Question

A small object has a mass of 2.9 10-3 kg and a charge of -36 µC. It is placed at a certain spot where there is an electric field. When released, the object experiences an acceleration of 2.49 103 m/s2 in the direction of the +x axis. Determine the magnitude and direction of the electric field.

Answer 1

200583.33 N/C direction opposite to that of the acceleration

Step-by-step explanation:

m = Mass of object =
`2.9* 10^(-3)\ kg`

a = Acceleration of the object =
`2.49* 10^3\ m/s^2`

q = Charge =
`-36\ \mu C`

E = Electric field

Here the inertia of the object will balance the electric force

`ma=Eq\\\Rightarrow E=(ma)/(q)\\\Rightarrow E=(2.9* 10^(-3)* 2.49* 10^3)/(-36* 10^(-6))\\\Rightarrow E=-200583.33\ N/C`

The magnitude of the electric field is 200583.33 N/C and the direction is opposite to that of the acceleration.