The electric force and potential due to an electric dipole

Question

The electric potential is zero on the equitorial plane of an electric dipole but there is an electric field at that plane anti parallel to the direction of dipole moment. Now what is the significance when we say that the electric potential is zero at a point in an electric field? Again, does the electric field exert force on a charge placed on that point where the electric potential is zero? Again, we know that the electric field is conservative in nature and a consequence of that the potential energy acquired by an electric charge is converted to the kinetic energy. But the electric potential is zero at a point means there will not be any potential energy at that particular point then how will the charge moves from a point where the electric potential is zero?

Yejus · Accepted Answer

Think about what the electric potential at a point means. It is the amount of work done in bringing a unit positive charge from infinity to that point without acceleration. A potential of $0$ means that you do net zero work in doing so. This emphatically does not mean that the force at that point is zero, or that it was zero everywhere along the path, or even that the work done was zero everywhere.
To get an intuitive idea of how the net work can be zero, think of how the potential affects work. Broadly speaking, if the potential is negative at a point, the force is attractive at that point; if it is positive, the force is repulsive. You do positive work against a repulsive force and negative work against an attractive one.
In the case of a dipole, there will be points where you do some positive work while transporting a positive charge part of the way, and negative work the rest of the way, such that they cancel each other out perfectly. Those points correspond to zero potential.
To answer your last point, if you leave a charge at a point with $V = 0$, it will move to a point with a lower potential, $V<0$.

Buraian · Answer

The potential being zero at a point is irrelevant. What really matters is the potential difference between two points, that is, suppose you have two points A and B separated by a distance of $Delta r$,  and the potential of B is greater than A by $ Delta V$ volts ( say), and both are separated by a distance of $Delta r$. Then,
$$ vec{E} approx - frac{ Delta V}{Delta r} vec{r}$$
where $r$ is the unit vector pointing from A to B, now observe that the rise over run slope of the voltage difference is what causes the electric field.
To illustrate this idea further, suppose whole space had a voltage of a $10000000V$, then there still would be no electric field in this space because we need a 'potential' difference for electric field.
Now, suppose you have a space where this is a very large potential and a small region of space with low potential, if you have a 'positive charge' then this charge would be sucked into the region of low potential, while the electrons would be pushed away into infinity. Now, for equilibrium condition (for + charge), we just need to find the point where the 'rise over run slope' or in a calculus way, where the derivative is zero.
I.e
$$ 0 = nabla  U(x,y,z)$$
Or, in a single variable case,
$$ 0 = frac{dU(x)}{dx}$$
This derivative would be a function of 'x', finding the value of 'x' for where the function is zero, we get the  equilibrium point.
For example,
$$ U(x) = 3x^2 -x +C$$
$$ frac{dU}{dx} = 6x - 1$$ (notice here that derivative was independent of C, the constant, another way to say it only depends on potential difference)
Now, institute the condition that derivative must be zero
$$ 0 = 6x -1$$
$$ x= frac16$$
So, for a positive charge, $ x =frac{1}{6}$$ is where it would be pushed towards for the system to acquire least energy state. However a negative charge particle, the story is opposite, it would be pushed away from this point because the force on a negative particle is opposite to that of positive one.

The electric force and potential due to an electric dipole

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