Consider an oil droplet of mass m and charge q. We want to determine the charge on the droplet in a Millikan-type experiment. We will do this in several steps. Assume, for simplicity, that the charge is positive and that the electric field between the plates points upward. a) An electric field is established by applying a potential difference to the plates. It is found that a field of strength E0 will cause the droplet to be suspended motionless. Write an expression for the droplet's charge q. Let g be the acceleration due to gravity. Express your answer in terms of the suspending field E0 and the droplet's weight mg.

Explanation:

Given:

Charge = q

Electric field strength =

weight of the droplet = mg

The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.

electric force on charged droplet,

This is balanced by the weight,

Equating the two:

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.