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

The particle is moving due to force $F = qE$ where q is charge of object, E is the electric field.

According to the second Newton's law,

$ma = qE$

$\displaystyle E =\frac{ma}{q} = \frac{3 \cdot 10^{-3} \cdot3.4 \cdot 10^{3}}{-37} = -0.276$ V/m.

Answer: $E = -0.276$ V/m (opposite to the direction of +x axis)