Question

What is the electric field at a location <-0.5, -0.1, 0> m, due to a particle with charge +4 nC located at the origin?

Answer 1

Answer:

138.4625 N/C

Explanation:

We have that the charge +4 nC is at the origin (0,0,0)

We have to find the electric field at point (-0.5,-0.1,0)

We know that distance between two points is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}=\sqrt{(0-(-0.5))^2+(0-(-0.1))^2+(0-0)^2}=0.5099m$

The electric field due to a point charge is given by $E=\frac{KQ}{R^2}=\frac{4\times 10^{-9}\times 9\times 10^{9}}{0.5099^2}=138.4625N/C$