Well ok I know that electric potential at equatorial point of a dipole is 0. It is a derived conclusion.

Q1. But how this can be explained "theoretically"?

Another question that electric potential at infinity is 0. So if I bring a positive charge from infinity to the equatorial point of the dipole then the work done by me to bring it is 0. How could this be possible if I am true?

Q2. If I am not then what is the work done in this case?

Please help


If you look at the electric filed lines you will note that a positive charge moving in the direction of the blue arrow does have a force on it but that force is always at right angles to its motion. Hence no work is done moving the charge.

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Q1. But how this can be explained "theoretically"?

I assume your question is about the concept of 'electric potential ' due to a distribution of charges and in the present case 'a dipole'.

The best way is to imagine an unit positive charge being carried/moved from infinity to a point on the equatorial line of the dipole.

Naturally your probe charge will be moving in the field of dipole charges , which are equal and opposite and

at any segment of the equatorial line say dy ,the effective field of forces or the net electric field is sum of the two forces acting due to the two charges of the dipole, which are opposite in character but equal in magnitude -

so if the effect of one is attractive then the effect of other one of the pair will be repulsive and

the segment dy is symmetrically placed with respect to the dipole charges ,therefore the two forces will be equal in magnitude but acting in such directions that their resultant will be normal to dy , and the displacement of unit positive charge being normal to net electric field can lead to a zero work done.

If work done during translation along all such segments be added together then the sum will be zero. so net work done will vanish and the potential will be zero.

The above displacement was done on equatorial line but in conservative force field of electric charges the path is not important -the end points are important -the end points are infinity and point P at the equatorial line and potential at both points are zero.

the electric potential difference between two points are independent of actual path traversed by the charge

therefore any curved path between infinity and the point P should give us the same potential difference-which in the presence case comes to zero. One can ask what is happening ?

Actually the potential due to one charge of the dipole is just equal and opposite to that of due to other charge on any point on the equatorial line,therefore the potential of a dipole vanishes on any point on the equatorial line. the above is due to symmetry of the charges of dipole and their opposite character.

  • $\begingroup$ First of all very very sorry for the late comment. What I want to say is that at any point of equatorial line the net force/electric field is non zero. So how can u say that the individual forces due to charges of dipole are Equal And Opposite? And Again sorry for this late comment $\endgroup$ – Perspicacious Jun 11 '16 at 16:07
  • $\begingroup$ <Naturally your probe charge will be moving in the field of dipole charges , which are equal and opposite and> actually the statement is about charges and not forces as force will have a direction and line of action and one can think of net force acting on a test charge . $\endgroup$ – drvrm Jun 12 '16 at 3:45
  • $\begingroup$ Well thanks for your answer. At last your answer satisfied me. Thanx for ur time and sorry for any inconvenience $\endgroup$ – Perspicacious Jun 12 '16 at 5:48

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