Answer
The magnitude of the electric field at point $S$ is $~\frac{7~k~q}{4~d^2}~$ and the electric field is directed in the negative x-direction.
Work Step by Step
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 $q$:
$E_1 = \frac{k~q}{(2d)^2} = \frac{k~q}{4~d^2}$
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 $2q$:
$E_2 = \frac{k~(2q)}{d^2} = \frac{2~k~q}{d^2}$
We can find the net electric field due to both point charges:
$E = E_1-E_2$
$E = \frac{k~q}{4~d^2}-\frac{2~k~q}{d^2}$
$E = -\frac{7~k~q}{4~d^2}$
The magnitude of the electric field at point $S$ is $~\frac{7~k~q}{4~d^2}~$ and the electric field is directed in the negative x-direction.
