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

Answer:

200583.33 N/C direction opposite to that of the acceleration

Explanation:

m = Mass of object = $2.9\times 10^{-3}\ kg$

a = Acceleration of the object = $2.49\times 10^3\ m/s^2$

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

E = Electric field

Here the inertia of the object will balance the electric force

$ma=Eq\\\Rightarrow E=\frac{ma}{q}\\\Rightarrow E=\frac{2.9\times 10^{-3}\times 2.49\times 10^3}{-36\times 10^{-6}}\\\Rightarrow E=-200583.33\ N/C$

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