Consider a uniformly charged ring of radius R. Find the pint on the axis where the electrie field is maximum.

Explanation

We have the following relation for the electric field intensity of ring on a point on its axis :
`E=1/(4piepsilon_0) (Qr)/(a^2+r^2)^(3//2)`
`E=Q/(4piepsilon_0)r/(a^2+r^2)^(3//2)`
let us assume the variable part of above relation as `y:y=r/(a^2+r^2)^(3//2)` , here if y is maximum , then E being proportional , will also maximum .
`(dy)/(dr)=((a^2+r^2)^(3//2)xx1-rxx3/2(a^2+r^2)^(1//2)xx2r)/(a^2+r^2)^3=0`
`rArr (a^2+r^2)^(3//2) xx1-rxx3/2 (a^2+r^2)^(1//2)xx2r=0`
`rArr a^2+r^2-3r^2=0 rArr r=pm a/sqrt2`
On substituting back in expression of electric field intensity , we get the following :
`E=Q/(4piepsilon_0)(a/sqrt2)/(a^2+a^2/2)^(3/2)`
`rArr E=Q/(4piepsilon_0) a/(sqrt2((3a)^2/2)^(3/2))`
`rArr E=Q/(4piepsilon_0)2/(3sqrt3a^2)`
`rArr E=2/(3sqrt3)1/(4piepsilon_0)Q/a^2`