# Three point charges of +2 microcoulomb,-3 microcoulomb and -3 microcoulomb are kept at the vertices 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 mid point(M) of side BC so that the charge at A remains in equilibrium? 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 = 101/2

=>  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.

@ somya: good work. keep posting!

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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.

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