Answer
The net electric field at point P is 0
Work Step by Step
We can write the general equation for the magnitude of an electric field due to a charge $q$:
$E = \frac{\vert q \vert}{4\pi ~\epsilon_0~r^2}$
The electric field due to $q_1$ points away from $q_1$ since $q_1$ is a positive charge.
The electric field due to $q_2$ points away from $q_2$ since $q_2$ is a positive charge.
These two charges have the same magnitude and they are the same distance from point P, so the electric fields due to these two charges have the same magnitude. Since the electric fields due to these two charges point in opposite directions, they cancel with each other.
The electric field due to $q_3$ points away from $q_3$ since $q_3$ is a positive charge.
The electric field due to $q_4$ points toward $q_4$ since $q_4$ is a negative charge.
We can find the magnitude of the electric field due to $q_3$:
$E_3 = \frac{\vert 3e \vert}{4\pi ~\epsilon_0~d^2}$
We can find the magnitude of the electric field due to $q_4$:
$E_4 = \frac{\vert 12e \vert}{4\pi ~\epsilon_0~(2d)^2}$
$E_4 = \frac{\vert 12e \vert}{4\pi ~\epsilon_0~4d^2}$
$E_4 = \frac{\vert 3e \vert}{4\pi ~\epsilon_0~d^2}$
$E_4 = E_3$
The electric fields due to these two charges have the same magnitude. Since the electric fields due to these two charges point in opposite directions, they cancel with each other.
Therefore, the net electric field at point P is 0.
