- #1

bluelensflares

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## Homework Statement

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.

## Homework 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.

## 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!