# Electric Field and two parallel plate capacitors

1. Jan 27, 2007

### chaoseverlasting

If you have two parallel plate capacitors side by side and the electric field in one is constant. If you vary the electric field in the other, will it interfere with the electric field that was constant?

2. Jan 27, 2007

### arunma

Theoretically no, since we generally consider the electric field to be entirely contained within the parallel plate capacitor. In practice, there are fringing effects. So the far end of the "constant" capacitor will have a slightly different electric field. Other than that, I can't think of anything that would change the field appreciably.

3. Jan 27, 2007

### Minjin

I just registered on this forum to ask a question about electric fields so I might as well ask in this thread rather than create a new one.

How is it possible to have a uniform electric field?

I know we simplify things to make them easier to understand and calculate but my teacher was pretty insistent (when questioned) that a parallel plate capacitor creates a uniform field. I just don't see how its possible. If field intensity varied inversely with distance, then yes. But field intensity varries inversely to the SQUARE of the distance and I don't see how you can combine the two graphs to form a straight line. I know you can get them close enough that it might as well be straight, but will it ever be straight? Anyone understand what I'm saying?

4. Jan 27, 2007

### cesiumfrog

That square thing is an artifact of the third dimension, and so disappears when you add a couple symmetries. If you have electric field lines emanating from just a point in space, the density of those lines (like the surface area of a sphere) is proportional to the square of the distance. For an (almost infinite) parallel plate capacitor (which for specific reasons will have a uniformly distributed electric charge), the field lines have no direction to spread out into, therefore their density remains constant at any distance.

5. Jan 27, 2007

### Minjin

Ok, I can certainly visualize that the lines will remain parallel and won't spread out as they go from one plate to the other. I just can't seem to convince myself that the intensity of the field doesn't go down. I kind of see it like a rope held horizontal. No matter how hard you pull on both ends, there's going to be some sag in the middle due to the weight of the rope. I'm not saying there's tension or gravity involved here , I'm just saying thats how I visualize what the graph of intensity across the gap looks like.

6. Jan 27, 2007

### arunma

Yes I do, and that's an excellent question. Before I begin, let me preface this by saying that technically, the electric field between the two plates of a parallel plate capacitor is not perfectly uniform, and I think that if you press your teacher, he will certainly say so as well. It's close to uniform, but as I said earlier, there are fringing effects at the edges of the plates.

The reason we say that the field between the plates is uniform is because we treat the plates as infinite sheets of charge. To a first approximation, this yields accurate results. So at this point you should be wondering: why does the electric field of an infinite sheet of charge not decrease with distance?

One way to answer that question is, "because the math works out that way." I realize that this isn't very intuitive, but if you really want to convince yourself that the electric field created by an infinite sheet of charge doesn't vary with distance, then you can calculate the electric field along the axis of a circular sheet of charge with finite radius r, and then let r go to infinity. Don't use Gauss' Law; calculate it directly from Coulomb's Law (if you use Gauss' Law, then you're beginning with the assumption that the field is uniform, and you won't gain the understanding of how this works).

Here's the intuitive explanation. Imagine that you were standing on an infinite sheet, and that you were able to increase your altitude arbitrarily. You would be unable to determine how far you were from the sheet! In real life, people are able to determine their altitude from the surface of the earth by two means: the varying texture of the earth, and the horizon. An infinite sheet has no horizon, and it has no variance in texture. As such, it is meaningless to even define your distance from a purely infinite sheet. The reason the electric field created by an infinite sheet is uniform is because if you were to place a test charge near the sheet, the charge wouldn't have any way of telling how close it is to the sheet.

A good analogy is the gravitational field near the surface of the earth. You know that gravity works like electric fields: it goes off as 1/r². But the gravitational field near the surface of the earth is uniform. A plane flying at 30,000 feet feels effectively the same gravitational force as a person standing on the ground. That's because from the perspective of a ground observer, the earth can be approximated as a flat surface, and the field is uniform. An electron between capacitor plates is like a person on the earth. Both are under the influence of infinite sheets of charge/mass, and so the field does not fall off as 1/r² as they move away from the sheets.

BTW, one other thing. You'll note that this whole time, I've been talking about one single infinite sheet, when in fact a set of capacitor plates is really two infinite sheets. Keep in mind that one plate is positively charged, and the other is negatively charged. The first plate has field lines emanating away from it, and the second plate has field lines coming towards it. So the electric field between the plates is equal in magnitude to twice the field of either plate by itself, and zero outside the plates (don't take my word for it, draw a quick sketch and see for yourself!). Otherwise, everything I've said about single infinite sheets is perfectly valid.

Well, I hope that helps!

Last edited: Jan 27, 2007
7. Jan 28, 2007

### chaoseverlasting

In practice, how could you set up a system of two parallel plate capacitors with one constant field and one in which you could vary the field?

I know $$E= \frac{\sigma}{2\epsilon}$$
$$E_{net}=\frac{q}{A\epsilon}$$
For a given potential difference, $$V=d \frac{q}{A\epsilon}$$
and hence $$C=\frac{A\epsilon}{d}$$

From this it seems that the capacitance of the capacitor is a purely geometrical property.
Therefore, $$dV=\frac{dq}{C}$$
and $$dE=\frac{-dV}{dr}=\frac{-dq}{Cdr}$$
Integrating this you would get the value of the electric field...right?

Last edited: Jan 28, 2007