Question

Two particles A and B,each having a charge Q,are placed at a distance d apart.Wher ahould a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force?what is the magnitude of this maximum force..?

Answer 1

see pic
let two equal charges , Q placed A and B.

and another charge q placed at C , CD distance from AB

according to question , AD = DC
and CD perpendicular upon AB

now , let AB = d => AD =DB =d/2
and CD = y ( let)

Fnet = sum component of forces acted by both charge particle In vertical direction .

Fnet = 2Fcos∅

where cos∅ = y/√(d²/4 +y²)

F = KqQ/(d²/4 + y²)

so,

Fnet =F = 2KqQy/(d²/4 + y²)^3/2

differentiate wrt y
dF/dy = 2KqQ{ (d²/4+y²)^3/2 -3/2y(d²/4+y²)½(2y)}/(d²/4+y²)³

dF/dy = 0

(d²/4+y²)½ { d²/4 + y² -3y² } = 0

d² = 8y²

y² = d²/8

y = ±d/2√2

force will be maximum at y = d/2√2
becoz here d²F/dy² < 0 at y = d/2√2

now ,
F = 2KQqy/( d²/4+y²)^3/2

=2KQq(d/2√2)/(d²/4 + d²/8)^3/2

=16KqQ/3√3d²