Determine a) the electric field strength at a point x at a distance 20cm from a point charge q= +6ucb) the electric force that acts on a point charge q=-0.20uc placed at point x.

A) The magnitude of the electric field generated by a point charge Q at distance r from the charge is

where k is the Coulomb's constant. In this problem, the charge is

and the distance is

Therefore the electric field is


b) The electric force acting on a charge q is given by

where E is the intensity of the electric field at the point where the charge q is located.
We have already calculated the magnitude of the electric field, E, at point x, therefore the force acting on the charge is

and the negative sign means the force points in the opposite direction of the electric field, since the charge is negative.