Two particles A and B having equal charges are placed at distance d apart. A third charged particle placed on the perpendicular bisector at a distance x will experience the maximum Coulomb’s force when :

Explanation

Horizontal force is balanced .
So, total vertical force is equal to
` F_eq=2Fsinetheta `
` F_(eq) =2xx1/(4piepsilon_0)(qtheta)/((d/2)^2+x^2)x/sqrt((d/2)^2+x^2) `
` 1/(2piepsilon_0) (qthetax)/[(d/x^2)+x^2]^(3/2) `
` For the maximum force (dF_(eq))/dx=0 `
` So, `
` (d theta )/(2piepsilon_0).[[1.(d/2)^2+x^2]^(3/2)-x3/2[(d.2)^2+x^2]^(1/2) 2x] `
` or (d/x^2)+x^2-3x^2=0 `
` or 2x^2=(d/2)^2 `
` x=d/(2sqrt2)`
` F=1/(2piepsilon_0 ) (qtheta.(d/(2sqrt2)))/[(d/2)^2+(d/2sqrt(2))^2]^(3/2) =3.08 (thetaq)/(4piepsilon_0d^2)`.