Two particles A and B , each carrying a charge Q , are located a distance d apart. At what point on the perpendicular bisector of AB must a particle of charge q be positioned to experience the maximum force, and what is the magnitude of this maximum force?

Step-by-step solution

Let the two charges Q be placed at (- d 2 , 0 ) and ( d 2 , 0 ) . The charge q is placed at (0, y) on the perpendicular bisector. The net force on q is directed along the y-axis and its magnitude is given by: F = 2 1 4 ₀ Qq ( d 2 )^2 + y^2 where = y ( d 2 )^2 + y^2 . Thus, F = 2Qq 4 ₀ y ( d^2 4 + y^2 )^ 3/2 . For maximum force, dF dy = 0 . d dy [ y ( d^2 4 + y^2 )^ -3/2 ] = 0 ( d^2 4 + y^2 )^ -3/2 - 3 2 y ( d^2 4 + y^2 )^ -5/2 (2y) = 0 d^2 4 + y^2 - 3y^2 = 0 2y^2 = d^2 4 y = d 2 2 . Substituting y = d 2 2 into the