Answer to Question #317317 in Physics for Hana

Question

A capacitor consists of two square metal plates, each measuring 5.00×10^2m on a side. In between the plates is a sheet of mica measuring 1.00×10^4m thick.

(a) What is the capacitance of this capacitor? If the charge in one plate is 2.00×10^8C.

(b) What is the potential difference?

(c) What is the electric field between the plates?

Give: Permittivity of Mica (E= 4.8×10^-11C²/N.m²)

Answer 1

(a) What is the capacitance of this capacitor?

$$C=\frac{\epsilon A}{d}\\ =\frac{4.8\times10^{-11}\times 5.00\times 10^{-2}}{1.00\times 10^{-4}}=2.40\times 10^{-8}\:\rm F$$

(b) What is the potential difference between plates

$$V=\frac{q}{C}\\ =\frac{2.00\times 10^{-8}}{2.40\times 10^{-8}}=0.833\:\rm V$$

(c) What is the electric field between the plates?

$$E=\frac{V}{d}\\ =\frac{0.833}{1.00\times 10^{-4}}=8330\:\rm V/m$$