Question

You have two square metal plates with side length of 16.50 cm. You want to make a parallel-plate capacitor that will hold a charge of 18.5 nC when connected to a 37.8 V potential difference. Determine the necessary separation in mm. Round your answer to three significant figures.

Answer 1

d = 3.44 *10^{-7} m

Step-by-step explanation:

given data:

length of metal plates = 16.50 cm

capacitor charge = 18.5 nC

potential difference = 37.8 V

capacitance of parallel plate capacitor

`C = (A\epsilon _(0))/(d)`

area of the individual plate

A=
`a^2 = (16.5*10^(-2))^2 = 272.25 *10^(-4)` m2

capacitance

`C = QV = 18.5 *10^(-9) *37.8 = 699.3 * 10^(-9) C`

separation between plates d is given as
`= (A\epsilon _(0))/(C )`

`d =  (272.25 *10^(-4) *8.85*10^(-12))/(699.3 *10^-9)`

d = 3.44 *10^{-7} m