Why would the oil rise in this situation? Given:

Two long coaxial cylindrical metal tubes (inner radius a, outer radius b)
  stand vertically in a tank of dielectric oil (susceptibility χe, mass density
  ρ). The inner one is maintained at potential V , and the outer one is
  grounded (see figure). To what height (h) does the oil rise, in the space
  between the tubes?

Why would the oil rise in the first place? I assumed the oil was placed in there, everything was set up, and only then did the inner tube get set to a specific potential. But I may be wrong.
Update: so it looks like the force that pushes the oil up is the electric force. I assume this comes from the battery.
 A: I assume that the outer tube has the inner radius $b$ and the inner tube has the outer radius $a$. When you apply a voltage $V$ between the tubes, the additional capacitance of the height $h$ of the tube filled with the liquid dielectric with susceptibility $\chi_e$, giving the relative permittivity $\epsilon =1+\chi_e$, is $$C=\frac{2\pi\epsilon_0\chi_e h}{\ln{(b/a)}}$$ Thus the energy of this additional capacitance $C$ is $$W=\frac{CV^2}{2}=\frac{\pi\epsilon_0\chi_e h V^2}{\ln{(b/a)}}$$ This energy $W$ corresponds to the work done by the battery to lift the liquid dielectric to height $h$. Thus the battery force on the liquid dielectric corresponding to a change in h is $$F_B=\frac{dW}{dh}=\frac{\pi\epsilon_0\chi_e V^2}{\ln{(b/a)}}$$ On the other hand, the counteracting hydrostatic pressure force is $$F_H=A\rho gh=\pi (b^2-a^2)\rho gh$$ where $A$ is the area of the lifted liquid between the tubes. This force increases with h. Thus the liquid rises upon application of the voltage $V$ to a height $h$ where $F_B=F_H$. This yields the height $h$ of the liquid $$h=\frac{\epsilon_0\chi_e V^2}{(b^2-a^2)\rho g\ln{(b/a)}}$$
