# Line Charge and Sheet Distribution

1. Jun 1, 2010

### dhruv.tara

When we consider the electric field from an infinite line charge we see that electric field falls off as a function of 1/r while it is constant in case of a infinite sheet... How come such differences arise when both the charge sources are infinite in nature? (I didn't get the explanation in the book)

2. Jun 1, 2010

### GRDixon

Although the total charge is infinite in each case, practically 100 % of that charge is located an infinite distance from your field evaluation point. The easiest way to appreciate the "why" of the difference is to use Gauss' law, plus symmetry considerations. If you have misgivings about the adequacy of Gauss' law, you will probably benefit from coming to grips with the idea that it is correct because of Coulomb's law. Coulomb's law, in turn, is generally accepted as axiomatic in classical theory.

3. Jun 1, 2010

### pgardn

It seems to me an infinite line of charge would have to fall of if you just think about the E-field lines emanating from it. They will all pop out radially from the line. If you look at the line head on you would "see" field lines emanating out radially. But a point in space would get some field lines from all parts of the wire, and through symmetry if you go to the left and right at some particular point, each little piece of paired charges an equal distance from the point of interest would cancel contributions from a left and right perspective but would add up out from the wire. As you move left and right further from the point of interest this of course dies off towards zero.

In an infinite sheet there is nowhere to go around this sheet where the field dies off, its an infinite plane. I will attempt a drawing.

In my drawing I take three little pieces of charge on a line of + charge and draw their E-fields on a point in space. Notice how the left C1 and the right C3 pieces canel each other out in the horizontal or x direction (left and right) but add together with C2 which is the closest piece of charge to that particular point. Now if you moved the P further away. All these vectors would get smaller and eventually approach zero.
But in the plane there is no die off. Any piece of charge on that infinite plane has a partner somewhere else on that plane where it cancels as before (symmetry). But there are always other pairs 360 degrees around the plane. So the field is uniform and there is no die off. The field may be weak in an infinitely charged plane due to a small amount of charge density, but it does not die off the further away you get. I could have drawn more field lines or fewer, but they would not die off as I can never reach an edge of the plane where the charge ends. One can never reach the end of the line of charge either, but as you get further away from it there are not other lines of charge stacked one on top of another like in a plane. In fact, you can think of a line of charge as reaching the edge of a very long long plane of charge but only the edge of the plane makes a contribution to the field.

Hope this helps. If you want to understand why its 1/r and prove the infinite plane has no die off mathematically you must use Gauss. But intuitively this is how I see it (the above and the picture). And intuition does not explain the 1/r instead of 1/r^2, like for a point charge, for me anyway.

Last edited: Jun 1, 2010
4. Jun 2, 2010

### dhruv.tara

Thanks, I needed an intuition only. I had got the mathematical forms before. Also the same intuition till an extent holds for the difference of 1/r and 1/r^2 compared to the infinite line charge and the point charge. This holds as in infinite line charge there is always some component that adds up to make the additive part of the field no matter where the location of interest lies. This amount of addition is the key difference for the infinite line charge field to fall slower than that of the point charge.
Since the additive component is theoretically 0 only at infinity (the charges at infinite ends of the rod have a purely vertical field) therefore the field is not constant and it falls. That's a little qualitative analysis of the difference.

Anyway thanks your explanation of the sheet charge distribution was a great help... :)

5. Jun 2, 2010

### pgardn

You are right. I just could not see that it would be 1/r v. 1/r^2 without the math. Heck, it could be 1/r^1/2... I just cant see it clearly without the math.

6. Jun 2, 2010

### dhruv.tara

Yea... That cannot be seen without maths. Its only that you can guess that it will fall slowly than 1/r^2 but it will also not be a constant (i.e. it will fall for sure)... I think rest will be the work of maths only...