Mastering Physics Solutions: Flux out of a Cube

Flux out of a Cube

Part A = Φ = q / (6ε0)
Part B = Φ = q / (6ε0)

Solutions Below:

A point charge of magnitude q is at the center of a cube with sides of length L.

Part A

What is the electric flux Φ through each of the six faces of the cube?
Use ε0 for the permittivity of free space.

According to Gauss’s Law, an inside charge is equal to the flux times the permittivity of free space. In other words, we have the the formula:

ε0Φ = q

Where ε0 is the permittivity of free space, Φ is the flux, and q is the inside charge. Since the problem asks about the flux at each of the 6 faces of the cube, we just divide this by 6 to get the answer. We can do this because each face of a cube is equal. If we were talking about a rectangular box or some other shape that wasn’t symmetrical, we would have to take that into consideration, but not for this problem!

ε0Φ = q
Φ = q / ε0

Φface = q / (6ε0)

We could also solve by knowing that the total flux is equal to the electric field at a particular point within an area, times the total area. So since each face of the cube is 1/6th it’s surface area, the total flux on a given face would be given by:

Φ = E * 1/6A

But then we would have to relate E to q, which is beyond the scope of this problem (but you can find it algebraically if you want).

q / (6ε0)

Part B

What would be the flux Phi1 through a face of the cube if its sides were of length L1?
Use ε0 for the permittivity of free space.

The answer is the same as for Part A. The length of the sides doesn’t matter because a cube has equal faces and each face is always 1/6th of the total surface area.

q / (6ε0)

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One Response to Mastering Physics Solutions: Flux out of a Cube

  1. Goliath says:

    Super! Thanks

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