An electric field with a magnitude of 6.0 × 10⁴ N/C is directed parallel to the positive y-axis. A particle with a charge q = 4.8 × 10^-¹⁹ C is moving along the x-axis with a speed v = 3.0 × 10⁶ m/s. The force on the charge is approximately

Question

Quizplus · Accepted Answer

The force on a charged particle moving in an electric field is given by $F=qE$, where $q$ is the charge of the particle and $E$ is the electric field. In this case, the force is $F=(4.8	imes10^{-19}	ext{ C})(6.0	imes10^{14}	ext{ N/C})=2.9	imes10^{-5}	ext{ N}$. Since the electric field is parallel to the y-axis, and the particle is moving along the x-axis, the force will be perpendicular to both, i.e. in the y-direction. Thus, the answer is B, 2.9 × 10$^{-5}$ N in the y direction.

An Electric Field with a Magnitude of 6