A whin wire ring of radius `r` carries charge `q`. Find the magnitude of the electric field strength on the axis of the ring as a function of distance `l` from its centre. Investigate the obtained function at `l gt gt r`. Find the maximum strength magnitude and the corresponding distance `l`. Draw the appoximate polt of the function. `E (l)`.

Explanation

From the symmetry of the condition, it is clear that, the field along the normal will be zero
`E_(n) = 0` and `E = E_(1)`
Now, `dE_(1) = (dq)/(4pi epsilon_(0) (R^(2) + l^(2))) cos theta`
But `dq = (q)/(2pi R) dx` and `cos theta = (1)/((R^(2) + l^(2))^(1//2))`
Hence
`E = int d E_(1) = int_(0)^(2pi R) (ql)/(2pi R) , (dx)/(4pi epsilon_(0) (R^(2) + l^(2))^(3//2))`
or `E = (1)/(4pi epsilon_(0)) (ql)/((l^(2) + R^(2))^(3//2))`
and for `l gt gt R`, the ring behaves like a point charge , reducing the field to the value,
`E = (1)/(4pi epsilon_(0)) (q)/(l^(2))`

For `E_(max)`, we should have `(dE)/(dl) = 0`
So, `(l^(2) + R^(2))^(3//2) - (3)/(2) l (l^(2) + R^(2))^(1//2) 2l = 0` or `t^(2) + R^(2) - 3 l^(2) = 0`
Thus `l = (R )/(sqrt(2))` and `E_(max) = (q)/(6 sqrt(3) pi epsilon_(0) R^(2))`