# Homework Help: Flux Through a Cube

1. Feb 7, 2010

### r34racer01

A cube, placed in a region of electric field, is oriented such that one of its corners is at the origin and its edges are parallel to the x, y, and z axes as shown to the left. The length of each of its edges is 1.3 m.

(a) What is the flux through the cube if the electric field is given by E = (3.00y)j ?
Φ = ? Nm^2/C

(b) What is the flux through the cube if the electric field is given by (-4.00)i + (6.00+3.00y)j ?
Φ= ? Nm^2/C

(c) For part (b), how much charge must be enclosed by the cube?
q = ? C

Φ = int(E * dA)

I get the feeling this should be really easy but so far its not. So for part a the field is going through the top and bottom surfaces and the formula I believe should just shrink down from E dot dA the just E*A so if we do it for just one surface it should be (3e-6)(1.3^2) = 5.07e-6 but since it goes through 2 faces parralel with the area vectors summing those two should make it equal to zero correct?

For part b it looks like the electric field vector doesn't even pass through the cube (4 to the left and 9 up from the origin?) So shouldn't the answer be zero?

For part c I assume that when I do get the right answer I can just set it equal to Qencl/ε and solve for Q right?

2. Feb 8, 2010

### gabbagabbahey

It's not clear to me what you mean when you say this? What face are you calling the top, and what face are you calling the bottom? The electric field you are given permeates all space, so what do you mean it is "going through the top and bottom"? Isn't it going through every surface?

The equation,

$$\int_{\mathcal{S}} \textbf{E}\cdot d\textbf{A}=EA$$

is generally only true if $\textbf{E}$ is uniform (constant) over the surface $\mathcal{S}$ and everywhere perpendicular to it.

Which two surfaces is $\textbf{E}$ perpendicular to? What is the value(s) of $\textbf{E}$ for each of those two surfaces?

3. Feb 8, 2010

### r34racer01

What i meant to say is that for pt a at least the field will be parallel to 2 side of the cube, the sides that are parallel to the yz-plane I believe. The field will be perpendicular to the other 4 sides so the flux there should be zero. The flux through the other 2 sides parallel will be equal and opposite, so total flux will be zero. But that's not the right answer so what is wrong with my logic?

4. Feb 8, 2010

### gabbagabbahey

You seem to be confused on the concepts of "perpendicular" and "parallel" when it comes to surfaces.

What direction would you say $d\textbf{A}$ points in for the face located at

(i)$z=0$
(ii)$z=1.3\text{m}$
(iii)$x=0$
(iv)$x=1.3\text{m}$
(v)$y=0$
(vi)$y=1.3\text{m}$

5. Feb 8, 2010

### r34racer01

For z = 0, that's the origin so I couldn't say.
z = 1.3, I'd say E is perpendicular to the area
x = 0, again can't say.
x = 1.3, I'd say that's also perpendicular to E.
y = 0, again can't say.
y = 1.3, that would be parallel, no?

I can see now why u might be confused by my logic, I think I was visualizing axes in my head different to the picture i've posted, so really in part a the E field points to the right correct?

6. Feb 8, 2010

### gabbagabbahey

What do you mean? I'm not asking you about the point x=y=z=0, I'm asking about the surface located at z=0 (its x and y values vary from 0 to 1.3m over the surface, but every point on the surface is at z=0)

7. Feb 8, 2010

### r34racer01

I'm not quite sure what you mean. At z = 0, 3 of the cube's sides intersect, so I'm not sure which side you're referring to.

8. Feb 8, 2010

### gabbagabbahey

I'm talking about the face/side where every point on the face/side is at z=0, not just the points along one edge of the face.

In the picture you've posted, that would be the bottom face (the top face would be the one at z=1.3m)

9. Feb 8, 2010

### r34racer01

Ok I finally made some progress here. For part a the answer is Φ = E*L^3. I kinda understand why its L^3 but I could use some clearing up. The answer is the same for part b. Part c is where I'm stuck now. I used the formula Q = Φ/epsilon = 7.447e11 but that's wrong. Can someone explain what went wrong here?

10. Feb 8, 2010

### gabbagabbahey

No, it isn't.

Well, how did you get this (incorrect) answer?