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I am grading questions from the lab manual provided by the professor to review with the students. I am trying to wrap my head around one of the questions about what an external electric field would do.

Assume two parallel plates, with a positive charge on the right, and negative/ground on the left.

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If there is an external field aligned with the capacitor plates (positive on right), that field will result in a greater negative charge on the left plate, thus raising the voltage, and thus the charge, correct?

Then in the opposite case, the external field has its positive source on the left side of the capacitor. It will repel positive charge on the right plate, thus decreasing the voltage potential between the plate, thus lowering the "capacitance".

Is "lowering the capacitance" the correct term, since "capacitance" is a fixed property of the component geometry? Or does capacitance also mean the amount of charge that is held by the plates? When voltage goes down, the amount of charge held goes down, thus "capacitance" goes down, too?

Thank you for your feedback.

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    $\begingroup$ Don't get this wrong... but if you are this unclear about the concepts, I don't think it's a good idea that you should be teaching them. On the other hand, I am willing to assume that you are really just a student who is trying to get help with homework. That will make me way less freaked out than the idea of a science teacher who doesn't know the subject, at all. $\endgroup$ – CuriousOne Feb 15 '16 at 23:11
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    $\begingroup$ How can the net charge on the right hand plate change as it is electrically insulated from its surroundings? $\endgroup$ – Farcher Feb 15 '16 at 23:19
  • $\begingroup$ To clarify, I am a teaching assistant, and I am trying to decode a single question from the lab manual. Given the inane question, what is the best answer? $\endgroup$ – MapMike Feb 16 '16 at 1:55
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Assuming you mean you have a full electrical circuit and you're wondering what an external field does to it:

The external electrical field would change the voltage, and thus the charge on the capacitor. This is not changing the capacitance.

Generally, though, capacitors are very small, and so the potential difference across the capacitor would be very small for reasonable electrical field magnitudes anyway.

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    $\begingroup$ Thank you. I understood capacitance to be constant, but every student answered that it could vary. I wanted to ensure that I was not misunderstanding something. $\endgroup$ – MapMike Feb 16 '16 at 4:27
  • $\begingroup$ If this were true then the reduction of the net electric field as introduced by a dielectric would not change the capacitance, which we know is not true. $\endgroup$ – Lone Wolf Jul 23 '17 at 23:22
  • $\begingroup$ @LoneWolf A dielectric reduces the net electric field (and thus voltage difference across the terminals) for a fixed charge and thus increases the capacitance. Applying an external electric field (which, if it helps, is equivalent to applying an external voltage) increases both the charge and voltage in a way that keeps the capacitance fixed. $\endgroup$ – Chris Jul 24 '17 at 19:05

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