Question

Three charges lie along the x-axis as in the Figure. The positive charge q1 = 16 µC is at x= 2.0
m, and the positive charge q2 =8.0 µC is at the origin. Where must a negative charge q3 be placed
on the x-axis so that the resultant electric force on it is zero?

Answer 1

Let third charge be placed in between $q_1 , q_2$ at a separation of r from $q_2$

Then, according to the question

$\frac{k q q_2}{r^2 } = \frac{k q q_1}{(2-r)^2 }$

$\frac{k q*8*10^{-6}}{r^2 } = \frac{k q*16*10^{-6}}{(2-r)^2 }$

then $\frac{1}{r^2} = \frac{2}{(2-r)^2} \implies r^2+4r-4 =0$

Solving it,

$r = 2(\sqrt{2}-1) = 0.83 m$ from origin.