Calculating Electric Flux of a Cube with Charges

[Ja4Coltrane]

Homework Statement

A cube has a charge Q at its center and has 6 charges q placed symetrically around the central charge. Each one is placed the same distance from the central charge down respective axises which are perpendicular to the planes of the cube which they pass through the center of (does that make sense?) This distance from the central charge is an unnecessary distance. I need to determine the electric flux through one of the six planes.

Homework Equations

Flux=int(E*dA)
Flux=q/(epsilon 0)

The Attempt at a Solution

Well, I'm having a little trouble (I assume my brain just isn't functioning well tonight). I could not see why the distance from the central charge doesn't matter, so I sort of ignored that and assumed that if that is true, it would be fine the put the six charges in contact with the planes.
So I basically added up all of the fluxes. I did Q/(6e) for the middle, q/(2e) for the charge in contact with the surface I care about. I had trouble with the one across from the surface I am fluxing and now realize why my answer for that was wrong. Once I get that one, I can subtract the percents of flux from the one across and the one in contact from 1 and divide that by 4 and multiply that value by q/e and then multiply it by 4.
I know there must be an easier way to do this!

 

[lippyka]

I'm sure I'm missing something simple, but can't figure out what it is. Can anyone help me? Thanks in advance!

 

[m2003]

Hello,

First, let's clarify some things about the problem. The distance from the central charge to the six charges does matter, as it affects the electric field at each point. However, for simplicity, we can assume that the distance is the same for all six charges, and that they are all placed at the same distance from the central charge. Also, it is important to note that electric flux is a scalar quantity, so we do not need to worry about direction when calculating it.

To calculate the electric flux through one of the six planes, we can use the equation Flux = ∫E⋅dA, where E is the electric field and dA is the differential area element of the plane. Since the charges are symmetrically placed around the central charge, we can use Gauss's Law to simplify our calculation. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (Flux = Q/ε0).

In this case, we can consider the cube as our closed surface. The charge enclosed by the cube is Q + 6q, since the central charge Q is inside the cube and the six charges q are placed around it. Therefore, the electric flux through one of the planes is given by Flux = (Q + 6q)/ε0.

If we want to calculate the flux through a specific plane, we can divide this value by 6, since there are six planes in total. So, the flux through one of the planes is given by Flux = (Q + 6q)/6ε0.

I hope this helps clarify the problem and provides a simpler solution for you. Let me know if you have any further questions.