# Net Electric Flux for a Cube

1. Sep 20, 2015

### Kurtis McIntosh

1. The problem statement, all variables and given/known data
Assume the magnitude of the electric field on each face of the cube of edge L = 1.07 m in the figure below is uniform and the directions of the fields on each face are as indicated. (Take E1 = 35.1 N/C and E2 = 25.3 N/C.)

A.) Find the net electric flux through the cube.

B.) Find the net charge in the cube.

2. Relevant equations
Electric Flux = Electrical Field * Area

3. The attempt at a solution

I don't really understand how to solve for the net electric flux in the problem. I thought you would just take the difference in forces for each direction, multiplied by the area, then solve to find the magnitude using the x and y fluxes, but I'm clearly missing something here. I've included a photo of the cube. I've also tried solving for the flux on each individual face and then finding the sum of the fluxes, but that doesn't seem to work either.

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2. Sep 20, 2015

### Staff: Mentor

Hi Kurtis, Welcome to Physics Forums.

Your Relevant Equation holds the key. The flux through a given area is given by the field strength multiplied by the area (for a uniform field over the area, you need to get fancier and do an integration if the field is not uniform). Note in the diagram that some fields are directed out of the face while others are directed into the face. That is to say, some flux is leaving the cube and some is entering it.

Assign a positive sign to flux leaving the cube and a negative sign to flux entering. Sum them up (algebraically).

If there is a net flux leaving or entering then there must be a charge inside the cube (See Gauss' Law).