25: Magnitude and Direction of Electric Fields

INTRO:
A small object A, electrically charged, creates an electric field. At a point P located 0.250m directly north of A, the field has a value of 40.0N/C directed to the south.

Part A:
What is the charge of object A?

SOLUTION:
givens:
E = 40 N/C 
r = 0.250 m

we can solve for q by rearranging the formula E = K ⋅ q/r²
→ q = E⋅r²/K
q = (40 N/C) ⋅(0.250 m)² ⋅ 4πε₀
→(40 N/C) ⋅ (6.25×10^-2 m²)⋅ 4π(8.854×10^-12 F/m)
q = 2.782×10^-10 C
however, we must also take into consideration the direction of the electric field, since the field is pointing toward the charge, the charge must be negative. Electric fields point away from positive charges
∴ q = -2.782×10^-10 C

Part B:
If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?

SOLUTION:
Here we just have to find the net electric field by adding the electric field from A to the electric field from B

E_A = 40 N/C
r_B = r_A + 0.25 m = 0.5 m

E_B = K ⋅ q/r_B²
→[-2.782×10^-10 C] / [4πε₀⋅(0.5 m)²]
→[-2.782×10^-10 C] / [4π(8.854×10^-12 F/m)⋅(0.25 m²)]
E_B = -10 N/C

what this means by -10 N/C is that it's pointing south, just as E_A is
so E_NET = -40 N/C + -10 N/C = -50 N/C 
aka 50 N/C pointing south