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I am reading Feynman's Lectures volume_II where he discussed the impossibility of the presence of stable equilibrium in an electrostatic field.

There are no points of stable equilibrium in any electrostatic field—except right on top of another charge. Using Gauss’ law, it is easy to see why. First, for a charge to be in equilibrium at any particular point $P_0$, the field must be zero. Second, if the equilibrium is to be a stable one, we require that if we move the charge away from $P_0$ in any direction, there should be a restoring force directed opposite to the displacement. The electric field at all nearby points must be pointing inward—toward the point $P_0$. But that is in violation of Gauss’ law if there is no charge at $P_0$, as we can easily see.

enter image description here

Fig. 5–1.If $P_0$ were a position of stable equilibrium for a positive charge, the electric field everywhere in the neighborhood would point toward $P_0$. Consider a tiny imaginary surface that encloses $P_0$, as in Fig. 5–1. If the electric field everywhere in the vicinity is pointed toward $P_0$, the surface integral of the normal component is certainly not zero. For the case shown in the figure, the flux through the surface must be a negative number. But Gauss’ law says that the flux of electric field through any surface is proportional to the total charge inside. If there is no charge at $P_0$, the field we have imagined violates Gauss’ law. It is impossible to balance a positive charge in empty space—at a point where there is not some negative charge. A positive charge can be in equilibrium if it is in the middle of a distributed negative charge. Of course, the negative charge distribution would have to be held in place by other than electrical forces

I am having problems with the explanation especially the bold lines. It can be comprehensible that to be at equilibrium at $P_0$, the force i.e. the field must be zero. There must be a restoring force in order to bring back the charge at $P_0$. Feynman then wrote that as there is no charge, it cannot have any net flux as is evident from the bold lines. Confusion! Where did the charge go that he was discussing so far? He only displaced the charge , previous at $P_0$ away from $P_0$. That doesn't mean the charge is no more inside the surface!! I am really not getting what/why he wrote if there is no charge at $P_0$. Can anyone help me clear this confusion?

Also, if the field is zero, doesn't it mean that flux is also zero? If not, why?

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    $\begingroup$ That is known as Earnshaw's theorem. $\endgroup$ – ACuriousMind Jun 18 '15 at 10:17
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    $\begingroup$ This link is worth intuitive. $\endgroup$ – user36790 Jun 18 '15 at 13:20
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    $\begingroup$ I've realised that my answer is far from convincing, so I've deleted it. The link you found is much better, and actually helped me get a clear understanding. $\endgroup$ – user3237992 Jun 19 '15 at 10:08
  • $\begingroup$ I suppose in a truly microscopic sense, one could have a zero net field at some location but a finite flux through the surface of a macroscopic shell enclosing the location of zero net field. Flux is considered over finite areas while I think the field itself can be (classically) defined at individual points. $\endgroup$ – honeste_vivere Jan 24 '16 at 14:35
  • $\begingroup$ Now I'm confused and curious... Could one set up some external field in such a fashion as to create a finite value for Gauss' law but have no local sources (i.e., charges) within the gaussian surface? I think the answer is yes, but it makes me curious as to what is the physical interpretation for the term we refer to as enclosed charge... $\endgroup$ – honeste_vivere Jan 24 '16 at 14:38
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He talks about imaginary situation when there is a point in space with stable equilibrium.

As Feynman says, it requires $E=0$ at some point $(P_0)$ and all $E$ vectors looking inwards (like local minima) around that point. Then let's use Gauss's law: compute flux. It will definitely be not zero, which means that such point does not exist.

What is doable is a spherical shell with some charge and same sign charge in its middle. You enclose your centered charge with surface. Now there is field from the shell and from the charge crossing it. Bo-ya, this field has non-zero flux through the surface.

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    $\begingroup$ As I recall, the electric field is zero at all points inside a uniformly charged spherical shell. So this doesn't give you a stable equilibrium any more than a single charge in free space. $\endgroup$ – ragnar Jun 18 '15 at 14:31
  • $\begingroup$ oh god, you are right (gauss's!) $\endgroup$ – Oct18 is day of silence on SE Jun 18 '15 at 14:33
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The answer from "introduction to electrodynamics" Have a look at that problem.

The answer-from solution manual- :A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV . But we know that Laplace’s equation allows no local minima for V . What looks like a minimum, in the figure, must in fact be a saddle point, and the box “leaks” through the center of each face.

enter image description here

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Where did the charge go that he was discussing so far ? The charge he was discussing so far is a test charge. Usually in electrostatics, a test charge is considered to be a very small charge such that it does not modify the field in question or the field which is being studied. The direction of motion of this hypothetical test charge gives the direction of lines of force.

Here Feynman is discussing the field produced by the presence of several static charges. Then he is considering a point P in this field. Please understand that "there is no charge at P". Then he says that if we calculate the divergence of the field from this point P and if the divergence comes out to be negative, then the point P will be an equilibrium point. (Then he says that the above situation is impossible since that violets The Gauss law)

He refers to placing a test charge at P to make us visualize the equilibrium situation. In the equilibrium situation the test charge would not move and if displaced by a little distance in any direction it would come back to the same position P again.

Mathematical description of Equilibrium: Divergence of the field at P is negative

Intuitive understanding of Equilibrium: A test charge at P, would not move and if displaced by a little distance in any direction it would come back to the same position P again.

So the idea of placing a test charge at P is only to make us visualize "the situation of equilibrium" by studying "the characteristics of the motion" of this charge.

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I am posting this because I believe none of the the other answers address the OP's question. The answer is already in the text by Feynman but somewhat implicitly.

We have a positive charge $q_+$ at the point $P_0$. We want to know if it is possible for $q_+$ to be in stable equilibrium.

Q: What do we mean by stable equilibrium?

A: A particle is in stable equilibrium happens at a point $P_0$ if it experiences no force at $P_0$ and if we make a small perturbation (i.e., we move the particle from the equilibrium point by a small distance) there is a force that makes it return towards the equilibrium point. Such a force is called a restoring force.

In electrostatics all forces can be expressed in terms of electric fields. A charge $q$ will experience a force

$$\overrightarrow {F_q} = q \overrightarrow E$$

For the particular case of our charge $q_+$ the force points towards the same side as the electric field.

We can now start with the assumption that $P_0$ is a point of stable equilibrium for $q_+$. This means that there is a restoring force in some neighborhood $V$ of $P_0$ inside of which, at every point $Q$, the electric field $\overrightarrow E$ points from $Q$ to $P_0$. If we apply Gauss's law to such a system we get

$$\oint\limits_{S=\partial V}\overrightarrow E\cdot\overrightarrow n ~\mathrm d a=\frac{q_+}{\varepsilon_0}$$

where $S$ is the boundary of the neighborhood $V$. But if we look at the left side of the equation we see that $\overrightarrow E$ points inwards whereas $\overrightarrow n$ points outwards (as per the convention for normal vectors to a closed surface). This means that the integrand is always negative and so the integral is as well.

A negative number cannot be equal to a positive number! Therefore, by reductio ad absurdum we conclude that $P_0$ cannot be an equilibrium point. Our starting assumption was wrong.


With the legwork we have done here we can also expand one of points Feynman makes. The text reads

except right on top of another charge

If there are several charges, and they are at different points in space we can always find a volume inside of which it is isolated and as we saw above, this can never be a point of stable equilibrium. But if two charges $q_1, q_2$ are right on top of one another then no matter how small a volume we take the right hand side of the equation will always be $(q_1+q_2)/\varepsilon_0$.

So, if our charge $q_+$ sits right on top of a negative charge $Q_-$, and $|Q_-|>q_+$, then is in stable equilibrium because Gauss's law does allow it this time (both sides are negative).

With this we can expand Feynman's claim to:

There are no points of stable equilibrium in any electrostatic field—except right on top of another charge of opposite sign and higher value.

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protected by Qmechanic Aug 29 '16 at 11:22

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