Apositive test charge of 5.00 e-5 c is placed in an electric field. the force on it is 0.751 n. the magnitude of the electric field at the location of the test charge is

In order to get the magnitude of the electric field at the location of the test charge, you have to use the formula: E=Fe/q.
Given E as electric field strength; Fe as electric force = 0.751 N and q as charge = 5*10-5 C. Therefore,
E= 0.751 N / 5 * 10-5 CE=15020 N C-1 or V m-1 Electric field at the location.

Biased means unfair but that's all ik