Force on Three Parallel Electrical Sheets

[rbrayana123]

Homework Statement

Consider three plane charged sheets, A, B and C. The sheets are parallel with B below A and C below B. On each sheet, there is a surface charge of uniform density:
-(4/3) * 10^-5 (Call α)
-(7/3) * 10^-5 (Call β)
-(3/3) * 10^-5. (Call γ)

(The density given includes charge on both sides of the sheet). What is the magnitude of the electrical force on each sheet, in N/m^2

Answers are 3.2pi, 1.4pi and 1.8pi.

Homework Equations

For an infinite plane, the electrical field strength is:

σ/2ε where ε is 8.85 * 10^-12 and σ is charge density.

Also, Force = Electric Field * Charge

The Attempt at a Solution

dF = EΩdA or
dF/dA = EΩ where Ω is the charge density of the plane whose force is being calculated

(Can someone verify if this line of reasoning is correct?):

The electric field for any patch of area on A should be the average of the electric fields above and below. This comes out to simply be the superposition of the electric fields generated by sheets B & C because the electric field generated by A cancel out on both sides.

Therefore, for a small patch of area on A, E:

E = (β+γ)/2ε

Finally:

dF/dA = (β+γ)α/2ε = 8pi

For Sheets B & C:
dF/dA = (abs(-α+γ))β/2ε = 1.4pi
dF/dA = (α+β)γ/2ε = 6.6pi

EDIT: So I did a quick test to check something. Apparently if I subtract the two density charge contributions for the electric field for Sheets A & C, I get the right answer but this has only made sense for Sheet B because the electric fields were in opposite directions. For Sheet A, the electric fields are both pointing down towards B & C while for Sheet C, the electric fields are both pointing up towards A & B.

Also, I'm curious as to why the answers have pi in it. All my calculations simply didn't require it. Is there another way to do this?

 

[ehild]

Your solution looks correct. I think pi is used to replace 1/ε with 4pik≈36 pi 10⁹ which makes calculations with hand possible.

ehild

 

[rbrayana123]

Thanks. Guess the book was wrong =)