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_{e}, density ρ). The inner one is maintained a potential V, and the outer one is grounded. To what height (h) does the oil rise in the space between the tubes? (Griffiths Introduction to Electrodynamics, problem 4.28).

First I calculated the capacitance of the system, as a function of the height of the oil, as C(h) = 2πε

_{0}(χ

_{e}h + l)/ln(b/a), where l is the total height of the capacitor.

Then I say: let the initial potential energy stored in the electric field be U

_{0}= 1/2 C(h=0)V

^{2}, and the final energy be U

_{f}= 1/2 C(h)V

^{2}.

The difference of energy has to be the potential energy gained by the system as gravitational, namely by the rising of the oil: ΔU = ρhπ(b

^{2}-a

^{2})g(h/2).

This gives, after some rearrangement, h = 2V

^{2}ε

_{0}χ

_{e}/ ln(b/a)ρg(b

^{2}-a

^{2}).

This is the solution given by Griffiths, except for the factor 2 I'm writing. He solves it by balance of force, saying that the electrostatic force has to be equal and opposite to the gravitational force, while I solve it by conservation of energy. What is wrong with my solution? Is energy not conserved?

I'm aware that the energy is not conserved in the ideal charge of an ideal capacitor with a voltage source, that is, the energy provided by the source is VQ, while the energy stored in the electric field is VQ/2. Does this also have to do with my solution being off by the same factor?