Question

A small object has a mass of 3.0 × 10-3 kg and a charge of -32C. It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of 2.4 × 103 m/s2 in the direction of the +x axis. Determine the electric field, including sign, relative to the +x axis.

Answer 1

Answer:

-0.0225 N/C

Explanation:

First of all, we can calculate the force exerted on the object, which is given by:

F = ma

where

$m=3.0\cdot 10^{-3} kg$ is the mass

$a=2.4\cdot 10^3 m/s^2$ is the acceleration

Substituting,

$F=(3.0\cdot 10^{-3}kg)(2.4\cdot 10^3 m/s^2)=7.2 N$

and the direction is the same as the acceleration (+x axis).

Now the electric force is given by

$F=qE$

where

$q=-32 C$ is the charge

E is the electric field

Solving for E, we find

$E=\frac{F}{q}=\frac{7.2 N}{-32 C}=-0.225 N/C$

and the negative sign means the direction is -x axis.