Why potential at equatorial point of dipole is zero?
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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
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.
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
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.
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I hope this helps you!
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