Accepted
at 2:35 p.m. Jun, 01, 2025
by
pkaps01
Author:
medish
Type of change:
Updated content
Rationale for change
added two clozes to denote the question wants two answers
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Gauss theorem for electric field between plates: \(E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0A}\)
Magnitude of potential difference between plates: \(\Delta V = Ed = \frac{Qd}{\epsilon_0A}\)
Therefore, capacitance (C) of a parallel plate capacitor \(= \frac{Q}{\Delta V} = \frac{Q}{\frac{Qd}{\epsilon_0A}}\)\(= \frac{\epsilon_0A}{d}\)
A = area of each plate
d = distance between plates
Magnitude of potential difference between plates: \(\Delta V = Ed = \frac{Qd}{\epsilon_0A}\)
Therefore, capacitance (C) of a parallel plate capacitor \(= \frac{Q}{\Delta V} = \frac{Q}{\frac{Qd}{\epsilon_0A}}\)\(= \frac{\epsilon_0A}{d}\)
A = area of each plate
d = distance between plates
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