Two particles A and B having charges ` 8.0 xx 10 ^(-6) C` and `-2.0 xx 10(-6) C` respectively are held fixed with a. separtion of 20cm. Where should a thired charged. particle be placed so that it dose not experience a net electric force?.

Explanation

As the net electric force on C should be equal to zero, the forces 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 charges of opposite signs, C cannot be between A and B also, A had larger magnitude of beloxx charge than B. Hence, C should be placed closer to B than A. The situationis shown in figure suppose BC=x and the charge on C is Q.
`vecF_(CA)=(1)/(4 pi epsilon_(0))((8.0xx10^(-6))Q)/((0.2+x)^(2))hat i`
and `vecF_(CB)=(1)/(4 pi epsilon_(0))((2.0xx10^(-6))Q)/(x^2)hat i`

`vecF_(C)+vecF_(CA)+vecF_(CB)=(1)/(4 pi epsilon_(0))[((8.0xx10^(-6))Q)/((0.2+x)^(2))-((2.0xx10^(-6))Q)/(x^2)]i`
But `|vecF_(C)|=0`
Hence, `(1)/(4 pi epsilon_(0))[((8.0xx10^(-6))Q)/((0.2+x)^(2))-((2.0xx10^(-6))Q)/(x^(2))]=0`
Which goves `x=0.2m`