# Electric Field of an Infinite Plane

1. May 7, 2010

### 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?

2. May 7, 2010

### 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).

Last edited: May 7, 2010
3. May 8, 2010

### 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?

4. May 8, 2010

### nickjer

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

5. May 8, 2010

### 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.

6. May 8, 2010

### 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.

7. May 8, 2010

### 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.