Surface charge densities on two conducting plates.

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Homework Statement


I have attached the problem with a diagram.


Homework Equations


flux = EA = Qenclo


The Attempt at a Solution


I am confused about how I can find the charge densities on each surface. What I think is that the 5C will spread to both surfaces of plate one so σ1=σ2=5C and σ3=σ4=-6C. However I realize there may be induction and also I am not sure if we can assume the plates are enclosed surfaces... Although they do go to infinity. To be honest, I really have no clue how to approach this question.

Is the electric field at C zero? Since these are conducting I would think that it would act like a faraday cage and shield out all the electric fields.
 

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  • #2
Spinnor
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You wrote,

"What I think is that the 5C will spread to both surfaces of plate one so σ1=σ2=5C and σ3=σ4=-6C."

Yes I think the charge will spread but as charge can't be created or destroyed, what happens I think is that the charges will spread to both sides of the plates such that (σ1+σ2)A=5C and (σ3+σ4)A=-6C where A is the area of the plate. Also the σ's are charge per area so what you wrote above can't be correct as both sides don't have the same dimensions.

You need to draw some Gaussian surfaces, there are only a few different ones possible, and use Gauss's law to relate the known quantities (5C and -6C) to the unknown quantities (σ1, σ2, σ3, and σ4).

Good luck!
 
  • #3
364
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You wrote,

"What I think is that the 5C will spread to both surfaces of plate one so σ1=σ2=5C and σ3=σ4=-6C."

Yes I think the charge will spread but as charge can't be created or destroyed, what happens I think is that the charges will spread to both sides of the plates such that (σ1+σ2)A=5C and (σ3+σ4)A=-6C where A is the area of the plate. Also the σ's are charge per area so what you wrote above can't be correct as both sides don't have the same dimensions.

You need to draw some Gaussian surfaces, there are only a few different ones possible, and use Gauss's law to relate the known quantities (5C and -6C) to the unknown quantities (σ1, σ2, σ3, and σ4).

Good luck!
If that's the case then I have:
σ1 = 2.5*10^-2 C/m2
σ2 = 2.5*10^-2 C/m2
σ3 = -3*10^-2 C/m2
σ4 = -3*10^-2 C/m2

as my surface densities. Also at C it is zero right? Our professor gave us a weird example the other day where there is a sphere with a charged surface (he did not say it was conducting or insulating) and inside is hollow. If we put a +q charge outside the sphere the inside will experience a electric field... Which didn't make sense to a lot of us. What we are guessing is that the sphere is not conducting.
 
  • #4
Spinnor
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If that's the case then I have:
σ1 = 2.5*10^-2 C/m2
σ2 = 2.5*10^-2 C/m2
σ3 = -3*10^-2 C/m2
σ4 = -3*10^-2 C/m2

...
I don't think that is right, they do sum to the right numbers though. One plate effects the other. Draw a cylindrical Gaussian surface whose top and bottom surfaces lie in the middle of the left and right plate. The flux out of that surface is zero? Which implies what.
 
  • #5
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I don't think that is right, they do sum to the right numbers though. One plate effects the other. Draw a cylindrical Gaussian surface whose top and bottom surfaces lie in the middle of the left and right plate. The flux out of that surface is zero? Which implies what.
So you mean the electric field inside each plate is zero? Which means sigma 2 and sigma 3 should cancel out and have a net of zero as well.
 
  • #6
Spinnor
Gold Member
2,170
368
Right, zero net flux zero net charge. That gives you one relationship, you need one or more more relationships which you will get using different Gausian surfaces.
 

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