Answer to Question #196047 in Electric Circuits for Plea

Question

A parallel plate capacitor is composed of two rectangular plates with length 5mm and width 3 mm. The thickness of the insulating material is 0.5 mm. Find the permittivity of the insulating material if the capacitance is 2 μF.

Answer 1

Given data

Capacitance of capacitor is $C=2\times 10^{-6} F$

Area of each plate is $A=(5\times 10^{-3})(3\times 10^{-3}) m^2$

Separation between the plates is $d=(0.5\times 10^{-3}) m$

The expression for the capacitance of parallel plate capacitor is

$C=\frac{A \epsilon}{d}$

The permittivity of the insulating material is

$\epsilon=\frac{Cd}{A}=\frac{(2\times 10^{-6} F)((0.5\times 10^{-3}) m)}{(5\times 10^{-3})(3\times 10^{-3}) m^2}=6.67\times 10^{-5} C^2/N.m^2$