More Questions about Conducting Shells

A conducting shell (ie, a hollow sphere) of radius a = 10cm is given a charge of Q =1 uC. Calculate the electric field outside the sphere. Calculate the potential at the centre.

r is the point of interest.

(I won’t type out my work as I hate writing out mathematical formuals)

So we know that E = 0 at the centre, since no charge is enclosed in the corresponding Gaussian surface.

At r > a, I get the equivalent expression for a point charge, as expected, and as indictated by the solution.

For the potential, I integrate from infinity to a. I get a value for V of approx. 90 kv. This is the correct answer.

I had no difficulty getting the right answer. However, I have terrible difficulty understanding the process and the results.

  1. The only reason I “knew” to integrate from zero to a was because that’s how its always done in the other questions. I don’t understand why. Why wouldn’t we integrate from infinity to the centre, zero? And please don’t respond with, “you’ll get a zero in the denominator!”. I want a physical interpretation of why we only integrate to a. Similarly, why can’t we integrate from the centre to a?

  2. Regardless of where your non-infinity limit is, you’re still brining in your test charge from infinity. Infinity is always infinity, regardless of how much you subtract from it - so why does this integration technique work in the first place? Again, I know how to do the math and that that math works - I want a physical interpretation. My understanding is that voltage is a measure of the energy required to move the particle from infinity to a given point. If it’s infinitely far away, why doesn’t it take an infinite amount of energy?

  3. How can there be a voltage at the centre when there is no electric field? I mean, if you establish a voltage differential across a wire, it establishes an electric field which causes charge to flow. Even in a non-conductor, if there is a voltage differential there’ll be an electric field (ie, inside a capacitor). Yet apparently, here we see a voltage without an electric field!

I have an electromag midterm on tuesday…this is a TRIVIAL problem and look how many questions I have~~!! :frowning:

First question: The voltage at the center of the charged sphere must be the same as at the surface of the charged sphere. This is because electric field is the derivative of the potential, so since there’s no E inside the sphere, there’s no change in V. Strictly speaking, you could do an integral from a to 0 to find the change in potential between those two points, but that’d be the integral of zero, since E = 0.

Second question: You do have to move the charge an infinite distance, but most of that distance is really easy. Work is force times distance, and although the distance is great, the force is miniscule. Your question really comes down to asking how an integral with infinite limits can be finite, and that really comes down to asking how the sum of an infinite series can be finite. In much the same way that 1/2 + 1/4 + 1/8 + 1/16 + … is finite, Int(1/r[sup]2[/sup],r=infinity…a) is finite.

Third question: The voltage is nonzero, but the voltage differential is zero. If it makes you feel any better, you can define the voltage inside the sphere to be zero, and just live with the voltage at infinity having some value. It’s only voltage differences which ever matter in physics; the voltage itself is basically just a constant of integration which can be set to an arbitrary value, so long as all of the differences remain the same. Some choices (like setting V(infinity) = 0) are just more convenient than others.