# Electric Flux through a wall

1. Jun 11, 2007

### krv77

1. The problem is:

A vertical wall (5.9m x2.5m) in a house faces due east. A uniform electric field has a magnitude of 150 N/C. This field is parallel to the ground and points 35degrees north of east. What is the electric flux through the wall?

2. Relevant equations:

Electric flux is found by using E cos of angle between electric field and normal.
E= f/q

3. The attempt at a solution

Well, I tried it this way and don't think it is right.

E= 150 N/C
_______
(5.9 x 2.5m) x cos 35 degrees

I think the answer will be radially outward as oposed to inward, but I am not too sure.

The answer choices we were given are

A. 7.9 x 10*5 radially outward
B. 3.2 x 10*5 radially outward
C.1.4 x 10*6 ......inward
D. 5.9 x 10*5....inward
E. zero

I don't see how it would be zero, and I think I have narrowed it down to either A or B.

Last edited: Jun 11, 2007
2. Jun 11, 2007

### G01

Assume the wall has infinite length and a finite thickness. There is a a flux going into the wall from the left equal to:

$$\Phi_e=EA\cos\theta$$

If this is the case, what does this tell you about the net flux?

Last edited: Jun 11, 2007
3. Jun 11, 2007

### nrqed

You are confusing things, here. In this equation, E is the magnitude of the electric field an d f is the magnitude of the force experienced by a charge q. This equation has nothing to do with flux.

Use the equation provided by GO1. As for the direction, the electric field is pointing north of east so if you picture the E field vectors as representing a flow, you will see that there is a net flow east theorugh the surface (so the flux is positive if you take the area vector of the wall to be pointing east). That should give you the answer.