To solve this problem we will apply the concepts of the potential difference such as the product between the electric field and the distance, then we will use two definitions of capacitance, the first depending on the Area and the second depending on the load to find the Area. Finally we will look for capacitance with the values already obtained in the first sections of this problem
PART A) Potential Difference is
[tex]V = Ed[/tex]
Here,
E = Electric Field
d = Distance
Replacing,
[tex]V = (5*10^6)(3.0*10^{-3})[/tex]
[tex]V = 15000V= 15kV[/tex]
PART B) Capacitance of the capacitor is
[tex]C = \frac{\epsilon_0 A}{d}[/tex]
Here,
A = Area
[tex]\epsilon_0[/tex] = Permittivity Vacuum
d = Distance
Rearranging to find the Area we have,
[tex]A = \frac{Cd}{\epsilon_0}[/tex]
We know at the same time that Capacitance is the charge per Voltage, then
[tex]C = \frac{Q}{V}[/tex]
Replacing at this equation we have that
[tex]A = \frac{Qd}{V\epsilon_0}[/tex]
[tex]A = \frac{(79*10^{-9})(3*10^{-3})}{(15000)(8.853*10^{-12})}[/tex]
[tex]A = 1.78mm^2[/tex]
PART C)
Capacitance is given by,
[tex]C = \frac{Q}{V}[/tex]
[tex]C =\frac{79*10^{-9}}{15000}[/tex]
[tex]C =5.26pF[/tex]
