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.
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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?
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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?
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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~~!!