Q. 313.9( 7 Votes )

Two particles A and B, each having a charge Q, are placed a distance d apart. Where should 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 :

Charge on particles A and B: q1= q2= Q
Separation between A and B : d

here ‘x’ is the distance at which particle C of charge ‘q’ is place on the perpendicular bisector of AB.
Formula used:
From the figure we can find sinθ :
The horizontal components of force cancel each other.
Total vertical component of force is: F’= 2Fsinθ
We use Coulomb’s Law:
Where F is the electrostatic force on b due to A, k is a constant .
k =
= 9× 109 Nm2C-2 and r is the distance between the two charges.
Here r =

Substituting we get,

The magnitude of maximum Force is:

Hence, the particle C must be placed at a distance of from the perpendicular bisector of AB so as to experience a maximum force of magnitude

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Superfast Revision of Potential & Potenial Energy (Theory + Interactive Quiz)47 mins
Check your knowledge of Electric Potential | Quiz45 mins
Relation between electric field and potential34 mins
Relation between Electric field & Potential45 mins
Interactive quiz on Relation between electric field and potential43 mins
Electric field inside cavity37 mins
Few Applications of Gauss's law54 mins
Test Your Knowledge of Electric Charge & Coulomb's Law47 mins
Electric potential inside solid and hollow sphere30 mins
Electric Field Lecture47 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses