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

$E=\frac{\sigma}{\epsilon_0}$

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:

$E=\frac{80.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=9040 V/m$

(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

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