a continuous function, which gives us the potential at r=a since the surfaces of the conductor are equipotential surfaces.

10. Oct 11, 2009

### kuruman

When you say "I checked and these still work for the original boundary equations", exactly what did you check? The correct boundary condition is

$$\Phi(a,\theta,\phi)=constant$$

where the constant is independent of theta and is equal to q/b. I don't see that happen here.

Here is how I would do the problem.

The standard image problem treats the grounded conducting sphere with the real charge outside the shell and the image inside. If you reverse the roles of image and real, you get the potential inside a grounded sphere with the real charge inside. Then

$$\Phi_{in}(a,\theta,\phi)=0$$

So first I would use an image charge that would make the sphere at r = a an equipotential at zero potential by reversing the standard problem. Now to raise the potential of the shell to q/b, I would add a second image charge q'' = q on the surface at r = b. This adds a constant term q/b to the potential inside, but does not affect the electric field.

$$\Phi(r,\theta,\phi)=\Phi_{in}(r,\theta,\phi)+\frac{q}{b}$$

11. Oct 11, 2009

### Brian-san

I took the the modified values for q', d' and but them into the original two relations I had, and they both simplified to q/b. So there must have been an algebra error in the first attempt.

Since the potential at r=a=b is q/b and is independent of theta, I took the choice of theta=0 for generating my relations for the image charge located outside the sphere.

For r=a, I had
$$\frac{2q}{a}+\frac{q'}{d'-a}=\frac{q}{b}$$

and for r=b,
$$\frac{2q}{2b-a}+\frac{q'}{d'-b}=\frac{q}{b}$$

Each of those was pulled from the general expression:
$$\Phi(r,\theta,\phi)=\left(\frac{q}{\sqrt{r^2+\frac{a^2}{4}-racos\theta}}+\frac{q'}{\sqrt{r^2+d'^2-2rd'cos\theta}}\right)$$

Unless I should be using r=-a instead of q=b for the second relation then
$$\frac{2q}{3a}+\frac{q'}{d'+a}=\frac{q}{b}$$

This one gives me
$$q'=\frac{-q(4b^2-8ab+3a^2)}{2b^2}, d'=\frac{a(4b-3a)}{2b}$$

This seems to make a bit more sense as now the image charge is proportional to -q. Is there a reason I can't use r=b as part of the boundary condition? There must be, as it gives different results for the image charge.

12. Oct 11, 2009

### kuruman

You're off the track. Please read my suggestion in posting #10 and do it that way. I did it myself after I posted and it works a charm. I am not posting the expression I got because I don't want to deny you the pleasure of getting it yourself.

13. Oct 11, 2009

### Brian-san

I resolved the boundary conditions for a zero potential at r=a and got an image charge of q'=-2q at d'=2a. If we put a third charge q''=q at r=b, then the potential expression should be

$$\Phi(r,\theta,\phi)=q\left(\frac{2}{\sqrt{4r^2+a^2-4arcos\theta}}-\frac{2}{\sqrt{r^2+4a^2-4arcos\theta}}+\frac{1}{\sqrt{r^2+b^2-2brcos\theta}}\right)$$

At the origin, this expression gives
$$\Phi(0)=\frac{q(b+a)}{ab}$$

Assuming this is right, I have a couple of problems with it (the method, not the answer). In the general expression, at r=a, the first two terms cancel because of the first image charge, but does not give q/b as the expected answer, instead it is
$$\Phi(a)=\frac{q}{b-a}$$

Also, is there any reason it cannot be done simply with one image charge? I ask because my first instinct was to simply try and find one image charge.

14. Oct 11, 2009

### gabbagabbahey

Why add the third charge?...If you wanted the potential to be zero at $r=a$, you would just add the image charge $q'=-2q$ at $r=2a$. In this case you want it to be equal to a constant $\frac{q}{b}$, so why not just add a constant:

$$\Phi(r,\theta,\phi)=q\left(\frac{2}{\sqrt{4r^2+a^2-4arcos\theta}}-\frac{2}{\sqrt{r^2+4a^2-4arcos\theta}}\right)+\frac{q}{b}$$

15. Oct 11, 2009

### kuruman

You have correctly noticed that your potential does not reduce to a constant when r = a. The third term should be just q/b, a constant, because the image charge is not a point charge, but a charge distribution uniformly smeared on the outer surface. It serves to raise the potential inside the shell to the desired value. Look at the last equation in posting #10.

Unless the sphere is grounded, I don't think you can do it with a single image charge. You can get the electric field with a single image charge, but the potential depends on the outer radius b and that's another independently adjustable parameter. In other words, you need one image charge to get the electric field right for all shells regardless of thickness and a second image charge (uniform distribution in this case) to get the potential for a shell of a particular thickness.

16. Oct 11, 2009

### Brian-san

That makes sense. Now is it because the charge is uniformly distributed over the r=b surface the reason we can just add the q/b constant to the potential expression, or is the distribution of the charge irrelevant to this fact? I just wasn't seeing why you could add a constant term instead of an additional point charge potential which I why I chose the final incorrect term in my expression.

Then is my final expression for the potential at the origin correct? The wrong final term shouldn't matter, as it reduces to q/b for r=0.

17. Oct 12, 2009

### kuruman

But the boundary condition is not q/b at r = 0. The boundary condition is q/b at r = a. You need to understand that the image charge method derives its legitimacy from the uniqueness theorem. Point charges are solutions of Laplace's equation and so are constants. Therefore, if you can find a sum of point charges and constants that satisfies the boundary conditions, you are guaranteed to have found the solution and that there shall be no other solutions before you. So here, we found first the point charges that produce zero potential at r = a, then added the constant q/b to the solution inside the sphere to get a match with the boundary condition that the inner surface of the shell sits at potential that is not zero but q/b.

That's the reasoning behind writing

$$\Phi(r,\theta,\phi)=\Phi_{in}(r,\theta,\phi)+\frac {q}{b}$$

where Φin(r,θ,φ) reduces to zero at r = a for any choice of angles θ and φ.

18. Oct 12, 2009

### Brian-san

I know the last term in the expression is wrong as it does not reproduce the proper boundary condition. So the potential expression I gave was incorrect with that particular term and needed to be replaced with q/b. However, I needed to calculate the potential at the origin, and was making sure that changing that final term had no impact on the answer, since it is q/b in both the correct and incorrect expression.

19. Oct 12, 2009

### kuruman

The correct expression is the one posted by gabbagabbahey

$$\Phi(r,\theta,\phi)=q\left(\frac{2}{\sqrt{4r^2+a^2-4arcos\theta}}-\frac{2}{\sqrt{r^2+4a^2-4arcos\theta}}\right)+\frac{q}{b}$$

Find what this becomes when r = 0 and you are done.

Last edited: Oct 12, 2009
20. Oct 12, 2009

### Brian-san

Using that correct formula gives a potential of
$$\Phi(0)=\frac{q(b+a)}{ab}$$

Which is the same as what I got, so that must be the right answer. Thanks.