What are the magnitude and the direction of the force on a -5.31-nC charge in an electric field of +2750 N/C î – 8830 N/C ĵ?

Question

The equation to find the magnitude of force is F=qE, the answer key for this practice problem is in micronewtons and gives that the answer is 49.1 micronewtons, however, I keep getting 32.3 micronewtons. I have an upcoming exam and I want to know what I am doing wrong. Please help!!

Answer 1

Thank you for asking the type of question that is not just "do my HW for me."

F = qE = -5.31 x 10^-9C*<2750,-8830>N/C = <-1.46 x 10^-5, 4.69 10^-5.> N

|F| = sqrt(F_x²+F_y²) = 4.91 x 10^-5N or 49.1 μN.

Direction is obtained from θ = arctan(F_y/F_x) + 180 degrees (F_x<0) (This gives the math angle CCl from x axis.