Answer:
electric field 7.64 10⁵ N / C
Explanation:
For this problem we can use Gauss's law, which indicates that the load within a Gaussian surface can be considered located in its center, therefore, as the point is much greater than the radius of the spheres we can take them as if it were a spot loading Let's calculate the field at the midpoint between them
r = 0.520 m / 2
r = 0.260 m
The electric field is a vector
Et = E1 + E2
E1 has a negative charge, so the force is attractive and E2 has a positive charge, the outside is repulsive, so the two forces point in the same direction on the positive test charge.
E1 = k Q1 / r²
E2 = k Q2 / r²
Et = E1 + E2
Et = k Q1 / r² + k Q2 / r² = k (Q1 + Q2) / r²
Et = 8.99 10⁹ (2.4 10⁻⁶ + 3.35 10⁻⁶) / 0.26²
Et = 7.64 10⁵ N / C
