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.

(a) 9040 V/m

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

where

is the charge surface density on each plate

is the vacuum permittivity

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

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

(b) 90.4 V

The electric potential difference between the two plates is given by

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

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explanation:

true is the correct answer

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