Charge density of capacitor plates

In summary, A parallel plate capacitor with two flat conducting plates in a vacuum, each of area A, separated by a small gap has one plate with a total charge of 2Q and the other with a total charge of -Q. The surface charge densities on the four horizontal metal surfaces can be expressed as sigma1 = Q/2A, sigma2 = 3Q/2A, sigma3 = -3Q/2A, and sigma4 = Q/2A. The electric field between the plates is constant, and the electric field immediately above and below the capacitor is also constant. The E field can be calculated using Gauss's Law and the divergence of D (and thus E) must be zero in the absence of charge
  • #1
Physgeek64
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Homework Statement


A parallel plate capacitor is made of two flat horizontal conducting plates in a vacuum, each of area A, separated by a small gap. One plate carries a total charge 2Q, the other a total charge −Q. Find the surface charge densities on the four horizontal metal surfaces in terms of Q and A. Find the electric field between the plates, and the electric field immediately above and below the capacitor.

Homework Equations

The Attempt at a Solution


I would have though ##\sigma_1 =\sigma_2=\frac{Q}{A}## and ##\sigma_3=\sigma_4=-\frac{Q}{2A}## but the answer is ## sigma_1=\frac{Q}{2A}## ##\sigma_2=\frac{3Q}{2A}## ##\sigma_3=\frac{-3Q}{2A}## ##\sigma_4=\frac{-Q}{2A}## but i cannot see how they have got this
 
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  • #2
3 equations, 3 unknowns, since the inner surface charges will obviously be equal and opposite.
2 equations should be obvious.
The 3rd equation hint: force the E field in either plate = 0.
 
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  • #3
rude man said:
3 equations, 3 unknowns, since the inner surface charges will obviously be equal and opposite.
2 equations should be obvious.
The 3rd equation hint: force the E field in eithr plate = 0.
Why should the two inner plates be equal and opposite?
 
  • #4
rude man said:
3 equations, 3 unknowns, since the inner surface charges will obviously be equal and opposite.
2 equations should be obvious.
The 3rd equation hint: force the E field in eithr plate = 0.
BTW the answer is wrong. If σ1 + σ2 = 2Q then σ3 + σ4 = -Q, not -2Q.
(I left out A in all the σ terms).
 
  • #5
Physgeek64 said:
Why should the two inner plates be equal and opposite?
Because the E field is constant between the plates. Use Gauss to verify.
 
  • #6
rude man said:
BTW the answer is wrong. If σ1 + σ2 = 2Q then σ3 + σ4 = -Q, not -2Q.
No, that's right. There's 2Q on one plate and -Q on the other plate
 
  • #7
rude man said:
Because the E field is constant between the plates. Use Gauss to verify.
The E field would still be constant if ##\sigma_2## is not equal to ##\sigma_3## ?
 
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  • #8
rude man said:
BTW the answer is wrong. If σ1 + σ2 = 2Q then σ3 + σ4 = -Q, not -2Q.
Physgeek64 said:
No, that's right. There's 2Q on one plate and -Q on the other plate
You just contradicted yourself.
 
  • #9
Physgeek64 said:
The E field would still be constant if ##\sigma_2## is not equal to ##\sigma_3## ?
No. Put a gaussian surface from the inside of one plate to any point between the plates. The E field is the same everywhere inside the plates.
Another argument would be that the divergence of D (and thus E) must be zero in the absence of charge.
 
  • #10
rude man said:
You just cotradicted yourself.
Sorry, have I? There is 2Q on one plate and -Q on the other
 
  • #11
rude man said:
No. Put a gaussian surface from the inside of one plate to any point between the plates. The E field is the same everywhere inside the plates.
Another argument would be that the divergence of D (and thus E) must be zero in the absence of charge.
Sorry if I am being really stupid but having ##\sigma_2## not equal to ##sigma_3## gives a uniform field in between the plates of strength ##\frac{\sigma_2}{2\epsilon_0 } + \frac{\sigma_3}{2\epsilon_0}## using gauss' law
 
  • #12
Physgeek64 said:
Sorry if I am being really stupid but having ##\sigma_2## not equal to ##sigma_3## gives a uniform field in between the plates of strength ##\frac{\sigma_2}{2\epsilon_0 } + \frac{\sigma_3}{2\epsilon_0}## using gauss' law
Maybe we're having problems defining which surface is which. You did not show a drawing or verbally describe which is which.
Which is why i said that if 1 and 2 are on the plate with charge 2Q then 3 and 4 must add up to -Q but your answer adds to -2Q.
 
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  • #13
rude man said:
Maybe we're having problems defining which surface is which. You did not show a drawing or verbally describe which is which.
Which is why i said that if 1 and 2 are on the plate with charge 2Q then 3 and 4 must add up to -Q but your anwer adds to -2Q.
Ahh I see the confusion now. I think then maybe ##\sigma_4=\frac{Q}{2A}##

I have define plates 1234 to be the sequence of planes met if we travel from the top of the capacitor to the bottom where the top capacitor plate has charge 2Q and the bottom charge -Q.
 
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  • #14
Physgeek64 said:
Ahh I see the confusion now. I think then maybe ##\sigma_4=\frac{Q}{2A}##
I think so too. Then their answer would be correct.
I have define plates 1234 to be the sequence of planes met if we travel from the top of the capacitor to the bottom where the top capacitor plate has charge 2Q and the bottom charge -Q.
PS I woud call the two plates "plates" but their surfaces "surfaces".
 
  • #15
Physgeek64 said:
Ahh I see the confusion now. I think then maybe ##\sigma_4=\frac{Q}{2A}##
I think so too. Then their answer would be correct.
I have define plates 1234 to be the sequence of planes met if we travel from the top of the capacitor to the bottom where the top capacitor plate has charge 2Q and the bottom charge -Q.
PS I woud call the two plates "plates" but their surfaces "surfaces".
EDIT: sorry, I didn't see you use the word "planes" I thought it was "plates". Perfectly OK to say "planes" but I still like "surfaces". We speak of "surface charge" not "plane charge".
 
  • #16
rude man said:
I think so too. Then their answer would be correct.PS I woud call the two plates "plates" but their surfaces "surfaces".
EDIT: sorry, I didn't see you use the word "planes" I thought it was "plates". Perfectly OK to say "planes" but I still like "surfaces". We speak of "surface charge" not "plane charge".
I'll use surfaces from now on :) I still don't see why you wouldn't have a uniform field with different charges on the inner surfaces
 
  • #17
Physgeek64 said:
I'll use surfaces from now on :) I still don't see why you wouldn't have a uniform field with different charges on the inner surfaces
Put a Gaussian cylinder between the plates. The contained charge is zero. If the E field were not uniform you would get a net non-zero flux emanating from the gaussian surfaces parallel to the plates but that is a violation of Gauss's theorem.
 

1. What is the definition of charge density?

Charge density refers to the amount of charge per unit area of a surface, typically measured in units of coulombs per square meter.

2. How is charge density related to the capacitance of a capacitor?

The charge density on the plates of a capacitor is directly proportional to the capacitance, meaning that as the charge density increases, so does the capacitance.

3. How does the distance between the plates affect the charge density?

The charge density on the plates of a capacitor is inversely proportional to the distance between the plates. As the distance increases, the charge density decreases.

4. Can the charge density of a capacitor be changed?

Yes, the charge density of a capacitor can be changed by adjusting the amount of charge on the plates or by altering the distance between the plates.

5. What is the SI unit for charge density?

The SI unit for charge density is coulombs per square meter (C/m2).

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