Answer:
0.07 V
Explanation:
The electric field between two parallel plates is uniform, and its magnitude is given by
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where
[tex]\sigma[/tex] is the magnitude of the charge density on each face
[tex]\epsilon_0 = 8.85419 \cdot 10^-12 C^2/Nm^2[/tex] is the vacuum permittivity
In this problem, we have
[tex]\sigma=69 pC/m^2 = 69\cdot 10^{-12} C/m^2[/tex]
So, the electric field is
[tex]E=\frac{69\cdot 10^{-12} C/m^2}{8.85419\cdot 10^-12 C^2/Nm^2}=7.79 N/C[/tex]
So now we can calculate the potential difference between the two plates, given by:
[tex]\Delta V=E d[/tex]
where [tex]d=9 mm=0.009 m[/tex] is the distance between the two plates. Substituting, we find
[tex]\Delta V=(7.79 N/C)(0.009 m)=0.07 V[/tex]
