electromagnetism - Does a conductor of total charge zero placed in a uniform external electric field experience net force?

The question I have in mind is: If we place a conductor (arbitrary shape) of total charge zero in a uniform external electric field $\textbf{E}_0$, does it experience any net force? Why (not)?

Now I will discuss the context of the question. I am working on Griffiths Introduction to Electrodynamics, Fourth Edition, p.112 Problem 2.59 (not homework problem, though). it says,

Prove or disprove (with a counterexample) the following

Theorem: Suppose a conductor carrying a net charge $Q$, when placed in an external electric field $\textbf{E}_e$, experiences a force $\textbf{F}$; if the external field is now reversed ($\textbf{E}_e \to - \textbf{E}_e$), the force also reverses ($\textbf{F} \to -\textbf{F}$).

What if we stipulate that the external field is uniform?

In general this is obviously not true. I will first limit myself to the case of $Q=0$.

One approach: when $\textbf{E}$ is reversed, the surface charged distribution $\sigma$ is also reversed (to cancel $\textbf{E}$), so the electrostatic pressure at every point, $\frac{1}{2\epsilon_0} \sigma^2\, \hat{\textbf{n}}$ stays the same. Consequently, $\textbf{F}$ stays the same rather than flips sign.

Another approach: there is an intuitive counterexample. A conductor is generally attracted to a point charge nearby; if the sign of the point charge is flipped, the conductor is still attracted rather than repulsed.

So the first question is easy, and the interesting one is "What if we stipulate that the external field is uniform?" I suspect that in a uniform external field the net force is zero, so that $\textbf{F} = 0 = -\textbf{F}$, but I can't think of a way to prove or disprove it.

Answer

I don't think it is that tough to analyse. If a conductor is present in a uniform electric field then there will be redistribution of charges to counter Electric Field inside the conductor (so that the net field inside the conductor is zero). However in uniform electric field this redistribution of charges will not cause any net force on the conductor. Why? Because the amount of +ve charge on the conductor is equal to the -ve charge. Hence F = q*E will be countered (or balanced) by equal and opposite force (-q*E). The geometry on conductor will not play any role at all. (Nature of coulomb in force.) So centre of mass will not experience any acceleration. What about torque? It turns out torque = r×F. Ahh... "r". Interesting.So will it experience any angular acceleration? :)