Question

Mr. Mullenmeister is photocopying lab sheets for his first-period class when the toner light on the onA particle of toner carrying a charge of the copying machine experiences an electric field of 1.2 * 10 ^ 6 * N / C as it's pulled toward the paper. What is the electric force acting on the toner particle? F=4.5=(40*10^ C)(1.2*106N/C)=4.8*10^ .5

Answer 1

Answer:

0.0048 N

Explanation:

The electric force acting on a particle is given by

$F=qE$

where

q is the charge

E is the magnitude of the electric force

F is the strength of the force

The force has the same direction as the field if the charge is positive, and the opposite direction if the charge is negative.

In this problem, we have:

$Q=4.0\cdot 10^{-9}C$ is the charge

$E=1.2\cdot 10^6 N/C$ is the electric field

Therefore, the electric force on the charge is

$F=QE=(4.0\cdot 10^{-9})(1.2\cdot 10^6)=0.0048 N$