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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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    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 But you don't know what $\textbf{E}$ is. $\textbf{E}$ was uniform before we threw in the conductor, but it is not uniform afterwards. – 4ae1e1 Dec 17 '13 at 22:35
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    @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
  • @WetSavannaAnimalakaRodVance That's right. Otherwise the question is simply too stupid... – 4ae1e1 Dec 17 '13 at 22:38
  • My advice to kevin is to consider Newton's first law. In particular: "An object that is at rest will stay at rest unless an external force acts upon it." – Brian Moths Dec 17 '13 at 22:38
  • @NowIGetToLearnWhatAHeadIs That's a great law, but not very helpful here. You need something at least powerful enough to explain why a uniform field is different from a random one—as I already showed the "theorem" is wrong in the case of random external fields. – 4ae1e1 Dec 17 '13 at 22:41
  • Well I am going to let you think about this point some more. It is an important one. I don't want to rob of the satisfaction of finding the rest of the answer for yourself. You can do it. – Brian Moths Dec 17 '13 at 22:44
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    @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 think you're right, Kevin, this is harder than it looks. Intriguing question. – Selene Routley Dec 17 '13 at 23:01
  • @WetSavannaAnimalakaRodVance Thanks. Any suggestions? – 4ae1e1 Dec 17 '13 at 23:02

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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  • 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
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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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  • 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
  • Ok so that is your doubt. True we can put a test charge close to conductor & see that it will not be uniform.But
    1. a charge can not exert force on itself due to its own electric field. Just like earth can not accelerate itself due to its own gravitational field.

    You might say that there is not one charge there are many electons and +ve nucleus they can influence each other. But since we are taking conductor as a single body, without any external force it can not accelerate or de-accelerate itself (a result from mechanics). Remember when we analyzse the motion of COM we ignore interal forces.

     – Manish Khokhar Dec 17 '13 at 23:14
  • " If a put an object on the table perhaps it can start to move of its own accord?" That's a violation of conservation of energy. – Manish Khokhar Dec 17 '13 at 23:16
  • Ahhh, you are right Manish, that would be not only a violation of conservation of energy, but also a violation of Newton's first law, which says that a body experiencing no external force cannot accelerate, or put another way, that a body cannot exert a force on itself. Thus as you say, we need only consider external forces when calculating the net force on an object. I see now. – Brian Moths Dec 17 '13 at 23:18
  • @NowIGetToLearnWhatAHeadIs Thanks for the kind help, but I am afraid that you (both) totally missed the point. First of all, the system is not closed, so talking about conservation of energy is nonsense: there is energy stored in the electric field, and the electric field can be influenced by the conductor. – 4ae1e1 Dec 17 '13 at 23:26
  • Ah, now I understand what you mean, Manish. You were right from the very beginning, but I was preoccupied with some complicated stuff so that I interpreted your explanation in a complicated and wrong way. Sorry about that. However, the reason I'm wrong is certainly not a problem of energy. (Or more accurately, you didn't get what I'm saying just like I didn't get yours. We were not on the same page.) – 4ae1e1 Dec 17 '13 at 23:31
  • ha ha. That happens with me as well. I ask some really strange question with my teacher which now I classify as "stupid". But it's process of learning. Btw that energy thing was for NowIGetToLearnWhatAHeadI not for you. It was not required at all. – Manish Khokhar Dec 17 '13 at 23:42