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The height of a dielectric material between two coaxial pipes

  • #1
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Homework Statement


This is the exercise 10.6 from Feynman lectures on Physics 2.
Two coaxial pipes of radii a and b(a<b) are lowered vertically into an oil bath. If a voltage V is applied between the pipes, show that the oil rises a height H.
Show that H=(V^2)(κ-1)ε_0/[ln(b/a)ρ(b^2-a^2)g]
where κ is the dielectric constant of the oil, ρ is the mass density of the oil, and g is the gravitational acceleration constant.

Homework Equations


the conservation law of energy: ΔWork(V=0 to V=V) = ΔU(electrostatic energy) + work on the oil ?
QV=(1/2)QV + (1/2)QV

The Attempt at a Solution


[/B]
I assumed a conductor of height L above the oil surface( L>>b), originally neutral, then charged to voltage V with charge Q.
The electric field btwn two pipes is E = Q/(L2πε_0r) and the potential difference is V=Q*ln(b/a)/(L2πε_0). C = L2πε_0/ln(b/a).

If the oil is attracted into the conductor to height y(variable), the voltage stays the same but the charge and the capacity change:
C1=2πε_0(L-y+κy)/ln(b/a), Q1=Q(L-y+κy)/L

The total work of a battery is Wb(work of battery)=Q1V-0 = (V^2)2πε_0(L-y+κy)/ln(b/a)
The electrostatic energy of the conductor is U=(1/2)Wb = (V^2)2πε_0(L-y+κy)/[2*ln(b/a)]
-(The work done by gravity) = (the work done on the oil) = -πρ(b^2-a^2)g*y = (V^2)2πε_0(L-y+κy)/[2*ln(b/a)]

I do not know what is wrong.
 

Answers and Replies

  • #2
rude man
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First thing is to ignore L It doesn't appear in the given answer. Concentrate on the height y above the oil surface, just that section.

EDIT: disregard the rest of this post; see post 3.

Make a list of energies supplied and absorbed. The battery is obviously the source of the energy to lift the oil against gravity. What then about the change in E field stored energy before vs. after V is applied?

I haven't done this problem myself but would approach it that way. I'll try to find the time and energy (LOL) to do it later.
 
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  • #3
rude man
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OK, forget energies and forget the battery completely (except of course that it applies a constant V to the cylindrical capacitor.)

Instead, equate the force with which the oil column is pulled upwards with the gravitational force acting on the filled oil column. That way I got the right answer. It's a pretty easy problem that way.

(Going with energies is not only messier but resulted in a tautology!).
 
  • #4
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First thing is to ignore L It doesn't appear in the given answer. Concentrate on the height y above the oil surface, just that section.

EDIT: disregard the rest of this post; see post 3.

Make a list of energies supplied and absorbed. The battery is obviously the source of the energy to lift the oil against gravity. What then about the change in E field stored energy before vs. after V is applied?

I haven't done this problem myself but would approach it that way. I'll try to find the time and energy (LOL) to do it later.
Without L, how can I calculate the capacity? I differentiate the electrostatic energy of the conductor to obtain electric force inside the conductor. U= (V^2)C/2 and C changes with the total length of the conductor and the height of the oil. Of course the force do not contain L.
 
  • #5
rude man
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Without L, how can I calculate the capacity? I differentiate the electrostatic energy of the conductor to obtain electric force inside the conductor. U= (V^2)C/2 and C changes with the total length of the conductor and the height of the oil. Of course the force do not contain L.
This is fine, you have seemingly distanced yourself from energy balance incl. battery energy, and are going with force balance: ∂U/∂y = mg. But I don't see the need for any L when you already have y and h? I am guessing your L is h? Anyway, proceed!
 

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