# I Question about a sheet of charge

1. Apr 7, 2016

### toesockshoe

Today my prof said that for a point charge the force is proportional to 1/r^2. For a line charge it is proportional to 1/r. For a sheet charge it is a constant and for a volume it is proportional to r. He also did some derivations but I dont see how this can be.

For example for a sheet charge how can a test charge that is a million meters away experience the same force that an object one millimeter away experiences? Then doesnt that mean if I make a sheet charge of say 100C, then I should be able to provide a resulting force on an object that is across the country ... because a sheet charge produces a uniform electric field regardless of position right? Obviously this doesn't happen because if someone create a sheet charge in Africa...a metal object here certainly doesnt not feel it all the way here in America. Same thing with a volume charge.... How can an object further away from the charge feel a stronger attraction than something closer to the charge?

2. Apr 7, 2016

### axmls

It's always important to understand what assumptions are made when deriving these kinds of equations. In particular, the assumption when deriving the electric field of a sheet of charge is that the sheet is infinite in extent. Obviously, this is not physical, but it is a good approximation when looking at the electric field close to the sheet. It's like the acceleration due to gravity. Certainly objects don't always accelerate with an acceleration of $g$ (after all, the force decreases as you get farther from the Earth), but when close to the surface, it is a good approximation to treat the gravitational field as a constant.

If the sheet is finite, then at far distances, it will look like a point charge. Then you can imagine what the formula will look like, however, from searching around, it is very difficult to get a closed form solution for a finite sheet of charge, hence the infinite sheet approximation.

3. Apr 7, 2016

Thank you.

4. Apr 7, 2016

### rcgldr

If the sheet is a finite disc, and the particle lies on the axis of the disc, the math for the field (multiply by charge of particle in the field to get the force) is shown here:

http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/electric/elelin.html#c3 [Broken]

Last edited by a moderator: May 7, 2017
5. Apr 7, 2016

### axmls

Ah, of course. My brain instantly jumped to a finite rectangular sheet. Yes, a disc has a very simple closed form along the axis through its center, and I'm sure the OP can confirm that we obtain the formula for an infinite sheet if we let the radius go to infinity.

Last edited by a moderator: May 7, 2017
6. Apr 7, 2016

### rcgldr

It the sheet is a rectangle and the particle lies on the central "axis" of the sheet, the sheet could be considered as an infinite number of lines of finite length, where the component of force perpendicular to the lines cancel, so only the component related to z needs to be considered. This is the math for a line, the math for a sheet would involve an integral based on the formula for a line:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html