Question

a uniform electric field E=15 N/C points downwards. A particle with charge q=-0.15 C is placed in the electric field. What is the magnitude and direction of the force on the particle?

Answer 1

Answer:

2.25 N upwards

Explanation:

Answer:

800 N/C to the right

Explanation:

The equation that relates force on a charge, electric field and charge is

$F=qE$

where

F is the force acting on the charge

q is the charge

E is the electric field

In this problem, we have

$E=15 N/C$ is the electric field

$q=-0.15 C$ is the charge

Substituting into the formula, we find the force

$F=(-0.15 C)(15 N/C)=-2.25 N$

Concerning the direction:

- The electric field and the force have same directions if the charge is positive

- The electric field and the force have opposite directions if the charge is negative

Here the charge is negative, so the electric field has opposite direction to the force: therefore, it must be upwards.