A circular copper ring of radius r, placed in vacuum, has a charge q on it. Find out the electric fields For what value of x would the electric field be maximum?

Explanation

For E to be maximum, `(dE)/(dx)=0`
Now, `(dE)/(dx)=q/(4piepsilon_(0))cdotd/(dx)[x(r^(2)+x^(2))^(-3//2)]`
`=q/(4piepsilon_(0))[1cdot(r^(2)+x^(2))^(-3//2)+xcdot(-3/2)cdot(r^(2)+x^(2)^(-5//2)cdot2x]`
`=q/(4piepsilon_(0))(r^(2)+x^(2))^(-5//2)[(r^(2)+x^(2))-3x^(2)]`
`=q/(4piepsilon_(0))(r^(2)+x^(2))^(-5//2)(r^(2)-2x^(2))`
For maximum E,
`(dE)/(dx)=0` or `r^(2)-2x^(2)=0` or, `x=pmr/(sqrt2)`
The `'pm'` sign implies that the electric field on the axis will be maximum at a distance `r/(sqrt2)` on either side of the ring.