Tilting the plates of a parallel plate capacitor (and other changes)

  • #1
guv
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Homework Statement
You are given a parallel-plate capacitor with square plates of area A and separation d, in a vacuum. What is the qualitative effect of each of the following on its capacitance? (a) Double the area of one plate only, (b) Slide the plates parallel to each other so that the area of overlap is 50%. (c) Tilt one plate so that the separation remains d at one end but is ##\alpha d, 0 < \alpha < 1## at the other.

Is it possible to make quantitative calculation of the resulting capacitance in each case?
Relevant Equations
##C = \frac{\epsilon_0 A}{d}##
(a) (b) have intuitive solution but the asymmetry in (c) is confusing.
 
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  • #2
guv said:
(c) is confusing
Probably intentional. Make a wild guess: up, down, same ?
 
  • #3
Sorry I should have made the discussion clearer. I am looking for possible quantitative solution. In the case of (c) it's qualitatively greater (editted), is it possible to calculate the new capacitance without doing numerical calculation?
 
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  • #4
They are moving the plates closer together...
There you go.
 
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  • #5
guv said:
I am looking for possible quantitative solution
You could try to make some assumptions about the electric field to set up a simple integral
 
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  • #6
guv said:
possible quantitative solution
Might get some clues by modelling each plate as two or three plates, all parallel, but offset (so side by side) at different distances from the corresponding section of the other plate.
Code:
——-— +
     ———- +
          ———— +
          ———— -
Etc.
If the angle between the plates is shallow then, in this model, each plate is much broader than its distance from its opposite number, so effectively the model is several capacitors in parallel.
 
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  • #7
Or the plates can be thought of as a thin "pie slice" from an annular pie
If you "ignore fringing" you can maybe argue that the electric field between the plates will be circumferential (?). For constant potential difference this determines the charge (re)distribution on the plates, and allows a solution within this ansatz.
 
  • #8
guv said:
Sorry I should have made the discussion clearer. I am looking for possible quantitative solution. In the case of (c) it's qualitatively greater (editted), is it possible to calculate the new capacitance without doing numerical calculation?
I think all the geometries assume no fringing effects. Your formula C = ##\epsilon##A/d makes that assumption.

By that token (a) and (b) should be obvious.
In (c) however, assuming the same absence of fringing effects, you can do an integration to find the resultant capacitance as suggested by @BvU in post 5..
 
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  • #9
guv said:
Homework Statement:: You are given a parallel-plate capacitor with square plates of area A and separation d, in a vacuum. What is the qualitative effect of each of the following on its capacitance? (a) Double the area of one plate only, (b) Slide the plates parallel to each other so that the area of overlap is 50%. (c) Tilt one plate so that the separation remains d at one end but is αd,0<α<1 at the other.

Is it possible to make quantitative calculation of the resulting capacitance in each case?
Relevant Equations:: C=ϵ0Ad

(a) (b) have intuitive solution but the asymmetry in (c) is confusing.
Why is (c) confusing? One side of the plates is at the original distance d while the opposing side is at the lesser distance αd. Like: \ /. Top is d, bottom is αd.
 
  • #10
rude man said:
Why is (c) confusing? One side of the plates is at the original distance d while the opposing side is at the lesser distance αd. Like: \ /. Top is d, bottom is αd.
See post #3.
 
  • #11
BvU said:
You could try to make some assumptions about the electric field to set up a simple integral
Isn't it an assumption about the charge distribution that's needed? I appreciate that in a sense that is the same, but thinking in terms of charge seems simpler to me.
 
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  • #12
haruspex said:
Might get some clues by modelling each plate as two or three plates, all parallel, but offset (so side by side) at different distances from the corresponding section of the other plate.
Code:
——-— +
     ———- +
          ———— +
          ———— -
Etc.
If the angle between the plates is shallow then, in this model, each plate is much broader than its distance from its opposite number, so effectively the model is several capacitors in parallel.
Very nice I can see how this will lead to a intuitive result. Thanks.
 
  • #13
You know how in calculating areas under curves we consider a constant height for a distance ##(\Delta x\to 0)##? You can do the same for the tilted capacitor if you want to do some calculations.
 
  • #14
archaic said:
You know how in calculating areas under curves we consider a constant height for a distance ##(\Delta x\to 0)##? You can do the same for the tilted capacitor if you want to do some calculations.
Without knowing the charge distribution?
 
  • #15
haruspex said:
Without knowing the charge distribution?
Huh? His formula doesn't mention charges. I was thinking ##\epsilon_0L\int_0^\ell\frac{1}{d(x)}dx##.
 
  • #16
archaic said:
Huh? His formula doesn't mention charges. I was thinking ##\epsilon_0L\int_0^\ell\frac{1}{d(x)}dx##.
Ok, you are suggesting the continuous version of my approximation in post #6. Probably where @guv is heading, based on post #12.
It isn't clear how accurate this is, though. As the angle increases, the charges within a plate interact more with each other. Hard to figure out the distribution now.
 
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  • #17
haruspex said:
Ok, you are suggesting the continuous version of my approximation in post #6. Probably where @guv is heading, based on post #12.
It isn't clear how accurate this is, though. As the angle increases, the charges within a plate interact more with each other. Hard to figure out the distribution now.
I found someone linking to this in stackexchange. https://arxiv.org/abs/1208.2984
And this on google. https://riunet.upv.es/bitstream/han...llel thick-plate...pdf?isAllowed=y&sequence=1
 
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  • #18
archaic said:
Good tesearch!
Section 2 (inclined plates) of your second reference says:
To evaluate the capacitance of the ideal capacitor, the equation proposed by Y. Xiang can be directly used [11]. However, assuming zero thickness and neglecting the edge-effect, ideal analytical expressions are easily obtained [14]:

[14] J.L. Manglano de Mas, Lecciones de Física, Artes Gráficas Soler S.A., Valencia, Spain, 1995.

This implies the logarithmic solution does not depend on the angle being small. However, as the angle increases the edge effects will be more pronounced.
 
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