Answer to Question #186164 in Physics for Pumpkin15

Question

What is the magnitude of the net electric field at (+1.0 m, 0) due to the three charges such as q1 = +10 μC, is at x = -1.0 m. The second charge, q2 = +20 μC, is at the origin. The third charge, q3 = - 30 μC, is located at x = +2.0 m

Answer 1

Magnitude of electric field due to point charge is $E = \frac{k q}{r^2}$, where $k = 9 \cdot 10^9 \frac{N m^2}{C^2}$is Coulomb's constant.

All three electric fields, created the by charges are directed in the positive x direction, hence the net field is $E = k \left( \frac{|q_1|}{r_1^2} + \frac{|q_2|}{r_2^2} + \frac{|q_3|}{r_3^2}\right)$, where $r_1 = 2 m, r_2 = 1 m, r_3 = 1m$.

Substituting these distances, obtain $E \approx 472.5 \frac{k V}{m}$.