Three charges lie along the x-axis as shown in the figure. The positive charge `q_(1)=15.0 muC` is at x = 20 m, and the positive charge `q_(2)=6.00muC` is at the origin. Where must a negative charge `q_(3)` be placed on the x-axis such that the resultant force on it is zero? (in mm)

Explanation

Since `q_(3)` is negative and both `q_(1) and q_(2)` are positive, the force `vecF_(31) and vecF_(32)` are both attractive. Let x be the co-ordinate of `q_(3)`.
We have
`F_(31)=(K|q_(3)||q_(1)|)/((2-x)^(2)),F_(32)=(K|q_(3)||q_(2)|)/(x^(2))`

Since the net force on the charge `q_(3)` is zero,
we have, `(K|q_(3)||q_(2)|)/(x^(2))=(K|q_(3)||q_(1)|)/((2-x)^(2))`
or, `(4-4x+x^(2))(6xx10^(-6)C)=x^(2)(15xx10^(-6)C)`
Solving this quadratic equation for x, we get x = 775 mm.