Answer
The magnitude of the electric field is $1.52\times 10^8~N/C$ and it is directed toward the $-15~\mu C$ point charge.
Work Step by Step
The electric field is directed away from a positive charge and toward a negative charge. We can find the magnitude of the electric field due to the two point charges:
$E = \frac{k~\vert q_1 \vert}{r^2}+\frac{k~\vert q_2 \vert}{r^2}$
$E = \frac{(9.0\times 10^9~N~m^2/C^2)(12\times 10^{-6}~C)}{(0.040~m)^2}+\frac{(9.0\times 10^9~N~m^2/C^2)(15\times 10^{-6}~C)}{(0.040~m)^2}$
$E = 1.52\times 10^8~N/C$
The magnitude of the electric field is $1.52\times 10^8~N/C$ and it is directed toward the $-15~\mu C$ point charge.
