Question

An electric field of 11.6 N/C directed due north is superposed on an existing electric field of 16.9 N/C directed due east. What are the magnitude and the direction of the resultant field?

Answer 1

The magnitude of the net electronic field

E=√(E_1^2+E_2^2 )=√(〖11.6〗^2+〖16.9〗^2 )=20.5 N/C

The direction

tan⁡θ=E_1/E_2 =11.6/16.9=0.868

θ=34.5° (north of east)