How Do You Calculate Total Flux Through a Cube with Non-Uniform Electric Field?

[joemama69]

Homework Statement

A cube has lengths L.

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

Find the total Flux through the cube.

Homework Equations

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

 

[HallsofIvy]

Homework Statement

A cube has lengths L.

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

Find the total Flux through the cube.

Homework Equations

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

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.