Work done in bringing a test charge from infinity to center of dipole

Question

Potential at Equatorial point of dipole is given 0 in my book(after putting angle = π/2 and cos(π/2) = 0 in formulae
V = (K p cos(x)/r2)
x = angle made my direction vector of point where potential is to be determined with center of dipole
p = dipole moment
R = distance between two charges

I am confused that potential at a point is work done per unit charge in bringing from infinity to that point and suppose if I bring a test charge along the axial line and stops at a point lying on Equatorial line then I have to do work since Electric field was there due to dipole.

Can you please explain me this?

Answer 1

Consider the movement of a charge $+q$ from infinity, the zero of potential and at first consider the effect that charges $+Q$ and $-Q$ have on charge $+q$ separately.

In relation to charge $+Q$, the work done in moving charge $+q$ from infinity to position $P$ is $\dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a}$.
Again in relation to charge $+Q$ no work is done moving charge $+q$ from position $P$ to position $O$.
So, in relation to charge $+Q$, the total work done in moving charge $+q$ from infinity to position $O$ is $\dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a}$.
Of course one could just move charge $+q$ from infinity to position $O$ and the work done in relation to charge $+Q$ would be $\dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a}$.
Now, in relation to charge $-Q$ the work done in moving charge $+q$ from infinity to position $O$ is $-\dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a}$.

So the total work done in moving charge $+q$ from infinity to position $O$ is $\dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a}- \dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a} = \bf 0$.

Answer 2

then I have to do work since Electric field was there due to dipole.

The work done depends on the direction of the field. If you move perpendicular to the field, no work is done.

If you were to draw field lines for your dipole, you would find the sample path you drew to be everywhere perpendicular to the field. No work is done along this path. If you were to pick some other path, you would find that some portions have positive work and other portions have negative work. It would still sum to zero.

An equivalent way to think of this is that the work done is equal to $q \Delta V$. How do you change the potential? You need to have some component of the path in the direction of the electric field.