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.
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
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:
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.