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.
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?
Best Answer
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
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: