Electrostatic energy after a metal piece is inserted between 2 capacitor plates

  • #1
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Homework Statement



2qbuagl.png


fig 1 : Area of each plate is S, separated by 2d, charge Q in the capacitors
fig 2 : uncharged conductors of area S, thickness d, inserted parallel between plates

What is the ratio of electrostatic energy in fig 2 to electrostatic energy in fig 1?

Homework Equations


Q = CV
C = Q/V = Q/Ed = ##\frac{Q \epsilon} {charge density * d}## = ##\frac { A\epsilon}{d} ##
energy stored in capacitor = ##\frac{1}{2} CV^2##
E = qV (in general)

The Attempt at a Solution


I think the electrons will be like this
16by32q.png



ratio = ##\frac{C_1 {V_1}^2 }{ C_2 {V_2} ^2}## = ##\frac{\frac { A\epsilon}{d} {V_1}^2 }{ \frac { A\epsilon}{d} {V_2} ^2}##

are the voltages the same between fig 1 and fig 2?
how to find the ratio?
 

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Answers and Replies

  • #2
gneill
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You might be better off working with the charge, Q, rather than voltages. What's the expression for the energy stored in a capacitor given the charge?
 
  • #3
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You might be better off working with the charge, Q, rather than voltages. What's the expression for the energy stored in a capacitor given the charge?
It is ## \frac{1}{2}\frac{Q^2}{C}##

Is electrostatic energy is 1/2 CV^2 or E = Vd
E = Q/Cd ?

C = permivity*A/d

Why am i better off working for Q than V ?
 
  • #4
gneill
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It is ## \frac{1}{2}\frac{Q^2}{C}##

Is electrostatic energy is 1/2 CV^2 or E = Vd
E = Q/Cd ?

C = permivity*A/d

Why am i better off working for Q than V ?
The electrostatic energy stored in a capacitor is ##E = \frac{1}{2} C V^2 = \frac{1}{2}\frac{Q^2}{C} ## .

In this case it may be simpler to work with Q rather than V because you are given that the charge on the capacitor is Q, and when you insert the metal strip it effectively transforms the device into two capacitors (in series) that will each have the same charge Q. So, you don't have to take the time or trouble of determining the voltages on them in order to get at the energy stored.
 
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  • #5
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Why am i better off working for Q than V ?[/QUOTE]
Because there is no way Q can change. Better to start with a quantity which will not change rather than the quantity which we are not sure of. If the original system were connected to a battery of emf E, we would have started with that as we would be sure it cannot change and there would be reason by which Q could change.
 
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  • #6
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Ratio is ##\frac{\frac{1}{2} \frac{{Q_2}^2}{C_2}}{\frac{1}{2} \frac{{Q_1}^2}{C_1}}##
(For series capacitors Q are the same to the total Q)
Ratio is C1/C2
Ratio is ##\frac{ \frac{permittivity_1{A_1}}{d_1}}{\frac{permittivity_2{A_2}}{d_2}}##
A1 = A2
Permittivity1 = Permttiivity2

Ratio is d2/d1 = 1/2 ÷ 1 = 1/2

I am not sure that :
For fig 1, what is the area? S or 2S (2 plates)
For fig 1 the total Q is Q or 2Q
For fig 2 the total Q is ?
 
  • #7
gneill
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You'll have to explain better what exactly your C1 and C2 are.

I am not sure that :
For fig 1, what is the area? S or 2S (2 plates)
The area is S for each capacitor in this problem. The plate area for a parallel plate capacitor is described by the area of one of the facing plate surfaces.
For fig 1 the total Q is Q or 2Q
What is called the charge on the capacitor is Q. Yes, there are equal and opposite charges on the pair of plates but that just makes the overall charge on the device neutral.
For fig 2 the total Q is ?
The capacitor charge on the device remains Q. Each pair of facing plate surfaces will have +Q and -Q on them.
 
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  • #8
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You'll have to explain better what exactly your C1 and C2 are.


The area is S for each capacitor in this problem. The plate area for a parallel plate capacitor is described by the area of one of the facing plate surfaces.

What is called the charge on the capacitor is Q. Yes, there are equal and opposite charges on the pair of plates but that just makes the overall charge on the device neutral.

The capacitor charge on the device remains Q. Each pair of facing plate surfaces will have +Q and -Q on them.
Thank you. C1 = the capacitance in fig 1, C2 = capacitance in fig 2
 
  • #9
gneill
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Thank you. C1 = the capacitance in fig 1, C2 = capacitance in fig 2
Okay, but you'll have to be a bit careful about how you determine what C2 is. When you insert the metal strip, two capacitors are formed where there was previously only one:

upload_2018-3-16_11-33-20.png
 

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  • #10
rude man
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I would invoke Gaussian surfaces for the before & after cases; that would yield E directly; then go energy = energy density x volume for all three volumes.
 
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  • #11
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Okay, but you'll have to be a bit careful about how you determine what C2 is. When you insert the metal strip, two capacitors are formed where there was previously only one:

View attachment 222113
As in the image you posted, is the distance for Ca = 1/2d = Cb? And the area is S for both because one of the facing plate's area is S ?
 
  • #12
gneill
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As in the image you posted, is the distance for Ca = 1/2d = Cb? And the area is S for both because one of the facing plate's area is S ?
Yes, in the original figure the separation between the plates of the starting capacitor was given as 2d. The inserted metal strip takes up d, leaving d to be apportioned between the two "new" capacitors. So their separation is d/2 each.

The plate area does not change, so it is S for every capacitor in the problem.
 
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  • #13
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Yes, in the original figure the separation between the plates of the starting capacitor was given as 2d. The inserted metal strip takes up d, leaving d to be apportioned between the two "new" capacitors. So their separation is d/2 each.

The plate area does not change, so it is S for every capacitor in the problem.
Thanks
 

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