80pts what is the magnitude and direction of the electric field 4.52 to the left of a 8.15 c negative charge? plz show work

A) The magnitude of the electric field generated by a single point charge is given by

where k is the Coulomb's constant, q is the charge and r is the distance at which we calculate the field.
Since we want to know the intensity of the field at distance of r=4.52 m from the charge of q=8.15 C, the intensity of the electric field is


b) Direction of the field
The charge that produces the field is negative, and we know that the electric field generated by a negative charge points toward the charge. Therefore, the direction of the electric field is towards the charge.