Question

Three point charges + 2 microcoulomb and minus 3 microcoulomb are kept at the world is is a b and c respectively of an equilateral triangle of side 20 cm what should be the sign and magnitude of the charge to be placed at the midpoint of the side BC so that they charge at a remains in equilibrium

Answer 1

 first of all just draw the figure and calculate the net force on A due to B and C. as A will attract both B and C. the net force will be towards M, the midpoint.it will come out 2.33 N keep this force Equal to the force between AM. hence the answer is 3.8 micro coulomb.


 

Force between charge A and B, FAB= -0.015k N

Force between charge A and C, FAC = -0.015k N

=> FAB = FAC = F

=> Resultant force, F' = [2F2 + 2F2cos 60]1/2 = 0.025k 

Force between A and M be F''

Now the forces must be equal in magnitude and opposite in direction for A to be in equilibrium. 

=> F' = F''

Distance between A and M, AM = 10[3]1/2

=>  k 2Q/300 = 0.025k

=> Q = 3.89 microcoloumb

Therefore, a positive charge should be kept at the mid poiny of BC for charge A to be in equilibrium. 

Answer 2

Answer:

Explanation:

Solution: The forces on the charge at A due to the charges at B and C are equal in magnitude and are given by

These are directed as in the figure shown above. Their resultant is given

and is directed along AM.

Suppose, a charge +q is placed at M for the equilibrium of the charge at A.

The force on the charge at A due to the charge q is given by

But (AM)2 = (AB)2 - (BM)2 = (0.2m)2 - (0.1m)2

                      =  0.03m2

The force F1 is directed as shown. For the equilibrium of the charge at A, we must have

F1 = F12