Electric flux of a cube due to a point charge

[The Blind Watchmaker]

Homework Statement

Homework Equations

Gauss's Law

The Attempt at a Solution

I simply used logic and observed the symmetry of the cube. If the charge is placed on the middle of the cube, the flux would simply be Q/∈₀. The face of the cube itself can be split into 4 squares, with the charge at the center. Thus, if it is located at the side, it is simply 1/4th of the previous flux. Thus flux = ¼ Q/∈₀. However, there is no calculation involved as this is pure logic. Is there any calculation available to show the result? Thanks!

 

[kuruman]

However, there is no calculation involved as this is pure logic. Is there any calculation available to show the result?

That's the way I would do it. One can do the calculation, but it would be quite lengthy and would involve two integrals, one over the top (or bottom) face and one over the left (or back) face. The net flux would be the sum of the two integrals times 2. This question illustrates the usefulness of Gauss's Law.

 

[Orodruin]

The ”actual computation” way would be to note that the divergence of the field is zero away from the charge to rewrite the integral as the integral over a quarter of a sphere (and two semicircular shaped planar surfaces at which the field is parallel to the surface). The integral over the quarter of a sphere is trivial.

 

[kuruman]

The ”actual computation” way would be to note that the divergence of the field is zero away from the charge to rewrite the integral as the integral over a quarter of a sphere (and two semicircular shaped planar surfaces at which the field is parallel to the surface). The integral over the quarter of a sphere is trivial.

I note that this method requires the use of Gauss's Law to justify the assertion that the flux through the quarter-sphere is the same as the flux through the cube.

Now that I think about it, the so-called "brute force" method of doing surface integrals over each face is also based on Gauss's Law for the derivation of the electric field due to a charge at the origin, $\vec E(\vec r) = kq\vec r/r^{3}$. Thus, the calculation of the flux may vary from trivial to involved depending on how and at what point one chooses to bring in Gauss's Law.

 

[Orodruin]

In general, I don’t think you will get very far in electrostatics without invoking Gauss’ law a few times. As already noted by the OP, the easiest way forward here is to invoke Gauss’ law in conjunction with symmetry. If possible, this will always give you a simpler computation (I would actually consider it a computation as well - even if you are not explicitly solving integrals). I just mentioned changing integration domain if one reallt feels the psychological need to do a surface integral.

 

[kuruman]

In general, I don’t think you will get very far in electrostatics without invoking Gauss’ law a few times.

I agree.