Answer:
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.
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.