Capacitor and Dielectric

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Homework Statement



A parallel plate capacitor with plate area A = 100 cm^2 and plate separation of d = 5mm is connected to a 120 Volt power supply and allowed to fully charge.

a) Calculate the capacitance, stored charge, electric field midway between the plates and potential energy stored.

The capacitor is disconnected from the power supply and then a sheet of glass of thickness b = 2mm with dielectric constant κ = 4.5 in place between midway between plates.

b) Calculate the electric field inside the dielectric and in the gaps between the dielectric and the capacitor plates.

c) Hence calculate the voltage across the plates, and thus the new capacitance.

d) Calculate the new stored energy and determine whether work was done or given out as the glass plate was inserted.


Homework Equations


C = ε0A/d
Q = CV
E = Q/ε0A
u = (1/2)CV^2

The Attempt at a Solution



I've been trying my luck at this question. So for part a) I obtained all that was asked for without taking recourse to integrals. It seemed too easy to be the correct method to go about the problem. I simply substituted what was given in the question into the formulas noted above, and then substituted the resultant quantities into other equations to get the remaining quantities asked for. I also said that the direction of the electric field is perpendicular to the capacitor plates, from positive to negative.

As for part b) where the dielectric is taken into account, I obtained the electric field strength inside the dielectric by simply considering that it would be the field strength E without the dielectric divided by the dielectric constant k. The direction of it being the same as in the case where there is no dielectric.

However I'm not sure on how to go about calculating the electric field between the dielectric and the capacitor plates?

And for the following parts, is it still simply a case of substituting the known values into the above equations to get the desired quantities?

Thanks!
 

Answers and Replies

  • #2
gneill
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I've been trying my luck at this question. So for part a) I obtained all that was asked for without taking recourse to integrals. It seemed too easy to be the correct method to go about the problem. I simply substituted what was given in the question into the formulas noted above, and then substituted the resultant quantities into other equations to get the remaining quantities asked for. I also said that the direction of the electric field is perpendicular to the capacitor plates, from positive to negative.
Yes, that's fine.
As for part b) where the dielectric is taken into account, I obtained the electric field strength inside the dielectric by simply considering that it would be the field strength E without the dielectric divided by the dielectric constant k. The direction of it being the same as in the case where there is no dielectric.
Sure.
However I'm not sure on how to go about calculating the electric field between the dielectric and the capacitor plates?
You used the field strength without the dielectric in order to calculate the field strength within the dielectric... It hasn't changed since you did that calculation.
And for the following parts, is it still simply a case of substituting the known values into the above equations to get the desired quantities?
Pretty much, yes.
 
  • #3
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You used the field strength without the dielectric in order to calculate the field strength within the dielectric... It hasn't changed since you did that calculation.

Thanks a lot for the reply! Ah I see, so since the dielectric affects the electric field within itself only (due to its own internal electric field), outside of it (i.e. between it and the plates) the case is exactly the same as in a), right?
 
  • #4
gneill
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That's the idea, yes.
 
  • #5
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That's the idea, yes.

Superb, thanks a lot. For the next bit where it asks to calculate the new voltage across the plates, would it be valid for me to consider that the charge calculated in part a) remains constant, and as such that'd allow the following relation to hold: Vd=V0/k where Vd is the new voltage across the plates, and V0 the original. And from there the capacitance may be calculated by C=Q/Vd
 
  • #6
gneill
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Superb, thanks a lot. For the next bit where it asks to calculate the new voltage across the plates, would it be valid for me to consider that the charge calculated in part a) remains constant, and as such that'd allow the following relation to hold: Vd=V0/k where Vd is the new voltage across the plates, and V0 the original. And from there the capacitance may be calculated by C=Q/Vd
No.
Inserting the dielectric creates three regions between the outer plates. Two have the same field as the original capacitance since the field outside a charged plate is tied to its charge density (and the charge on those plates hasn't changed). You know the width of those regions from the geometry of the setup. You have the "new" field strength for the region filled with the dielectric, and you also know its width.

So sum the potential changes implied by the fields and their widths. Go from there.
 
  • #7
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No.
Inserting the dielectric creates three regions between the outer plates. Two have the same field as the original capacitance since the field outside a charged plate is tied to its charge density (and the charge on those plates hasn't changed). You know the width of those regions from the geometry of the setup. You have the "new" field strength for the region filled with the dielectric, and you also know its width.

So sum the potential changes implied by the fields and their widths. Go from there.

So I've separated the system into the three regions and marked on it the separation widths and field strengths. However I'm not clear on what you mean by summing the potential differences and the widths..
 
  • #8
gneill
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So I've separated the system into the three regions and marked on it the separation widths and field strengths. However I'm not clear on what you mean by summing the potential differences and the widths..
Three regions: Two air gaps and a dielectric region:
Fig1.gif


I suppose you could merge the two air gap regions into a single region (pushing the dielectric up or down to fill one gap).

If you have a field E in a region of width d, with the field oriented along the path of d, then what is the potential difference across d?
 
  • #9
rude man
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So I've separated the system into the three regions and marked on it the separation widths and field strengths. However I'm not clear on what you mean by summing the potential differences and the widths..
V = ∫E ds = E1*d1 + E2*d2 + E3*d3. You know d1 and d2, also E1 and E2, so you can easily compute V.
 
  • #10
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Three regions: Two air gaps and a dielectric region:
View attachment 79365

I suppose you could merge the two air gap regions into a single region (pushing the dielectric up or down to fill one gap).

If you have a field E in a region of width d, with the field oriented along the path of d, then what is the potential difference across d?

Sorry for the lateness in my reply. Ah yes I think I see what you mean now. Is it that I now invoke the relation that V = -E*d (multiplication since the field E is oriented along the path d and so there is no angle between them)? Thus I'd find V in each of the three regions (or two if I go about it as you mentioned) and then sum them up to get the total voltage across the two plates.
 
  • #11
rude man
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Sorry for the lateness in my reply. Ah yes I think I see what you mean now. Is it that I now invoke the relation that V = -E*d (multiplication since the field E is oriented along the path d and so there is no angle between them)? Thus I'd find V in each of the three regions (or two if I go about it as you mentioned) and then sum them up to get the total voltage across the two plates.
Right. I posted too late!
 
  • #12
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Ah thank you rude man, you got in there just before me! Much appreciated though! Good to see I'm on the right track.
 

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