Asmall particle has charge -2.00 uc and mass 1.50×10-4 kg. it moves from point a, where the electric potential is v(a) = 200 v, to point b, where the electric potential v(b) = 920 v is greater than the potential at point a. the electric force is the only force acting on the particle. the particle has a speed of 4.30 m/s at point a.

1.)what is its speed at point b?
2.)and is it moving faster or slower at b than at a?

.

Explanation:

Since no external force is acting on the system.

Therefore, Total energy remains constant before and after.

So, Total energy of system= energy due to potential applied+kinetic energy

(Here v=velocity ,V=potential ,q=charge and m=mass).

Putting values .

We get,  .

At point B charged particle is moving faster as compared to point A.

Hence, it is the required solution.

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.