# Homework Help: Potential of Two Charged Spheres

1. Feb 20, 2010

### hover

1. The problem statement, all variables and given known data.
A metal sphere with radius r_a is supported on an insulating stand at the center of a hollow, metal spherical shell with radius r_b. There is charge + q on the inner sphere and charge - q on the outer spherical shell. Take the potential V to be zero when the distance r from the center of the spheres is infinite.

Calculate the potential V(r) for r < r_a. (Hint: The net potential is the sum of the potentials due to the individual spheres.)

2. Relevant equations
E=Kq/r^2 and integrating with respect to r

3. The attempt at a solution

I'm thinking that since this is a conductor, the net charge inside the smaller metal sphere will be zero. Since the net charge inside the conductor is zero, the electric field will be zero and if that is zero then there can't be a voltage right?

2. Feb 20, 2010

### ehild

What do you think is the difference between the term "voltage" and "electric potential" ?

ehild

3. Feb 22, 2010

### hover

I didn't think that there was a difference. I thought that voltage was the measure of electrical potential.

4. Feb 22, 2010

### collinsmark

Just because the electric field inside the small, metal sphere is zero, does not mean that electric potential is zero (with respect to infinity). The electric field in-between the spheres is not zero, and you need to integrate over this part of the path too.

But like the hint says (in the problem statement), the easiest way to do this problem is to find the potentials (with respect to infinity) of each sphere separately, and add the results together. (Don't forget that they are oppositely charged). The result is not zero.

5. Feb 22, 2010

### hover

So are you saying that I should integrate r_b to r_a? It would be silly to integrate from r_a to r since the change of the voltage is zero inside the smaller metal sphere.

6. Feb 22, 2010

### collinsmark

But your path must extend all the way out to infinity (since it starts at infinity). You're correct that

$$-\int _a ^0 \vec E \cdot \vec {dl} = 0.$$

But you still have

$$-\int _{b} ^r \vec E \cdot \vec {dl}$$

for b > r > a, to integrate over, and the electric field in that region is not 0.

And you also have

$$-\int _{\infty} ^r \vec E \cdot \vec {dl}$$

for r > b, too, which may or may not be zero (I'll let you figure that out).

Hypothetically, you could integrate all of the above, after adding the electric fields together, but I believe your instructor is asking you to do as your "hint" says, and integrate both spheres' electric fields separately, over the path starting from infinity, to some r, and at maximum, ending at the respective sphere's surface. Then add the potentials together. Solving the problem the way the hint says is usually easier, because it is generally easier to add scalar fields together (such as potential) instead of vector fields (such as electric field).

7. Feb 22, 2010

### collinsmark

Looking at things another way, ask yourself, "how much energy would it take to move a unit charge from infinity to the region inside the small metal sphere?" True, the electric field inside the small metal sphere is zero. But there is(are) a region(regions) outside the small metal sphere sphere where the electric field is not zero. And you would need to move the charge through that(those) region(s) to get the unit charge inside the small metal sphere, and that would take a non-zero amount of work.

8. Feb 23, 2010

### hover

I think I see what you mean now. Starting out at infinity, move the imaginary unit charge closer to the sphere. Outside r_b the sphere appears to have a net charge of zero so no electric field and no voltage. Then once you go inside the sphere (r_b), the charges no longer at net zero. This means an electric field is present and from r_b to r_a there will be a change in voltage. Once we go past r_a we are now inside a conductor which will have no change in voltage. So I should basically take the bounds to be from r_a to r_b I think. Is my logic correct?

9. Feb 23, 2010

### collinsmark

Yes, you've got it!

But there are two ways of going about solving the problem. You could first add the electric fields together, and then solve for the potential (by integrating the combined electric fields). Or, you could solve for the potential of each electric field separately (via integration) and then add the potentials together.

In this particular problem, either method is really no more difficult than the other. But in the future, when you are confronted with more complicated scenarios, you will likely find that the latter method is easier (the latter method being the one where you solve for the individual potentials, and add together the potentials).

For example, if the spheres didn't share a common center, it would be a headache to add together the electric fields. Electric fields are vector fields. You would have to add all the components together separately, and perhaps deal with all the trigonometry and such that goes with vectors. Potentials on the other hand are scalar fields. When adding those together you don't have to worry about separate components.

As a mater of fact, in the future, when calculating the electric field of some complicated setup, sometimes the easiest approach is to calculate the potentials of the separate components in the setup, add the potentials together, and then take the -gradient of that get the electric field of the whole. Strange as this may sound, it can sometimes be easier than adding together the electric fields directly.

In this particular problem a hint was given, that suggests adding together the individual potentials. So I think you instructor would prefer you do it that way.