Question

0.17C charge is placedbat the origin and a - 0.08C is placed 4m to the left of it. What is the magnitude and direction of the electric field 10m meters to the right of the origin?

Answer 1

The net electric field at the point 10 m meters to the right from the origin is the vector sum of two electric fields due to two charges:

$$E_{net}=E_1-E_2=k(\dfrac{|q_1|}{r_1^2}-\dfrac{|q_2|}{(r_1+r_2)^2}),$$ $$E_{net}=9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot(\dfrac{|0.17\ C|}{(10\ m)^2}-\dfrac{|-0.08\ C|}{(14\ m)^2})=1.16\cdot10^7\ \dfrac{N}{C}.$$

The sign plus means that the electric field directed to the right.