An electromagnetic wave has an electric field strength of 145 V/m at a point P in space at time t. A parallel-plate capacitor whose plates have area 0.109 m2 has a uniform electric field between its plates with the same energy density as the answer to part a. What is the charge on the capacitor?

I found the Electric density to be 9.30e-8 but I cant find the relation that I could use to calculate charge.

Gauss:

The amount of electric field coming out of a volume is proportional to the charge inside.

In the case of parallel plates
E = sigma/eo
where sigma is the charge per unit plate area
Multiply sigma by the area to get the total charge. It + on one plate, - ont he other of course.

To find the charge on the capacitor, you need to use the equation relating electric field strength and capacitance. The relationship between electric field (E), capacitance (C), and charge (Q) for a parallel-plate capacitor is given by the equation:

Q = C * E

In this case, you are given the electric field strength (E = 145 V/m) and the area of the plates (A = 0.109 m^2).

First, let's determine the capacitance (C) of the parallel-plate capacitor using the equation:

C = ε₀ * (A / d)

where ε₀ is the permittivity of free space (8.854 x 10^-12 F/m) and d is the separation distance between the plates.

To find the separation distance (d), we use the energy density of the electric field, which is given by:

u = (1/2) * ε₀ * E^2

Rearranging the equation, we find:

E = sqrt( (2 * u) / ε₀ )

Using the given energy density (u = 9.30 x 10^-8 J/m^3), we can solve for E:

E = sqrt( (2 * (9.30 x 10^-8) J/m^3 ) / (8.854 x 10^-12 F/m) )
= 487811.41 V/m

Now we can find the separation distance (d):

d = ε₀ * (A / C)
= (8.854 x 10^-12 F/m) * (0.109 m^2 / (487811.41 V/m))
= 1.97 x 10^-14 m

Finally, we can calculate the capacitance (C):

C = ε₀ * (A / d)
= (8.854 x 10^-12 F/m) * (0.109 m^2 / (1.97 x 10^-14 m))
= 4.90 x 10^-12 F

Now, substituting the value for electric field strength (E = 145 V/m) and capacitance (C = 4.90 x 10^-12 F) into the equation Q = C * E:

Q = (4.90 x 10^-12 F) * (145 V/m)
= 7.10 x 10^-10 C

Therefore, the charge on the capacitor is 7.10 x 10^-10 Coulombs.