Answer :

Given:


Charges: q1=-1.0×10-6 C, q2=2.0×10-6 C



Let the third charge q be at a distance x cm from q2. As the forces must cancel, case II is not possible. Moreover, for forces to cancel, q must be near the smaller charge. Hence, only Case III is possible.



________________________________________________


Formula used:


By Coulomb’s law, the electric force is given by:



Where ϵ0 is the permittivity of free space


k is the electrostatic constant


q1 and q2 are the magnitude of charges


r is the distance of separation between the charges


__________________________________________________


Force on q due to q1 is given by:



Force on q due to q2 is given by:



Now,








For Case III, x must be greater than 10. Hence, x = 34.14 cm


Thus, q should be placed 34.1 cm from the larger charge on the side of the smaller charge.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
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
caricature
view all courses
RELATED QUESTIONS :

(a) Describe briePhysics - Board Papers

Repeat the previoHC Verma - Concepts of Physics Part 2

There is another Physics - Exemplar

Does the charge gPhysics - Board Papers