# Decay of Electric Field between two plates

1. Feb 5, 2012

### ZedCar

1. The problem statement, all variables and given/known data

I was looking at
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elesht.html
(scroll down a little on the link)

2. Relevant equations

3. The attempt at a solution

Would it be correct to say that the decay in the electric field between the plates is directly proportional to the distance between them? (No just proportional)

2. Feb 5, 2012

### vela

Staff Emeritus
What decay?

3. Feb 5, 2012

### ZedCar

The decay in the magnitude of the electric field? In the way that as the field radiates out in a cylindrical way from a wire the field would be less the further it is from the wire.

4. Feb 5, 2012

### vela

Staff Emeritus
No, the electric field of an infinite plate is a constant. It doesn't depend how far away from it you are. The magnitude doesn't change.

Obviously, the plates of a capacitor aren't infinite, but it's a good approximation for typical geometries. If you separate the plates enough, then the approximation is no longer very good and you would expect the field strength to fall off. To find exactly how it varies with distance is probably quite difficult.

Last edited: Feb 5, 2012
5. Feb 5, 2012

### ZedCar

Okay, thank you.

The question I have is;
So it would appear that the answer to the 'decay with distance' part of the question for all three is, "it does not".

6. Feb 5, 2012

### vela

Staff Emeritus
No, that's not right. The field falls off as you move away from a sphere and, as you mentioned earlier, a wire.

If you look at the problems and examples in your course, you'll probably find that all the problems involve a spherical distribution of charge (includes point charges), a line of charge, or a sheet of charge — or some combination of those. They're the building blocks for solving problems, so your instructor probably just wants you to know the basic properties of these charge distributions.

7. Feb 5, 2012

### ZedCar

Ok, thank you. Yes, I understand now.

Sorry, I'd misinterpreted your 2nd last post.