# Irodov Solutions : Problem 3.12

A thin non-conducting ring of radius R has a linear charge density given by λ=λₒcosΦ where λₒ is a constant and Φ is the azimuthal angle. Find out the magnitude of electric field intensity
(a) at the centre of the ring and
(b) on the ring's axis as a function of the distance x from its centre. Investigate the obtained function at x>>R.

Firstly we should choose an axis going along one of the diameters of the ring and taking it as the reference axis (i.e. the azimuthal angle Φ would be measured with respect to this axis.) the nature of charge distribution should be visualised. As stated in the problem, the linear charge density λ for the ring varies according to the law λ=λₒcosΦ, this implies λ≥0 for -π/2 ≤ Φ ≤ π/2 and λ<0 for π/2 < Φ < 3π/2.
The above shown figure is an approximate representation of the charge distribution along the circumference of the ring (x axis is the reference). The intersection points of x-axis with the ring correspond to the maximum charge density (equal to λₒ as Φ=0) and those of y-axis are at zero charge density (as Φ=π/2).
It should also be noted that though the charge density is continuously varying from point to point but still the distribution of charge on both sides of the x-axis is fully identical (The same is not the case with the y-axis) i.e. the x-axis is an axis of symmetry and therefore at any point on the x-axis the resultant electric field vector would be directed along the x-axis itself.

### (a) Electric Field at Centre of the Ring

The above discussion already concluded that the field vector at the centre O is along the x-axis. So here our steps should be :
1. Choose an elementary charge dq on the ring.
2. Approximating the charge dq as a point charge, write the elementry field dE produced by it at the centre.
3. Take the component of dE along the x-axis and integrate it for the entire ring charge distribution. We get the resultant field at the centre of the ring.

### (b) Electric Field on the ring's axis

Observe the diagram shown below. The ring lies in xy plane with its centre at the origin and we wish to find field strength at point P lying on the z-axis at a distance x from the origin.

Let dE be the infinitesimal field at point P produced by the small charge element S. As the vector dE is in 3D, we should take its components along x, y and z axes. On integrating the component along a particular axis we get the component of the resultant field along that axis.
The component of dE along z-axis is dEcosƟ and the other rectangular component is dEsinƟ which is obviously perpendicular to the z-axis. The normal component dEsinƟ is parallel to the line joining O and S and it can further be broken along x and y axes.

Thus the z and y components of the resultant field vanish. So the resultant electric field at point P (in fact at any point on the ring's axis) would be along the x-direction.

This is the required expression of resultant field at point P. We may take some special cases here:

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