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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?

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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, 𝑉=𝐾𝑄/𝑅, 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.

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  • $\begingroup$ 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. $\endgroup$
    – Mohammad Alshareef
    Commented May 12, 2019 at 12:03
  • $\begingroup$ 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. $\endgroup$
    – Mohammad Alshareef
    Commented May 12, 2019 at 12:17
  • $\begingroup$ I was forgetting the wording of the question, and am rewriting previous comment. $\endgroup$
    – Philip Wood
    Commented May 12, 2019 at 12:19
  • $\begingroup$ Thank you very much. $\endgroup$
    – Mohammad Alshareef
    Commented May 12, 2019 at 12:21
  • $\begingroup$ (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. $\endgroup$
    – Philip Wood
    Commented May 12, 2019 at 12:27

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