What is the surface charge density on the face of the penny?

[Mason Smith]

Homework Statement

The electric field strength just above one face of a copper penny is 2230 N/C. What is the surface charge density on this face of the penny?

Homework Equations

Electric field of an infinite plane of charge = η/(2*ε₀)

The Attempt at a Solution

I used the above equation, and the result was 3.95 x 10^-8C/m². However, this is not the correct answer. Where am I going wrong in my thinking? This homework problem comes from the textbook chapter about Gauss's Law, yet I do not see how Gauss's Law can tie into this question.

 

[Mason Smith]

To anyone who needs help with this problem, I believe that I have found a solution.
The equation for surface charge density is η = Q/A. Since the electric field is just above the face of the penny, the penny can be modeled as an infinite plane of charge E_plane = η/(2ε₀). Combining these equations will result in E_plane = Q/(A⋅2⋅ε₀). Rearranging this equation to find the surface charge density will be (Q/A) = (E_plane⋅2⋅ε₀). However, this will be the surface charge density for both sides of the penny. In order to find the surface charge density of only one side of the penny, divide E_plane⋅2⋅ε₀ by 2.

 

[kuruman]

You didvide by 2 too many times. If η is the surface charge density on one side, and you construct the standard Gaussian pillbox sticking out both sides of the penny, then the total electric flux through the pillbox is Φ_E = 2E_paneA. The total charge enclosed is Q_encl. = 2ηA (ηA per side).
By Gauss's law, 2E_planeA = 2ηA/ε₀ which gives η = ε₀E_plane.