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.
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.
Best Answer
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.