Consider two point particles that have charge +e, are at rest, and are separated by 2.7 x 10-15 m.

1) how much work was required to bring them together from a very large separation distance? kev

2) if they are released, how much kinetic energy will each have when they are separated by twice their separation at release? kev

3) the mass of each particle is 1.00 u (1.00 amu). what speed will each have when they are very far from each other? m/s

Explanation:

q1 = q2 = e = 1.6 x 10^-19 C

d = 2.7 x 10^-15 m

1. Work done is equal to the potential energy stored between the two charges.

The formula for the potential energy is given by

By substituting the values

U = 8.53 x 10^-14 J

So, the work done is 8.53 x 10^-14 J.

(2) d = 2 x 2.7 x 10^-15 m = 5.4 x 10^-15 m

So, the potential energy is

By substituting the values

U = 4.27 x 10^-14 J

(3) mass, m = 1 u = 1.67 x 10^-27 kg

So, the kinetic energy is equal to the potential energy.

Let v be the velocity.

v = 7.14 x 10^6 m/s

the answer would be   dignified - apex

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.