# Symmetry and electric fields

1. May 28, 2013

### Nikitin

Hey. Let's say you have an infinitely wide and long disk with a thickness h. Inside the disk, there is a constant charge density ρ0.

Why would the electric field lines be perpendicular to the disk?

Can somebody explain how symmetry and such generally affect electric field distribution of a charged object?

thanks! :)

2. May 28, 2013

### WannabeNewton

Hi Nikitin. Assume that at some point on the disk, the electric field vector points at an angle away from the perpendicular. Note that the system (infinite disk) exhibits rotational symmetry about this point so if I rotate by $\pi$ radians, the electric field must remain unchanged. However, this rotation will take the electric field vector and make it point in the opposite direction but the rotation must leave the electric field at this point unchanged so it must simultaneously point in the original direction as well, which is a contradiction. Thus, the electric field vector must point along the perpendicular in order to avoid this contradiction.

3. May 28, 2013

### Nikitin

hmm, that's a very smart trick! thanks allot!

But what if the object is not a disk, but a random, very large "flat" surface with a thickness h and constant charge density?

If the surface is large enough, shouldn't the field at its centre be approximately constant? Why?

4. May 28, 2013

### WannabeNewton

Do you mean like an infinite sheet? An infinite sheet has no absolute center i.e. if you pick any point on the sheet, the sheet will look rotationally symmetric about that point. Consequently, the same argument from before applies.

5. May 28, 2013

### Nikitin

Okay, thanks!

6. May 28, 2013

### WannabeNewton

Good luck in your studies friend!