It's been a while since I've studied electrostatics and is my first time posting here, so please forgive me if I'm missing something basic.
According to my reference book, the electric potential of a dipole along its perpendicular bisector is zero. Assuming a point P near a dipole such that the distance between the point P and the two end-point charges $q$ and $-q$ is constant, implying that the angle between the line of distance $r$ joining P and the midpoint of $q$ and $-q$ is $\theta=90^\circ$. Now varying this distance $r$ while keeping the angle constant, the potential energy remains zero and no work is done while traveling along the perpendicular bisector.
And we know that the electric field is non-zero along the perpendicular bisector and will definitely vary when r varies. How does this make sense when potential difference between two points divided by the displacement between them is supposed to give the negative of the electric field component in that direction?
Best Answer
Please regard this answer as complementary to Farcher's.
The potential of a point P may be defined as the work needed per unit 'test' charge to bring a test charge to P from infinitely far away.
Suppose we choose to bring the test charge to P by taking it towards the dipole along the perpendicular bisector of the dipole. No work is done on the test charge ($q$) because we are not pushing it against any opposing force: the electric force $q\vec E$ is at right angles to each increment of displacement, because along the perpendicular bisector the electric field is always parallel to the axis of the dipole.
Work done over displacement $d\vec s$ is $dW=q\vec E.d\vec s=0$.
The same result (zero potential at P) is obtained whichever route we take the test charge from infinity to P.