Question

A small object of mass 3.82 g and charge -16.5 µC is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. What are the magnitude and direction of the electric field?

Answer 1

Answer:

2271.16N/C  upward

Explanation:

The diagram well illustrate all the forces acting on the mass. The weight is acting downward and the force is acting upward in other to balance the weight.since the question says it is motionless, then indeed the forces are balanced.

First we determine the downward weight using

$W=mg\\g=9.81m/s^{2}$

Hence for a mass of 3.82g 0r 0.00382kg we have the weight to be

$W=0.00382kg*9.81m/s^{2}$

$W=0.0375N$

To calculate the electric field,

$E=f/q\\E=0.0375/16.5*10^{-6} \\E=2271.16N/C$

Since the charge on the mass is negative, in order to generate upward force, there must be a like charge below it that is  repelling it, Hebce we can conclude that the electric field lines are upward.

Hence the magnitude of the electric force is 2271.16N/C and the direction is upward