- #1

henryli78

- 11

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## 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, [itex]\Phi_{net}[/itex] = [itex]\oint \vec{E}\cdot d\vec{A}[/itex]

2. [itex]\Phi_{net}[/itex] = [itex]\frac{q_{enc}}{\epsilon_{0}}[/itex]

3. [itex]\vec{E}[/itex] = [itex]\frac{kQ}{r^{2}}[/itex]

## 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 [itex]\frac{4ke}{d^{2}}[/itex] where k is Coulomb's constant and [itex]e\ =\ 1.602176462(63)\ \times\ 10^{-19}\ C[/itex]

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

However, the answer in my textbook says it is actually [itex]3.01 \times10^{-9} N*m^2/C[/itex]. 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 [itex]\Phi_{net}[/itex] = [itex]\frac{1.6 \times 10^-19}{\epsilon_{0}}[/itex] 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!!