The Electrical energy of a conducting cylinder

  • #1
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1. The problem stsatemesnt, all variables and given/known data
A cylinder conductor of length ##l##, inner radius ##a## and outer radius ##b## is half-filled with a liquid dielectric as shown in the picture. The cylinder is attached to a battery of constant voltage ##U## and then separated from it. Imagine that the dielectric is leaking and calculate the change of electrical energy when there is no dielectric present. ( meaning the difference between the beginning state and the one were there is no dielectric)
There is a picture below.
IMG_2019.JPG

Homework Equations


3. The Attempt at a Solution
I'm gonna guide you through my work and thinking so you can correct me and help me finish the problem.[/B]
##ΔW_e=W_{e2}-W_{e1}## ( The state 2 is when its empty)
##W_{e1}=\frac{1}{2}\frac{Q_1^2}{C_1}## ( ##C_1,Q_1## are the total charge and capacitance of the system 1)
Because there is a dielectric in the first system i have two electric fields, the one above the ##h## and the one below going radial.
##E_1=\frac{Q'_1}{2 \pi ε_0 r}##
##U_1=\frac{Q'_1}{2 \pi ε_0}\ln (b/a)##
##E_2=\frac{Q'_2}{2 \pi ε_0 ε_r r}##
##U_2=\frac{Q'_2}{2 \pi ε_0 ε_r} \ln (b/a)##
Then i express the linear charge densities ##Q'_1,Q'_2##
##Q'_1=\frac{2 \pi ε_0 U_1}{\ln (b/a)}##
##Q'_2=\frac{2 \pi ε_0 ε_r U_2}{\ln (b/a)}##
##Q_{total}=Q'_1*(l/2) + Q'_2*(l/2)##
Something does not seem right here, can you help me with what i got wrong?
 
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Answers and Replies

  • #2
BvU
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Hi,

I'm missing the relevant equations in section 2 :smile: like ##Q_{\rm total}\ ## conserved, ##U_1 = U_2 ## and such...
And what about ##\ C_1, C_2\ ## ?
 
  • #3
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Hi,

I'm missing the relevant equations in section 2 :smile: like ##Q_{\rm total}\ ## conserved, ##U_1 = U_2 ## and such...
And what about ##\ C_1, C_2\ ## ?
It's not stated but im guessing that ##Q## must be conserved and ##C## and ##U## are changing. :) The whole work is my guess so i dont know the answer if anything is conserved xD I was thinking someone can provide an insight.
 
  • #4
rude man
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I would forget about E fields. Just compute C1 and C2, then take the difference in 1/2 CV^2. Yes, Q is conserved; removing the dielectric does not remove any charge. So you know that V1 and V2 will be different also.
 
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  • #5
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I would forget about E fields. Just compute C1 and C2, then take the difference in 1/2 CV^2. Yes, Q is conserved; removing the dielectric does not remove any charge. So you know that V1 and V2 will be different also.
But to compute C i need to have V and to have V i need E
 
  • #6
rude man
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But to compute C i need to have V and to have V i need E
No you don't. C is a property of the geometry etc etc and is independent of V, E or Q. (We are not yet dealing with nonlinear capacitors at your stage of the game).
 
  • #7
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No you don't. C is a property of the geometry etc etc and is independent of V, E or Q. (We are not yet dealing with nonlinear capacitors at your stage of the game).
##C=\frac{Q}{V}## by definition. What do you mean its not dependant on ##Q## or ##V##?
 
  • #8
rude man
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##C=\frac{Q}{V}## by definition. What do you mean its not dependant on ##Q## or ##V##?
##C=\frac{Q}{V}## by definition. What do you mean its not dependant on ##Q## or ##V##?
OK take two large parallel plates. Area = A, separation = d. What's the capacitance?
 
  • #9
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OK take two large parallel plates. Area = A, separation = d. What's the capacitance?
##C=\epsilon_0\frac{A}{d}##
I see your point, it is the conductors geometric property.
For the cylindrical system i then have two capacitances ##C_1=\frac{2\pi\epsilon_0L/2}{ln(b/a)}##, ##C_2=\frac{2\pi\epsilon_0\epsilon_rL/2}{ln(b/a)}## Now i need to know the ##V_1,V_2## for the system in case it's half filled and for that i need ##E##. When i find that what is my ##C## for the case where its half filled? I have two C's?
 
  • #10
rude man
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##C=\epsilon_0\frac{A}{d}##
I see your point, it is the conductors geometric property.
For the cylindrical system i then have two capacitances ##C_1=\frac{2\pi\epsilon_0L/2}{ln(b/a)}##, ##C_2=\frac{2\pi\epsilon_0\epsilon_rL/2}{ln(b/a)}## Now i need to know the ##V_1,V_2## for the system in case it's half filled and for that i need ##E##. When i find that what is my ##C## for the case where its half filled? I have two C's?
You don't need E.
You have two capacitors in "parallel", do you see that? Call them CA and CB. So C = CA + CB. Say CA has the dielectric, initially. You can compute CA1; also CB1 = CA2 = CB2.
You know V1, = same for CA and CB. You then can find Q = V1(CA1 + CB1).
For V2 you invoke the fact that Q does not change. This enables you to compute V2.
Finally, as I said, energy = 1/2 Δ{CV2}, or = 1/2 Q ΔV.
I can say no more.
 
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  • #11
BvU
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All a matter of writing down the relevant equations, including the conservation relationships ...
I'm missing the relevant equations in section 2 :smile: like ##Q_{\rm total}\ ##conserved, ##U_1 = U_2## and such...
(not to rub it in, but to make clear the ingredients of a solid problem solving approach ... )
It's not stated but I'm guessing that ##Q## must be conserved and ##C## and ##U## are changing.
No guessing. Physics. Where would Q go ?
 
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  • #12
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Dielectric polarization can make Q change at conductor surfaces. The net change in Q is 0 globally.
 
  • #13
rude man
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Dielectric polarization can make Q change at conductor surfaces. The net change in Q is 0 globally.
Don't think so. The dielectric does not assume net charge. All charge stays on the metal surfaces.
 

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