# Flux through a cube of side length $l$

Let's say I have a cube sitting in the first octant with one corner at the origin and of side length $l=0.1$meters. The question from the text is as follows, "The electric field is uniform, has magnitude $E=4\times10^3 N/C$, and is parallel to the $xy$-plane at an angle of $53.1^{\circ}$ measured from the positive $x$-axis towards the positive $y$-axis. Calculate the flux through each of the cube's faces."

I am having trouble with the wording, "...parallel to the $xy$-plane at an angle of $53.1^{\circ}$ measured from the positive $x$-axis towards the positive $y$-axis," and how this is pictured.

I now want to calculate the flux through each of the six cube faces but I know this part solely depends on the validity of the angle in the calculation $\Phi=EA\cos\phi$ which depends on the accuracy of the picture.

I know the net flux through the cube is zero however, so I know the sum of the six faces' flux is zero.

Can anyone help with the picture and wording here? Thanks in advance!

• If you write the electric field vector in unit vector notation $\textbf{E}=E_{x}\hat{\textbf{x}}+ E_{y}\hat{\textbf{y}}+E_{z}\hat{\textbf{z}}$. In this case you know $\textbf{E}$ is parallel to xy plane, so $E_{z}=$constant and you can use the given angle to find the other two components. – Alberto Navarro Sep 17 '18 at 7:04
• @coreyman317: My best guess is that $\mathbf{E}=2400\mathbf{\hat{i}}+3200\mathbf{\hat{j}}$. As you pointed out, since the E-field is uniform, $\Phi=EA\cos\theta$, where $\theta$ is the angle between $\mathbf{E}$ and the unit normal vector $\mathbf{\hat{n}}$ perpendicular to each side of the cube. Of course, you can simply use $\Phi=\mathbf{E}\cdot\mathbf{\hat{n}}A$, where $\mathbf{\hat{n}}$ is the unit vector perpendicular to each face of the cube, and $A=(0.1\mathrm{m})^2$. – Winter Soldier Sep 17 '18 at 7:15