# At each point on the surface of the cube

At each point on the surface of the cube shown, the electric field is parallel to the z axis. The length of each edge of the cube is 3.0 m. On the top face of the cube the field is $\vec E=-34 \vec k$ N/C and on the bottom face it is $\vec E=+20 \vec k$ N/C. Determine the net charge contained within the cube.

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

#### Solution:

We can figure out the flux through the top surface and the bottom surface of the cube. The flux through the top surface is:

$\Phi=(E)(A)$

(Where $\Phi$ is flux, $E$ is electric field, and $A$ is the area of the face)

$\Phi=(-34)(3^2)=-306$ $\dfrac{N\cdot m^2}{C}$

The flux through the bottom surface is:

$\Phi=(E)(A)$

$\Phi=(20)(3^2)=-180$ $\dfrac{N\cdot m^2}{C}$

(Our answer is negative because we are looking at the bottom face, which has a negative -k component since our area vector is facing down)

The net flux can be calculated by adding up the bottom and top face. Note that the other faces do not have a flux going through their surfaces since the electric field is parallel to the z-axis.

Total flux = -306 + (-180) = -486 $\dfrac{N\cdot m^2}{C}$

We can now use the following equation to figure out the charge enclosed.

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

(Where $\Phi$ is the net flux, $\epsilon_0$ is permittivity of free space and q is the charge enclosed)

$q=(\Phi)(\epsilon_0)$

$q=(-486)(8.85\times10^{-12})$

$q=-4.3\times10^{-9}$ C

$q=-4.3\times10^{-9}$ C