# Electricity and magnetism - planar symmetry question: plastic sheets

1. Jan 5, 2009

### thehammer

1. The problem statement, all variables and given/known data
Three plastic sheets that are large, parallel and uniformly charged are placed side-by-side. The electric field strength between sheets 1 and 2 is 2x105; the electric field strength between sheets 2 and 3 is 6x105. What is the ratio of the charge density on sheet 3 to that on sheet 2. Either side of the plates at the far end the field strength is 0.

2. Relevant equations
I don't know how to use Latex properly and don't have the time right now to find out so I'll just have to state the names of some of the equations.

Gauss' law.
E=$$\sigma$$/2$$\epsilon$$0.

3. The attempt at a solution
Okay. So what I did is firstly draw three sheets in a row. For an electric field to arise, the charge on sheet 2 must be different to both sheet 1 and 3. This implies that charge on sheets 1 and 3 must have the same sign.

Assuming the charge on the central plate to be negative, I drew two arrows going from the 1st plate to the 2nd plate for the fields that result from the positive and negative charged plates. I did the same for the 2nd and 3rd.

Consequently, three equations can be formed,

For the 1st and 2nd plates:

E = E(+) + E(-) = 2x105 = ($$\sigma$$/2)$$\epsilon$$0 + $$\sigma$$/2$$\epsilon$$0.

For the 2nd and 3rd plates:

E = E(+) + E(-) = 6x105 = ($$\sigma$$/2)$$\epsilon$$0 + $$\sigma$$/2$$\epsilon$$0.

Outside of the 3rd plate:

E = E(+) - E(-) = 0

---

I'm not sure about where to go from here. If the question or my answer isn't clear I could try to take a picture of the diagram in the book along with the question. I have an exam in electricity and magnetism in less than 2 weeks so I'm a little concerned. None of my friends that I have spoken to know what to do about this question either!

2. Jan 6, 2009

### Carid

thehammer,

I'll take a shot at answering this but if I'm in error I'll be happy to learn better.

The central sheet serves as support for both electric fields.
Its total charge must be the same as the total charges on the two outside sheets.
Since the sheets are all the same size its charge density must be higher than that on the outside ones.

So

charge on sheet 2 = charge on sheet 1 + charge on sheet 3
charge density on sheet 2 = charge on sheet 2 / identical sheet area
therefore by substitution
charge density on sheet 2 = (charge on sheet 1 + charge on sheet 3) / identical sheet area
whereas
charge density on sheet 3 = charge on sheet 3 / identical sheet area
now
the ratio you want is = charge density on sheet 3 / charge density on sheet 2
eliminating the identical sheet areas this simplifies to
the ratio you want is = charge on sheet 3 /(charge on sheet 1 + charge on sheet 3)
relating these to the electric fields
the ratio you want is = electric field on 3 /(electric field on 1 + electric field on 3)
the ratio you want is = 6*10e5 / (2*10e5 + 6*10e5 ) = 6/8
the ratio you want is = 0.75