Electric Flux, Gauss's Law Problem

[henryli78]

Homework Statement

A proton is a distance d/2 directly above the center of a square of side d. What is the magnitude of the electric flux through the square?

Homework Equations

1. Electric flux, \Phi_{net} = \oint \vec{E}\cdot d\vec{A}
2. \Phi_{net} = \frac{q_{enc}}{\epsilon_{0}}
3. \vec{E} = \frac{kQ}{r^{2}}

The Attempt at a Solution

I tried to use Equation - 3 first to calculate the net electric field and then from there, use the value of the electric field and multiply it by the area of the square.

I solved for the electric field from Equation 3 to be \frac{4ke}{d^{2}} where k is Coulomb's constant and e\ =\ 1.602176462(63)\ \times\ 10^{-19}\ C

Then I used Equation - 1 and because there is only one surface, the area is just equal to {d^{2}}. Thus, I calculated \Phi_{net} \ =\ \frac{4ke}{d^{2}}\times {d^{2}}\ =\ 4ke\ = 5.8 \times 10^{-9} N*m^2/C.

However, the answer in my textbook says it is actually 3.01 \times10^{-9} N*m^2/C. They used equation two and assumed that if the proton was contained in a cube and the square was one of the faces of the cube, the net electric flux of the cube would be \Phi_{net} = \frac{1.6 \times 10^-19}{\epsilon_{0}} and thus the electric flux for the square is 6 times less that.

Can someone point out how my method is flawed and where I went wrong in my logic? Thanks!

 

[nasu]

This problem was discussed a little wile ago.
See this:
https://www.physicsforums.com/showthread.php?p=4520522#post4520522

 

[henryli78]

Ahh I forgot to search first. But is it possible if someone could point out where I have a flaw in my logic? I understand the solution's concept but I get why my answer is different.

 

[nasu]

The electric field changes from point to pint along the surface. It is not a constant, as you assumed.
This is also discussed in that thread.

 

[henryli78]

Ok thank you very much! I had a hard time really understanding my error but now I get it. Thanks!