Question

A small object has a mass of 3.0 × 10^-3 kg and a charge of -37C. It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of 3.4 × 10^3 m/s2 in the direction of the +x axis. Determine the electric field, including sign, relative to the +x axis.

Answer 1

Given

mass is $m= 3.0 × 10^{-3} kg$

Charge is $q=-37 C$

Acceleration is $a=3.4 × 10^3 m/s^2$

According to Newton's second law

$F=ma$

And $F=qE$

From the above two equations,

$qE=ma$

The electric field strength is

$E=\frac{ma}{q}=\frac{( 3.0 × 10^{-3} kg)(3.4 × 10^3 m/s^2)}{-37 C}=-0.276 N/C$

Direction is along negative x axis.