# Homework Help: Flux through Cube

1. Apr 27, 2010

### joemama69

1. The problem statement, all variables and given/known data
A cube has lengths L.

E(x) = {a(x+L)2i - (ayL)j where a is constant

Find the total Flux through the cube.

2. Relevant equations

3. The attempt at a solution

i know the flux is 0 through the z = 0 and z = L planes but the other sides and the top & bottom i am unsure how to find.

I am confused about the x,y, & L variables. It seems that x & y = L. When I integrae over the area am i doing a double integral for x & y from 0 to L

2. Apr 27, 2010

### HallsofIvy

There is NO "L variable", L is a constant, not a variable.

If you are integrating the flux over the top and bottom, yes, you are integrating x and y from 0 to L.
In particular, [/itex]\vec{n}dS[/itex] for the top and bottom are $\vec{k}dxdy$ and $-\vec{k}dxdy$ respectively. Since your given function has no $\vec{k}$ component, yes, the flux through those is 0.

Now consider the face x= 0, in the yz-plane. A unit (outward) normal is $-\vec{j}$
and $\vec{n}dS= -\vec{j}dydz$.

$E\cdot \vec{n}dS= -(x+ L) dydz$ but we are in the face x= 0 so that is just $-Ldy dz$ and the flux is $\int_{y=0}^L\int_{z=0}^L -L dy dz$ which is just -L times the area of the face, $-L^3$

On the face x= L, everything is the same except that the outward normal is $\vec{k}$ rather than $-\vec{k}$ and x+ L is now 2L. The flux through that face is $\int_{y= 0}^L \int_{z= 0}^L 2L dydz$ which is 2L times the area of the face, $2L(L^2)= 2L^3$.

On the face y= 0, in the xz-plane an (outward) normal is $-\vec{j}$ and $\vec{n}dS=-\vec{j}dxdz$.

$E\cdot\vec{n}dS= -a(0)L(-1) dxdz= 0$ so the flux through that face is 0.

Finally, on y= 1, the (outward) normal is $\vec{j}$ and $\vec{n}dS= \vec{j}dxdz$.

$E\cdot\vec{n}dS= a(L)(L) dxdz$ so the flux through that face is $\int_0^L\int_0^L aL^2 dxdz$ which is $aL^2$ times the area of the face: $(aL^2)(L^2)= aL^4$.

The flux through the entire region is the sum of those.