Electricity and magnetism - planar symmetry question: plastic sheets

[thehammer]

Homework Statement

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.

Homework 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.

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 1^st plate to the 2^nd plate for the fields that result from the positive and negative charged plates. I did the same for the 2^nd and 3^rd.

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!

 

[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

 

[thomate1]

Dear student,

Thank you for sharing your approach to this problem. It seems like you have a good understanding of the setup and the equations involved. However, there are a few things that could be clarified.

Firstly, when you say "charge on sheet 2 must be different to both sheet 1 and 3", do you mean that the charge on sheet 2 must be of opposite sign to the charges on sheets 1 and 3? It is important to be clear about the signs of the charges, as this will affect the direction of the electric fields and the equations you set up.

Secondly, in your equations, you have used the electric field strength (E) to represent both the field between sheets 1 and 2 and the field between sheets 2 and 3. This may lead to confusion. It would be clearer to use subscripts to differentiate the fields (e.g. E12 for the field between sheets 1 and 2, and E23 for the field between sheets 2 and 3).

Now, to solve the problem, you can use Gauss' law to relate the electric field to the charge density (\sigma). For a planar symmetry, the electric field is constant and perpendicular to the sheets, so the equation becomes E = \sigma/\epsilon0. You can also use the fact that the electric field between the sheets is 0 outside of the 3rd sheet to set up a system of equations. For example, for the 1st and 2nd sheets, you can write:

E12 = \sigma1/\epsilon0 + \sigma2/\epsilon0 = 2x105

E23 = \sigma2/\epsilon0 + \sigma3/\epsilon0 = 6x105

And for the 2nd and 3rd sheets:

E23 = \sigma2/\epsilon0 + \sigma3/\epsilon0 = 6x105

Eout = \sigma3/\epsilon0 - \sigma2/\epsilon0 = 0

From here, you can solve for the charge densities on the 2nd and 3rd sheets (\sigma2 and \sigma3) and then take the ratio to find the answer to the question.

I hope this helps. Good luck on your exam!

Best,