Ok! This really makes hard to think, how does a single point determine the object's equilibrium? If an object is displaced from its equilibrium-position & if the equilibrium is stable, the object again comes back to the initial postition. My book says it is due to centre of gravity & couple constituted by the gravity & normal reaction force. But, how do this coupling and COG help in bringing back the equilibrium state? Can anyone help me visualize this??
[Physics] How does centre of gravity influence the equilibrium
equilibriumfree-body-diagramnewtonian-mechanicsstatics
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In a sense, you're both right. You're correct in that the force required is independent of the height of the centre of gravity, while your friend is correct that a lower CoG makes the box more stable. This may seem contradictory, but it's not, and here's why: the force required to move the box is independent of the height of the centre of gravity, but the work done to the box is not. The stability of the box is essentially defined as how far do you need to top it for it to topple over. This winds up being how far do you need to tip it for the CoG to pass over the pivot point so that it starts pulling the box down onto its side. If the CoG is very high, then it doesn't need to be tipped as far as if it's very low. And how far you tip the box is measured by the work done to it, not just the force that's applied. So while the magnitude of force needed to move the box is independent of the height, how long that force needs to be applied to get the box to tip over is not.
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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I am going to present an example which helped me understand this!
Assume two surfaces as shown below:
Assume a droplet of water to be placed in a stable state on both of them. (i.e. at the bottom of the left one, and the top of the right one.
In the left one, if you push the water droplet by a small amount, the water droplet would tend to flow back into the point where you pushed it away from. Therefore it is in a Stable Equilibrium state. (i.e. the object comes back to its initial position.) this happens because the gravitational force always tries to bring a body to a lower potential energy position.
On the contrary, in the surface represented on the right, the slightest push would cause the droplet to go further down the surface, under gravitational acceleration. i.e. It is in an unstable equilibrium state.
Here, the gravitational force causes the droplet to return/move further away from its equilibrium state, and that is how it returns to its initial position in the left hand representation.
I hope this has been clear enough.