# Electric Flux, Gauss's Law Problem

## 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!!