Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure below, in which q = 2 nC, d = 12 cm, and the positive x-axis points to the right?

Answer 1

For +2nC charges, kQ = 8.99*10^9 * 2*10^-9 = 18

Electric field strength E = kQ / r^2
Electric field direction is radially away from + charges

At the dot:
r1 = 6cm = 0.06m
so E1 = 18/ (0.06)^2 N/C = 5000 N/C @ direction angle 0.
E1 = 5000i + 0j

r2 = sqrt(6^2 + 12^2) cm = 13.42 cm = 0.1342 m
so E2 = 18 / (0.1342)^2 = 999.5 N/C
Angle is 90 - arctan(6/15) = 68.92 deg
E2 = 359.3 i + 932.6j

So the net field strength at the dot is
E = E1 + E2 = 5359.3 i + 932.62j

E = ((5359.3)^2+(932.62)^2)^0.5 =5439.8 N/C

tan (theta) = 932.62/5359.3

theta = 9.87 degrees above horizontal