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A Charged Conductive Plate with a Charged Plane

  • #1
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Homework Statement


In the "Before" scenario, there is conductive plate in electrostatic equilibrium with uniform surface charge density +η (the plate has some thickness but the width and length are significantly larger). In the "After" scenario, an infinite plane of negative charge with fixed uniform surface charge density -η is placed parallel below the plate. The plate is allowed to re-establish electrostatic equilibrium.
HW1_zps4gdkwkzp.png

a) Find the electric field value at three points, (1) above the plate, (2) between the plate and plane, (3) below the plane.
b) The surface charge densities of the two sides of the plate.
Only η and ε0 are known values.

Homework Equations


Econductor.surface = η/ε0
Eplane = η/2ε0

The Attempt at a Solution



I'm assuming, since the plate has a finite length, is in electrostatic equilibrium, and we aren't given width and length, that we use the electric field for a conductive surface.

I originally thought of the plate having positive on both the top and the bottom, but then I started thinking of the positive charge in the plate being attracted towards the plane leaving the top negative, or would it just be neutral? Honestly, I've spent so long on this problem, I've started to confuse myself.

Currently, assuming that both the top and bottom of the plate are positive, I have for part a:
(1) Etotal = η/ε0 (UP) + η/2ε0 (DOWN) = η/2ε0 (UP)
(2) Etotal = η/ε0 (DOWN) + η/2ε0 (DOWN) = 3η/2ε0 (DOWN)
(3) Etotal = η/ε0 (DOWN) + η/2ε0 (UP) = η/2ε0 (DOWN)

But when I started looking at part b, I started second guessing myself and I have found absolutely nothing anywhere to help clarify my confusion.

Originally, I was thinking that for part b:
ηTop = Etotal.1ε0
ηBottom = Etotal.2ε0
But that didn't get me anywhere, since I was still in terms of η
 

Answers and Replies

  • #2
18
1
Find out the charges first on each surface first.
Make an imaginary Gaussian cuboid with 2 of its faces "inside" the plates. Using that ,we find that all the charge gets deposited on the outermost surfaces,in this case the cumulative charge of the plates is 0 hence the outer surfaces of both the plates have 0 charge. Since charge can't travel from 1 plate to another,the Inner surface of the upper plate has eta charge and the lower plate gets negative eta charge.
I suggest proceeding from here.
Note;instead of applying gauss law, we can reason that the negative charge of the lower plate attracts the positive charge from the upper plate.
 
  • #3
Titan97
Gold Member
451
18
Is the bottom plate conducting as well?
 
  • #4
18
1
Yes,it must be.all the charge on the bottom plane rushed upwards,leaving the lower surface neutral and the upper surface(of the plane) negatively charged.
 
  • #5
Titan97
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  • #6
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I was assuming that since the plane is infinite and has a "fixed uniform surface charge density -η", that it would not be considered a conductor and the charge wouldn't move, regardless of what was placed next to it.
So would the plate be neutral on top and positive on bottom, since all the charge is attracted to the plane, while the plane remains negative on both sides?
 
  • #7
18
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If we consider the plane to be very thin,[lets say an atom thick], it can be treated as a conductor and that leads us to conclude that the charge is free to move either to the upper side of the plane or to the lower side.

If due to some reason, the assumption of taking the plane as conducting is wrong, the plane will posses an eta charge on both the surfaces
 
  • #8
Titan97
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@vidit jain I am not sure if thin plates can be considered as conductors. Do you have any reference to support your statement?
 
  • #9
18
1
Since it's a plane , it has a thickness approaching 0. Assume the plane to be z=0. Now all the charge on the plate is confined to z=0+dx(just above 0) and z=0-dx(just below 0).
Since the positive charge from the plate attracts the charge on the plane, the charge would be concentrated towards the 0+dx portion. Thus it will be concentrated on the 'upper surface' of the plane.

My main point is that since the plane has a negligible thickness, it doesn't really need to 'conduct charge' from upper surface to lower surface. Also, the net charge on the plane will remain 'fixed '.

Ps. Please use the terms 'plate' and 'plane' carefully.
 
  • #10
TSny
Homework Helper
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Let ηTopand ηBottom be the unknown charge densities on the top and bottom surfaces of the plate.

What is the expression for the electric field at point 1 due to just ηTop?
 
  • #11
18
1
(ηtop)/2ε
 
Last edited:
  • #12
TSny
Homework Helper
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(ηtop)/ε
This is not correct.

Since this is Thio's thread, it will be best to let him/her attempt to answer first. But you are very welcome to contribute to the discussion once he/she offers an answer.
 
  • #13
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I missed a 2 in the denominator. was that the error?
If not, what would be the correct expression?
 
  • #14
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I figured that I should share my professors response, so you don't keep going around in circles.

If we denote the top surface of the plate A, then the area of the plate Aplate=2A
Then +η=Qplate/Aplate, therefore Qplate=+2ηA
IMPORTANT: No matter how the initially uniform charge distribution on the plate is changed by the presence of the plane of charge, the total charge on the two surfaces of the plate must still sum to Qplate=+2ηA
Thus: ηtopA + ηbottomA = +2ηA

Essentially, we have three infinite sheets of charge of uniform densities: ηtop, ηbottom, and -η.

So we were incorrect in saying that the charge of the plate moved at all.
At some point during my thinking process, I had an inkling that the surface area would affect it, but it got lost in the conductor argument.
 
  • #15
18
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Essentially, we have three infinite sheets of charge of uniform densities: ηtop, ηbottom, and -η.
I agree with this, and this is essentially what I wanted to say. But in order to find the electric fields at various positions, we need to find ηtop and ηbottom individually. That may be achieved using gauss law.

So we were incorrect in saying that the charge of the plate moved at all.
The charge of the plate will always be confined to the plate but it may move within the surfaces,I.e., it can move from the top to the bottom of the plate and vice versa.

Since the negative charge of the plane tried to attract the positive charge from the upper plate, ηbottom becomes 2 times η and that leaves the upper surface of the plate neutral.
 
  • #16
TSny
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Since the negative charge of the plane tried to attract the positive charge from the upper plate, ηbottom becomes 2 times η and that leaves the upper surface of the plate neutral.
But how do you know for sure that all the charge that was initially on the upper surface of the plate moves to the lower surface of the plate when the plane is brought near? The purpose of my question in post #10 was to provide a nudge in a direction that allows Thio to derive how much charge ends up on the upper and lower surfaces of the plate.
 
  • #17
18
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But how do you know for sure that all the charge that was initially on the upper surface of the plate moves to the lower surface of the plate when the plane is brought near?
This may be proved using gauss law. But this must be left for the OP to demonstrate as per the PF guidelines.
 
  • #18
TSny
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This may be proved using gauss law.
Yes, that's a nice way.

Or, you can find the values of the field in the three regions first as asked in part (a). Then determine the charge densities on the surfaces of the plates. This approach uses just the equations that Thio listed in his "relevant equations" section.
 

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