Net Flux through a cube with a varying electric field only along the Y-axis.

[BeRiemann]

Homework Statement

An E-field is given as [itex]\vec{E}[/itex]_y = b[itex]\sqrt{y}[/itex][itex]\hat{j}[/itex] V/m. Find the net flux through a cube with vertex at (0, a, 0) and side lengths a. (A picture is attached, but it is essentially the cube that would typically be at the origin, shifted along the y-axis by a units) (You can use the divergence theorem or evaluate the flux directly)
I would like to know both methods, as the next few questions specify.

Homework Equations

Net Flux = (charge)/([itex]\epsilon[/itex]₀) = integral(Divergence dot E-field)dV

The Attempt at a Solution

The dot of divergence and E-field yields (b/2)(y^(-1/2)). This is where I'm lost, as I'm not sure how to integrate with respect to volume if the field is only along the y-axis. I've played around with it a little bit and got the answer ((b*a^3)/2)((1/sqrt(a)) - (1/sqrt(2a))) but I do not think this is correct. Any help or hints is appreciated.

 

[SammyS]

I don't see any attachment. Is another vertex at (0, 2a, 0) ?

Yes, the divergence of E is (b/2)(y^(-1/2)), but that is just a scalar. It has no direction after taking the divergence. Integrate this over the volume of the cube.

 

[BeRiemann]

Yes, another vertex is at (0,2a,0), I don't have a scanner with me to upload the visual. The other vertexes are at (0,a,a), (0,2a,a), (a,a,0), (a,2a,0), (a,a,a), (a,2a,a).
If it's just a scalar is my answer simply the integral at (0,a,0) minus at (0,2a,0) where v = a^3? I think that's what I did.

 

[SammyS]

Yes, another vertex is at (0,2a,0), I don't have a scanner with me to upload the visual. The other vertexes are at (0,a,a), (0,2a,a), (a,a,0), (a,2a,0), (a,a,a), (a,2a,a).
If it's just a scalar is my answer simply the integral at (0,a,0) minus at (0,2a,0) where v = a^3? I think that's what I did.

No. It is simple, but not that simple.

Do you know how to write a volume integral in rectangular coordinates: x, y, z ?

Otherwise you could use dV = a²dy .

 

[BeRiemann]

He's not keen on us using triple integrals yet, so I'll have to use the substitution. So in this case would the answer actually be (a^2)(b)((-sqrt(2a)) + (sqrt(a)))?

 

[SammyS]

Isn't the integral from y=a to y=2a ?

I get the opposite sign.

 

[BeRiemann]

Ah, I took the convention that flux out is negative and flux in is positive.