Answer
At the point $x = 0.586~m$, the electric field $E = 0$
Work Step by Step
In the region $0 \lt x \lt 1.00~m$, the electric field due to each point charge is non-zero and they point in opposite directions. The net electric field is zero when the magnitude of the electric field due to each point charge is equal.
The electric field due to the point charge $20.0~nC$ is directed in the positive x-direction. We can find the magnitude of the electric field due to this point charge:
$E_1 = \frac{k~q_1}{x^2}$
The electric field due to the point charge $10.0~nC$ is directed in the negative x-direction. We can find the magnitude of the electric field due to this point charge:
$E_2 = \frac{k~q_2}{(1.00-x)^2}$
We can find the point $x$ where the magnitude of the electric field due to each point charge is equal:
$E_1 = E_2$
$\frac{k~q_1}{x^2} = \frac{k~q_2}{(1.00-x)^2}$
$(1.00-x)^2~(20.0~nC) = x^2~(10.0~nC)$
$(1.00-x)^2~(2) = x^2$
$(1.00-x)~\sqrt{2} = x$
$(\sqrt{2}+1)~x = \sqrt{2}$
$x = \frac{\sqrt{2}}{\sqrt{2}+1}$
$x = 0.586~m$
At the point $x = 0.586~m$, the electric field $E = 0$.
