A particle with a charge of 2.0 x 10^-8 C is acted on by a force of magnitude 4.0 x 10^-6 N when it is placed in a particular point in an electric field.

Question

What is the electric field magnitude at the point?

Answer 1

Magnitude of electric field is newton per coulomb.

Electric Field = Force/charge = 4*10^-6 / 2*10^-8 = 2*100= 200 N/C

Answer 2

The magnitude of the electric force F acting on a charge q is related to the electric field E by F = qE, so the electric field magnitude is E = F/q, so with the values given

E = (4.0 x 10^-6 N)/(2.0 x 10^-8 C) = 2.0 x 10² N/C

Answer 3