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?
Best Answer
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 $\dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a}- \dfrac{1}{4\pi\epsilon_0} \dfrac{Qq}{a} = \bf 0$.