Question

Two parallel plates 19 cm on a side are given equal and opposite charges of magnitude 2.0 ✕ 10^−9 C. The plates are 1.8 mm apart. What is the electric field (in N/C) at the center of the region between the plates? (Enter the magnitude.)

Answer 1

E = 5291.00 N/C

Step-by-step explanation:

Expression for capacitance is

`C = (\epsilon  A)/(d)`

where

A is area of square plate

D = DISTANCE BETWEEN THE PLATE

`C = (\epsilon*(19* 10^(-2))^2)/(1.5* 10^(-3))`

`C = 24.06 \epsilon`

`C = 24.06* 8.854* 10^(-12) F`

`C =2.1* 10^(-10) F`

We know that capacitrnce and charge is related as

`V = (Q)/(C)`

`= (2\tiimes 10^(-9))/(2.* 10^(-10))`

v = 9.523 V

Electric field is given as

`E = (V)/(d)`

=
`(9.52)/(1.8*10^(-3))`

E = 5291.00 V/m

E = 5291.00 N/C