Answer
At the point $~x = 1.24~d~$, the electric field $E = 0$
Work Step by Step
In the region $x \lt 0$, the electric field due to each point charge is non-zero and points in the same direction. Therefore, the net electric field is non-zero. There are no points in the region $x \lt 0$ such that $E = 0$.
In the region $0 \lt x \lt 3d$, 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. There is one point in the region $0 \lt x \lt 3d$ such that $E = 0$.
In the region $x \gt 3d$, the electric field due to each point charge is non-zero and points in the same direction. Therefore, the net electric field is non-zero. There are no points in the region $x \gt 3d$ such that $E = 0$.
The electric field due to the point charge $q$ is directed in the positive x-direction. We can find the magnitude of the electric field due to the point charge $q$:
$E_1 = \frac{k~q}{x^2}$
The electric field due to the point charge $2q$ is directed in the negative x-direction. We can find the magnitude of the electric field due to the point charge $2q$:
$E_2 = \frac{k~(2q)}{(3d-x)^2} = \frac{2~k~q}{(3d-x)^2}$
We can find the point $x$ where the magnitude of the electric field due to each point charge is equal:
$E_1 = E_2$
$\frac{k~q}{x^2} = \frac{2~k~q}{(3d-x)^2}$
$(3d-x)^2 = 2x^2$
$3d-x = \sqrt{2}~x$
$(\sqrt{2}+1)~x = 3d$
$x = \frac{3d}{\sqrt{2}+1}$
$x = 1.24~d$
At the point $x = 1.24~d$, the electric field $E = 0$
