Problem related to energy in a capacitor I have a cylindrical capacitor and the inner conductor has slightly been move as shown in the figure.

The outter conductor is a shell, and the inner one is massive. There is a constant electric potential difference $V_0$ between them. I'm asked to calculate the force acting on the inner conductor using terms of energy.
I know that the force can be calculated as $$\mathbf{F}=-\mathbf{\nabla}U$$where $U$ is the energy stored in the capacitor. Also, if the two conductors were aligned, then
$$C=\frac{2\pi\epsilon L}{\ln(b/a)}$$
where $\epsilon$ is the permittivity of the dielectric inside the capacitor, $L$ is the length of the capacitor, $a$ is the inner radius and $b$ the outter one. And so, the energy would be
$$U=\frac12 V_0 \frac{2\pi\epsilon L}{\ln(b/a)}$$
So I have two doubts about this:


*

*In this case, where the two conductors are not aligned, would the expression of $U$ be the same but changing $L$ by $L-\Delta l$? (Where $\Delta l$ would be the distance that the inner conductor was displaced.) I'm not sure about this because the expression I used for finding $U$ is valid only if the electric field is confined inside the capacitor; if the two conductors are not aligned, I don't know if this is still true.

*How could I find the force using the energy? Because the energy I found is constant, so I can't use the expression $\mathbf{F}=-\mathbf{\nabla}U$ to get to a correct result.

 A: In your expression for $U$, find $dU = KdL$, where $K$ is all the other constants taken together. Note that $L$ is the paramrter of length in the direction in which the inner cylinder is taken out. Suppose the force on the cylinder is $F$ and it is pulled out by a small $dL$. Assume no change in charge density of the cylinders. Then, by using the principle of virtual work, $$F.dL = dU\implies F=K$$ That is, all those other constants in $U$ is the answer.
Since you seem to be unfamiliar with the principle of virtual work, read it here.
A: The expression for the potential energy U of a capacitor at a voltage V is U = (C·V^2)/2. (Your formula for the potential energy of the capacitor is incorrect.) If you change the capacitance at constant voltage V0  by changing the effective length l = L- Δl of the capacitor work has to be done in moving the charge back through the voltage source. The force F, according to your capacitor formula, is the negative derivative of the potential energy F = -(dU/dl) = -(dC/dl)·(V0^2)/2 = πϵ·(V0^2)/ln(b/a). The coaxial capacitor expression can also be used when the inner conductor is pulled out as long as the capacitance remains large compared to the capacitance of the leads including the pulled out inner conductor. Then the electric flux of the outer fields can be neglected.
