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A capacitor is connected to a battery with constant voltage. If you approach one of the plate to the other, what happens to electric potential?

I already have an answer to this question but it's different from what my teacher answered.

From the law $V = \frac {KQ}{R}$ we know that if we increase the charge, we increase the electric potential, and when we approach the plate to the other, we increase the capacitance of the capacitor so more charges will flow to the capacitor, so the electric potential will increases. However I am still believing that the voltage remains the same because every point on the capacitor will increase in electric potential, so the difference will still be the same.

But my teacher told me that the electric potential is still the same. Who is right?

MarianD
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Mohammad Alshareef
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1 Answers1

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I think that the question-setter meant to ask you what happens to the potential difference between the plates. The answer is then contained in the first sentence: "A capacitor is connected to a battery with constant voltage..."

The equation you quote, =/, gives the potential at a distance R from the centre of a spherically symmetric charge distribution. It is therefore not applicable to a parallel plate capacitor, which is what I believe the question-setter had in mind.

Philip Wood
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  • Does that mean that the charge doesn't have any effect on the electric potential , What happens to the electric potential on the plate its self. – Mohammad Alshareef May 12 '19 at 12:03
  • The potential of the positive plate becomes positive , do you mean The potential of the positive plate becomes more positive , and the same for negative statement. – Mohammad Alshareef May 12 '19 at 12:17
  • I was forgetting the wording of the question, and am rewriting previous comment. – Philip Wood May 12 '19 at 12:19
  • Thank you very much. – Mohammad Alshareef May 12 '19 at 12:21
  • (a) "Does that mean that the charge doesn't have any effect on the electric potential" No; the magnitude of charge on each plate is proportional to the pd between them. But the pd between them is, in this case, fixed by the battery. (b) "What happens to the electric potential on the plate its self?" The potential (relative to infinity) of the mid-plane between the plates stays zero (by symmetry). Since the pd between the plates doesn't change (because of the battery), the potentials relative to infinity of each plate do not change as the plates are brought closer together. – Philip Wood May 12 '19 at 12:27
  • Have you taken into consideration that the capacitor remain connected while approaching ? – Mohammad Alshareef May 12 '19 at 12:38
  • Yes. My original answer hinges on this, and it's what I meant in the comment above when I said "But the pd between them is, in this case, fixed by the battery." and "Since the pd between the plates doesn't change (because of the battery)..." – Philip Wood May 12 '19 at 12:47
  • I don't want to annoy you so This is my last comment . Assume I plane (a little bit thick plane) is charged by +1 μc ,(relative to infinity) you need some work to bring charge from infinity to the mid of the plate ,Now If you charge this plate more with +100 μc ,Then of course you need more work to bring it to the mid of the plane so the electric potential must be higher in the plane , but as you mentioned that it still the same. (the potentials relative to infinity of each plane do not change as the plates are brought closer together). – Mohammad Alshareef May 12 '19 at 12:57
  • No annoyance; I can see that you are trying to understand some quite tricky ideas. One thing I've noticed is that you don't seem to be taking full account of the capacitor consisting of TWO plates (with equal and opposite charges). When I talked about the mid plane between the plates I meant the plane, parallel to the plates, in the middle of the gap between the plates, not the middle of either plate! – Philip Wood May 12 '19 at 13:08
  • Oh , does still the same because if you increase the positive charge you also increase the negative charge so the positive charge increase the electric potential on positive plate and the negative charge decrease the electric potential on positive plate , so it still the same. – Mohammad Alshareef May 12 '19 at 13:15
  • You are correct in saying that if you move the plates (still connected to the battery) closer together, the positive plate will gain more positive charge and the negative plate will gain more negative charge. But this won't affect the potential difference between the plates. This is because (going back to first principles) the work done as a test charge is taken across the gap is the still same because, although the electric field strength is increased, the distance the test charge has to go is decreased! Or we can say that the battery maintains a constant pd ! – Philip Wood May 12 '19 at 13:23
  • Thank you very much , I love you ^_^. – Mohammad Alshareef May 12 '19 at 13:26