Question

Answer the following question.
A thin circular ring of radius r is charged uniformly so that its linear charge density becomes λ. Derive an expression for the electric field at a point P at a distance x from it along the axis of the ring. Hence, prove that at large distances (x >> r), the ring behaves as a point charge.

Answer 1

Let consider a thin circular ring of radius r with charge density as `lambda`

We need to find the electric field due to this charged ring at a point on the axis of the ring at a distance x from its centre.
Let us consider a small charge element `(dx)` on the ring having small charge `dq`

`dq = lambdadx`

the electric field due to this charge element at the point P is given by 

`d"E" = 1/(4piepsilon_0) (dq)/((r^2+x^2)`
`d"E" = 1/(4piepsilon_0) (lambda dx)/((r^2+x^2)`

Electric field at the point P will have two components one in the vertical direction and the other one in the horizontal direction.

`d E cos theta` along the horizontal direction.

`d E sin theta` along the vertical direction.

The vertical components will cancel out the effect of each other due to the presence of the diametrically opposite element.
So the horizontal component of the electric field will survive at the point P. 
From the figure we have the value of  `cos theta = x/sqrt(r^2+x^2)`

Now the integration of the horizontal component `dE cos theta` will be carried out.
`dE cos theta = (lambdax dx)/(4piepsilon_0(r^2+x^2)3/2`

Since the value of `dq = lambda dx`
`dE cos theta = (xdq)/(4piepsilon_0(r^2+x^2)3/2`

Now integrating the above equation and taking x and r quantities as constants we get 

`E_x = int_  dE cos theta = int_  (xdx)/(4piepsilon_0(r^2+x^2)3/2`

`E_x = int_  (xdq)/(4piepsilon_0(r^2+x^2)3/2) = (x"Q")/(4piepsilon_0(r^2+x^2)3/2` where Q is the total charge on the ring.

Here `E_x` is the value of the total electric field at the point P 

Special case:   

when x >> r, the denominator of the above equation gets modified in the following way:

`r^2 + x^2 ≈ x^2`

`E_x = (x"Q")/(4piepsilon_0(x^2)3/2) = (x"Q")/(4piepsilon_0x^3) = ("Q")/(4piepsilon_0x^2)`

So at large distances (x >> r), the ring behaves as a point charge.