A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.

Explanation

Let us consider a circular loop of radius a held perpendicular to the plane of paper.
Let whole of loop carries a charge q. charge on a small element of loop MN of length dl is
`dq=(q)/(2pia)dl` . . . (i)

Electric field at P due to charge element MN is
`dE=(1)/(4piepsi_(0))(dq)/(r^(2))`
`=(1)/(4piepsi_(0))(dq)/((x^(2)+a^(2)))` along PX
dE can be resolved into two rectangular components. dE cos `theta` acting horizontally along PZ and `dE sin theta` acting along PY.
For a pair diagrammatically opposite element M., N., dE cos `theta` acts along PX and hence addd up but `dE sin theta` along PY. cancel with dE sin `theta` along PY.
`therefore sumdE sin theta=0` for full loop and dE cos `theta` addd up
`therefore E=sumdEcostheta`
`=sum(1)/(4piepsi_(0))(dq)/((x^(2)+a^(2)))xx(x)/((x^(2)+a^(2))^(1//2))`
or `E=int(1)/(4piepsi_(0))xx(x)/((x^(2)+a^(2))^(3//2))xx(q)/(2pia)dl` [Substituting the value of dq from (i)]
or `E=(1)/(4piepsi_(0))xx(x)/((x^(2)+a^(2))^(3//2))xx(q)/(2pia)intdl`
But `intdl=l=2pia`
`therefore E=(1)/(4piepsi_(0))xx(x)/((x^(1)+a^(2))^(3//2))xx(q)/(2pia)xx2pia`
or `E=(qx)/(4piepsi_(0)(x^(2)+a^(2))^(3//2))` . . (ii)
Special case. when `x gt gt a ` i.e., P lies far off, then `a^(2) lt lt x^(2)` and hence can be neglected.
`therefore E=(qx)/(4piepsi_(0)(x^(2))^(3//2))=(qx)/(4piepsi_(0)x^(3))`
or `E=(1)/(4piepsi_(0))xx(q)/(x^(2))`
which is the expression for E at a distance x from a charge q. therefore circular loop of charge behaves as a point when P is at a very very large distance from the loop.