A point charge `q` is located at the centre fo a thin ring of radius `R` with uniformly distributed charge `-q`, find the magnitude of the electric field strength vectro at the point lying on the axis of the ring at a distance `x` from its centre, if `x gt gt R`.

Explanation

The electric potential at a distance `x` from the given ring is given by
`varphi (x) = (q)/(4pi epsilon_(0) x) - (q)/(4pi epsilon_(0) (R^(2) + x^(2))^(1//2))`
Hence, the field strength along x-axis (which si the net field strength in 'our case'),
`E_(x) = - (d varphi)/(dx) = (q)/(4pi epsilon_(0)) (1)/(x^(2)) - (qx)/(4pi epsilon_(0) (R^(2) + x^(2))^(3//2))`
`= ((q)/(4pi epsilon_(0)) x^(3) [(1 + (R^(2))/(x^(2)))^(1//2) -1])/(x^(2) (R^(2) + x^(2))^(3//2))`
`= ((q)/(4pi epsilon_(0)) x^(3) [1 + (3)/(2) (R^(2))/(x^(2)) + (3)/(8) (R^(4))/(x^(4)) + ...])/(x^(2) (R^(2) + x^(2))^(3//2))`
Neglectign the higher power of `R//x`, as as `x gt gt R`.
`E = (3q R^(3))/(4pi epsilon_(0) x^(4))`
Note : Instead of `varphi (x)`, we may write `E(x)` directly using `3.9`