Electric flux through gaussian cube

[cscott]

Homework Statement

I have a cube with four faces parallel to the field and two perpendicular.

The field is non uniform, given by E = 3 + 2x^2 in the +x direction.

The Attempt at a Solution

So I get $$\phi = \int \vec{E} \cdot d\vec{A} = \int EdA = A \int E = A \int (3+2x^2)$$

But how do I evaluate this with no dx?

 

[robphy]

Homework Statement

I have a cube with two faces parallel to the field and two perpendicular.

The field is non uniform, given by E = 3 + 2x^2 in the +x direction.

The Attempt at a Solution

So I get $$\phi = \int \vec{E} \cdot d\vec{A} = \int EA = A \int E = A \int (3+2x^2)$$

But how do I evaluate this with no dx?

Slow down... you need to justify each of these steps.
First, a cube has six faces.
Second, what precisely do you mean by the expression after the second equals-sign (and how do you justify it?) [Hint: it's wrong.]
$$\phi = \int \vec{E} \cdot d\vec{A} = \int EA$$
For each integral, what is the region of integration?

 

[cscott]

Sorry, I meant to say four parallel faces.

and I just realized that the perpendicular faces would have constant E over them since all points on both faces are at the same x value.

I said $$\int \vec{E}\cdot d\vec{A} = \int EdA$$because the angle between E and A vectors is 0

Would this make sense?
$$\int \vec{E}\cdot d\vec{A} = \int EdA = E \int dA = EA$$ and I'm given the dimensions of the cube as well as it's position along the x axis.

 

[robphy]

Technically, you have an integral over the closed surface of the cube... and by breaking the surface up conveniently into six faces, you have the sum of six fluxes through each face (an open surface), four of them being zero. So, what you write actually refers to one of the two remaining faces (which have the same form, but not same value). Note that you need to take the directions of the area-vectors (taken from the original cube) into account.

 

[cscott]

So I basically get +EA and -EA because of the antiparallel area vectors for the two faces?

 

[robphy]

Yes, although each E is "E on that surface used to calculate the flux".

 

[cscott]

Alright, thanks a lot.

 

[mywill1409]

Hi, i am having the same problem. But when you do the calculation of the remaining two faces, will they be zero since the area vectors opposite to each other?

I know this is a non uniform field. And i was thinking perhaps the field can move in any directions even in the negative x,y,z directions. If i were to calculate the net charged enclosed, i should calculate the e-flux first the use EA = q/Epsilon right?

 

[mywill1409]

In addition my cube sits at a distance 'a' from the x-axis. Is there any integration involved? This part confused me the most

 

[mywill1409]

Since the field is [3(N/C)+4(N/C m^2)x^2]i^, and x is in meters. So the field's magnitude should be 7 N/C right?

a=b=0.612m
c=0.7965m

What shall i do next?