What is the Electric Flux through a Vertical Wall Facing East?

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SUMMARY

The electric flux through a vertical wall measuring 5.9m x 2.5m facing east, subjected to a uniform electric field of 150 N/C at an angle of 35 degrees north of east, can be calculated using the formula Φ = EA cos(θ). The area A of the wall is 14.75 m², leading to a calculated electric flux of approximately 7.9 x 105 N·m²/C radially outward. The direction of the electric field indicates a net positive flux through the wall, confirming that the correct answer is option A.

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krv77
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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. Homework 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:
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Assume the wall has infinite length and a finite thickness. There is a a flux going into the wall from the left equal to:

[tex]\Phi_e=EA\cos\theta[/tex]

If this is the case, what does this tell you about the net flux?
 
Last edited:
krv77 said:
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. Homework Equations :

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



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.
 

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