A positive test charge of C R 8.7 is subjected to a force of 8.1 x 10-6 N in a direction making an angle N24 northeast. What is the magnitude of the electric field strength and its direction at the site of the test charge?

E = 0.93 10⁻⁶ N/C, 24º NorthEast

Explanation:

The electric force is related to the electric field by the relation

          F = q E

where the bold letters indicate vectors and q is a scalar electric charge.

Using this equation the electric force is in the same direction as the electric field.

The magnitude of the field is

          E = F / q

the charge they give us is q = 8.7 C

let's calculate

          E = 8.1 10⁻⁶ / 8.7

          E = 0.93 10⁻⁶ N / C

In summary, the magnitude of the electric field is E = 0.93 10⁻⁶ N/C and its direction is 24º NorthEast