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Question

# Two particles A and B having charges 8×10−6C and −2×10−6 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=14π∈0(8.0×10−6)Q(0.2+x)2^iand →FCB=−14π∈0(2.0×10−6)Qx2^i→FC=→FCA+→FCB=14π∈0[(8.0×10−6)Q(0.2×10−6)−(2.0×10−6)Qx2]^iBut |→FC|=0Hence 14π∈0[(8.0×10−6)Q(0.2×x)2−(2.0×10−6)Qx2]=0Which gives x=0.2m

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