Question

Two large, parallel, conducting plates are 1.00 cm apart and having electric charges of equal magnitudes (80.0 nC/m^2) and opposite sign on their facing surfaces. (a) Find the electric field in the space between the two plates; (b) Find the electric potential difference between the two plates.

Answer 1

(a) 9040 V/m

The magnitude of the electric field between two parallel, oppositely charged plates is given by

`E=(\sigma)/(\epsilon_0)`

where

`\sigma` is the charge surface density on each plate

`\epsilon_0 = 8.85\cdot 10^(-12)` is the vacuum permittivity

In this problem, the magnitude of the charge density on each plate is

`\sigma = 80.0 nC/m^2 = 80.0\cdot 10^(-9) C/m^2`

Substituting into the formula, we find the magnitude of the electric field:

`E=(80.0\cdot 10^(-9))/(8.85\cdot 10^(-12))=9040 V/m`

(b) 90.4 V

The electric potential difference between the two plates is given by

`\Delta V=Ed`

where

E is the magnitude of the electric field

d is the separation between the two plates

In this problem, we have

E = 9040 V/m

d = 1.00 cm = 0.01 m

Substituting into the formula, we find

`\Delta V=(9040)(0.01)=90.4 V`