# Three infinite sheets creating two regions - find the potential difference

1. Feb 4, 2013

### bluelensflares

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

Hi all, I have this problem that was on a recent exam but I did not know how to make sense of it.

So, suppose you have three infinite sheets, each layered on top of each other each separated by a distance d. So the first is d above the second, and the second is d above the third. The first has a surface charge density of n, the second sheet n, and the third sheet -n. The first part, which I believe I got correct, was to calculate the electric fields of the two inner regions. The second part, which tripped me up and made me panic, was to calculate the difference in potential between the two regions.

2. Relevant equations

E(for an infinite sheet) = n/(2e), where e is the permittivity constant, 8.85E-12.

E(net) = E1 + E2 +E3...

V = Es, for two infinite sheets, where s is the distance from the negative sheet.

3. The attempt at a solution

So for the first part, calculating the fields, I just added them up and watched the directions: so for the top region, E = (n/2e) + (-n/2e) + (n/2e), since they don't depend on distance, and found the top region to be E = n/2e.

The bottom region was just E = 3n/2e, since they all have fields that point downward at the bottom sheet.

The second part I was stuck - since V = Es, where s is the distance from the negative sheet, I tried to calculate ΔV = Vf- Vi by subtracting the potentials. The problem is, we're talking about regions, not well-defined points... So I was at a loss about what to use for s. Also, instantaneously, it goes from the top voltage to the bottom voltage as you cross the sheet... So I was wondering how to do this. I ended up putting the difference in the fields, and getting ΔV = n/e... but I'm not sure about this.

Thanks!

2. Feb 4, 2013

### Simon Bridge

What is the relationship between the electric field and the potential - in general?

Notes:
Technically there's four regions - but you don't care about the regions outside the outermost sheets?

Were you careful to take into account that the electric field is a vector, when you did the superposition?

3. Feb 4, 2013

### Dick

E=n/2e is just giving you the magnitude of the E field. The direction is different depending on which side of the sheet you are on, right? Suppose z is the coordinate defining the separation between the sheet and the sheet is located at z=b. Pick $\hat z$ to be a vector pointing in the positive z direction. Then above the sheet the E field is $\frac{{\hat z}n}{2e}$ and below it's $\frac{{-\hat z}n}{2e}$. Can you work out the potential function for that sheet? Hint: it will involve |z-b|. Once you have the potential function for a single sheet you can just add the potential functions for all of your three sheets.

4. Feb 4, 2013

### ehild

You got the electric field correctly, and you have the proper formula for the potential difference ΔV=Es where s is the distance between the two plates. Here s=d . The potential decreases in the direction of the electric field. Is plate 1 at higher or lower potential than plate 2? What is the electric field between them?

Remember, the potential difference was defined by the work done on unit positive charge. All points of one plate are at the same potential. No work is done when a charge is moved along a plate. Also the work does not depend on the actual path taken between two points.

So the work is the same if you move that unit charge from a point of one plate to any point of the other one - equal to the work between opposite points.

ehild