Jan 22, 2018

$0.158 m$

#### Explanation:

Field due to charge $q 1$ at point $X$ is $E 1 = 9 \cdot {10}^{9} \cdot \left(- 2\right) \cdot {10}^{-} \frac{6}{\frac{10}{100}} ^ 2$ i.e $- 18 \cdot {10}^{5} \frac{N}{C}$

So,if field has to be zero at point $X$,the field$\left(E 2\right)$ due to $q 2$ must be equals to and opposite in direction to $E 1$

so, we can write,$E 1 + E 2 = 0$

now, $E 2 = 9 \cdot {10}^{9} \cdot 5 \cdot {10}^{-} \frac{6}{r} ^ 2$ (where, $r$ is the distance of $q 2$ from $X$)

so,equating both,we get,

$r = 0.158 m$

Charge at $q 2$ is obviously positive,or else direction would have been same for $E 1$ and $E 2$