A charge of q = +7.03 μC is located in an electric field. The x and y components of the electric field are Ex = 6.37E+3 N/C and Ey = 8.39E+3 N/C, respectively. What is the magnitude of the force on the charge?

Question

ash2326 · Answer

Resultant E =$\sqrt{Ex^2+Ey^2}$
$$E=\sqrt{6.37^2+8.39^2}\times 10^3 N/C$$
$$E=10.534 \times 10^3 N/C$$
Force on the charge= Electric field on the charge $\times$ magnitude of charge
$$F=10.534\times 10^3N/C\times 7.03μC $$
$$F=10.534\times 10^3N/C\times 7.03\times 10^{-6}C $$
we get
Force on charge
$$F=7.4055 \times 10^{-2} N$$