A proton is a distance d/2 directly above

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?

Image from: J. Walker, D. Halliday, and R. Resnick, Fundamentals of physics: [extended], 10th ed. United States: Wiley, John & Sons, 2013.

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Let us assume that this is one side of a cube. Gauss’s law states the following:

\Phi=\dfrac{q}{\epsilon_0}

(Where $\Phi$ is the flux, $q$ is the charge in the enclosed Gaussian surface, and $\epsilon_0$ is the permittivity of free space.)

If we take the total flux through a closed Gaussian surface and divide it by 6, we get the flux through one side.

Thus, we can write the following:

\Phi=\dfrac{q}{6\epsilon_0}

Remember that the charge of a proton is $1.6\times10^{-19}$ C.

\Phi=\dfrac{1.6\times10^{-19}}{(6)(8.85\times 10^{-12})}

$\Phi=3.01\times 10^{-9}$ $\dfrac{N\cdot m^2}{C}$

\Phi=3.01\times 10^{-9}

\dfrac{N\cdot m^2}{C}

A proton is a distance d/2 directly above 2