# 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.

• Isn't the formula F=qE pretty much just the whole answer? – Brian Moths Dec 17 '13 at 22:18
• @NowIGetToLearnWhatAHeadIs I have fixed my question (limiting the question to the $Q=0$ case). Now could you please elaborate your idea a little bit? Thanks. – 4ae1e1 Dec 17 '13 at 22:31
• Ok break it in to two steps. 1. Write the external force $dF$ on each infinitessimal charge $dq$ in terms of the electric field. 2. Integrate. – Brian Moths Dec 17 '13 at 22:33
• @NowIGetToLearnWhatAHeadIs I think Kevin is thinking about the case where the field from the charges in the conductor cannot be neglected given his simple example of induced charge yielding an always attractive force. Kevin, could you confirm this: otherwise its simply as NowIGet.. says. – Selene Routley Dec 17 '13 at 22:37
• @NowIGetToLearnWhatAHeadIs As suggested above, this is not a homework problem. So it would be really nice to share your idea—and prove your idea is correct at the same time. If your only suggestions are F=qE and Newton's first law, well, sorry, but you haven't suggested anything. I'm not a newbie in E&M or physics in general, I promise—at least I finished one year of QM with almost perfect scores. – 4ae1e1 Dec 17 '13 at 22:49

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? :)

• Sorry, but I found the answer not very convincing (or I didn't get the right point). The problem is, the charge redistribution necessarily influence the electric field, so that after the redistribution the net field is not uniform any more. – 4ae1e1 Dec 17 '13 at 23:01
• Yes true if you put a test charge then close to the electric conductor then it will not be uniform. Buta charge can not exert force on itself due to its own electric field. – Manish Khokhar Dec 17 '13 at 23:04
• How do you know a charge (or object in general) can't exert a force on itself? If a put an object on the table perhaps it can start to move of its own accord? – Brian Moths Dec 17 '13 at 23:06
• We are not talking about the self-action of a point charge. We are talking about the force of a charge distribution, so for instance, the negative charges on one side of the conductor can exert force on the positive charges on the other side. – 4ae1e1 Dec 17 '13 at 23:06
• @NowIGetToLearnWhatAHeadIs You are assuming that a potato is a point particle. A conductor is not a point particle, but rather necessarily a collection of particles. Classical electrodynamics does not solve the problem of self-action (of point charges) very nicely, but we are not facing a self-action problem here, because we have no point charges here. – 4ae1e1 Dec 17 '13 at 23:09

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.

• It would have been clearer if you didn't call it an "object" but rather a "system." Anyway, sorry I didn't understand you earlier. – 4ae1e1 Dec 17 '13 at 23:44