Question

Why is potential due to dipole cylindrically symmetric?

Answer 1

The magnitude of the electric field intensity due to a point charge is given by

E=14πϵoQR2E=14πϵoQR2

It is clear from the above formula that at all point equidistant from Q, electric field intensity is same. Since all the equidistant points lie on the surface of a sphere, we speak of spherical symmetry.

However in case of a dipole the symmetry is not cylindrical but circular about the axis of the dipole.


Electric field due to a short dipole at general point is given by

E=P4πϵor33cos2θ+1−−−−−−−−√E=P4πϵor33cos2θ+1

Field is same at at all points where rr and θθ are constant. Thus if you rotate the position vector rrof point PP about the axis of the dipole, all such points (constant rr and θθ) lie on a circle. And hence the circular symmetry.

Notice that that the field at point QQ is not same as that at PP because it has different set of rr and θθ. Although PP and QQ are at the cercles of same radii i.e. on a cylindrical surface, the field is different. Hence no cylindrical symmetry