A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.00×10−4C is projected directly at q1. Ignore gravity. When q2 is 4.00m away, its speed is 800m/s. What is its speed when it is 0.200m from q1?

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

Kinetic energy

Potential energy =

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

                      +

                    =(1280-337.5)J

                    =942.5 J

Total energy of a system remains constant.

Therefore,

E +

  m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.