(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