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.

an obstruction

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.