(a) 9040 V/m
The magnitude of the electric field between two parallel, oppositely charged plates is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma[/tex] is the charge surface density on each plate
[tex]\epsilon_0 = 8.85\cdot 10^{-12}[/tex] is the vacuum permittivity
In this problem, the magnitude of the charge density on each plate is
[tex]\sigma = 80.0 nC/m^2 = 80.0\cdot 10^{-9} C/m^2[/tex]
Substituting into the formula, we find the magnitude of the electric field:
[tex]E=\frac{80.0\cdot 10^{-9}}{8.85\cdot 10^{-12}}=9040 V/m[/tex]
(b) 90.4 V
The electric potential difference between the two plates is given by
[tex]\Delta V=Ed[/tex]
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
[tex]\Delta V=(9040)(0.01)=90.4 V[/tex]
