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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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