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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.

4ae1e1
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2 Answers2

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The force on the conductor must be zero. We will solve the problem in two steps. First, we will write down the external force $d\mathbf{F}$ on each infinitessimal charge $dq$ in terms of the external field $\mathbf{E}_{ext}$ and then we will integrate $d\mathbf{F}$ to get the total force.

Note we need only consider the external force (i.e., the force from the external field), since an object cannot exert a force on itself. This is a result of Newton's laws. It can also be proven from the coulomb force law: $\int \mathbf{E} dq = \int \mathbf{E} \rho(x) d\mathbf{x} = \int \int \frac{\mathbf{x}-\mathbf{y}}{|\mathbf{x} - \mathbf{y}|^3} \rho(\mathbf{y}) \rho(\mathbf{x}) d\mathbf{y} d\mathbf{x}=0$, where $\rho$ is the charge density, and the last equality is by the antisymmetry of the integrand under interchange of $\mathbf{x}$ and $\mathbf{y}$.

Moving on to step 1, using the law $\mathbf{F} = q \mathbf{E}$, we find the force $d \mathbf{F}$ is $\mathbf{E}_{ext} dq$.

Now let's do step 2, $\mathbf{F} = \int d\mathbf{F} = \int \mathbf{E}_{ext} dq = \mathbf{E}_{ext} \int dq = Q \mathbf{E}_{ext} = 0$. The last equality is true because $Q=0$. Thus the force is zero.

Brian Moths
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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? :)

Manish Khokhar
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