Electric Field of an Infinite Plane

[magda3227]

1. We are given a static system of conductors and charge. We know that x=y is a plane of equal potential (for example, a system that can give such result is an infinite plate in that plane). Which of the following fields can represent an electric field of such system?

a. \vec{E} = az\hat{x} - az\hat{y}
b. \vec{E} = a\hat{x}
c. \vec{E} = a\hat{x} - a\hat{y}
d. \vec{E} = ay\hat{x} - ax\hat{y}
e. \vec{E} = a\hat{x} + a\hat{y} + a\hat{z}


The attempt at a solution:
I figured that since the plane extends infinitely in the z direction, no field can be present there. This is also because of the first axiom of electrostatics relating to the flux being outward normal to the plane. Therefore, choice e is eliminated. In fact, the z axis should have no effect on the field. a is also eliminated.
Also, choice b has no dependency on the y axis, but the E field should be partially dependent on it...b is eliminated.
We also know that the electric field due to an infinitely charged plane is constant. I want to say that the answer is therefore c...I am lead me to believe that choice d can be eliminated, but I'm not sure. I'm stuck between c and d.
Any suggestions?

 

[nickjer]

Some quick things you should know:

You must have,

\vec{\nabla}\times\vec{E} = 0

if there even exists a potential.

Also, you know the electric field will be perpendicular to an equipotential. So take the cross product of the electric field with a vector perp to the x=y plane (cross products of parallel vectors should be 0).

 

[magda3227]

If we say that x=y, then the method of taking a cross product between a vector perpendicular to the plate and the E-field works for both c and d, doesn't it? Or did I just miss something when I did it?

 

[nickjer]

It can't work for both. Just work them out and you will see.

 

[Antiphon]

The electric fied points away from the plane. The unit normal to the plane is given by "c" with arbitrary constant a. C is the correct answer.

 

[nickjer]

It is best if he used the cross product to show the answer is (c) and not (d) since the E-field is perp to the plane, and not give him the answer right away.

 

[Antiphon]

I agree. I only confirmed it a different way for him since he had already deduced the correct answer but not by the cross product or the unit normal methods.