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Question

Two particles A and B having charges 8×106C and 2×106 respectively are held fixed with a separation of 20cm. Where should a third charged particle C be placed so that it does not experiences a net electric force?

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Solution

As the net electric force C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have changes of opposite signs, C cannot be between A and B.
Also A has larger magnitude of charge than B. Hence, C should be placed closer to B that A. The situation is shown in figure. Suppose BC=x and the charge on C is Q
FCA=14π0(8.0×106)Q(0.2+x)2^i
and FCB=14π0(2.0×106)Qx2^i
FC=FCA+FCB=14π0[(8.0×106)Q(0.2×106)(2.0×106)Qx2]^i
But |FC|=0
Hence 14π0[(8.0×106)Q(0.2×x)2(2.0×106)Qx2]=0
Which gives x=0.2m
1029653_1013491_ans_ae2c8574ebe84343ae162d9825a3f68c.png

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