Two particles , each having a charge `Q`, are fixed at `y = d//2` and `y = -d//2`. Where should a particle of charge `q` be placed on `x`-axis from origin so that it experiences maximum force and what is it equal to ? Sketch variation of electric force experienced by `q v//s x`.

Explanation

Force on `q` to each `Q` will be same in magnitude
`F_(0) = (1)/( 4 pi in_(0)) (Qq)/(((d^(2))/(4) + x^(2))`
Resultant force on `q`,
`F = 2 F_(0) cos theta`
`= 2. (1)/( 4 pi in_(0)) (Qq)/(((d^(2))/(4) + x^(2))) (x)/(((d^(2))/(4) + x^(2))^(1//2))`
`= (Qq)/(2 pi in_(0)) (x)/((d^(2))/(4) +x^(2))^(3//2)`
For `F` to be maximum,
`(dF)/(dx) = 0`
`(Qq)/(4 pi in_(0)) (((d^(2))/(4) + x^(2))^(3//2) . 1 - x.(3)/(2)((d^(2))/(4) + x^(2))^(1//2) . 2x)/(((d^(2))/(4) + x^(2)))`
`(d^(2))/(4) + x^(2) - 3x^(2) = 0`
`x^(2) = (d^(2))/(8)`
`x = +- (d)/( 2 sqrt(2))`
`F_(max) = (Qq)/( 2pi in_(0)) (d)/(2 sqrt(2)) (1)/((d^(2))/(4) + (d^(2))/(8))^(3//2)`
`= (Qq)/( 2 pi in_(0)) (d)/(2 sqrt(2)) (1)/((3d^(2))/(8) .(sqrt(3) d)/(2 sqrt(2)))`
` = (4 Qq)/( 3 sqrt(3) pi in_(0) d^(2))`
`F v//s x` graph : when `q` is at right of `O` , force on it will be along ` + x` axis (assuming `+ve`) , graph will above x-axis.
When `q` is the left of `O`, force will be along-x-axis (assuming `-ve`) , graph will be below x-axis.
`x = 0 , F = 0`
`x = +- (d)/(2 sqrt(2)) , F = F_(max)`
`x rarr prop , F rarr 0`