Respuesta :
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 [tex]=\frac{1}{2} mv^2[/tex]
Potential energy =[tex]\frac{Kq_1q_2}{d}[/tex]
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
[tex]=\frac{1}{2} mv^2[/tex]+ [tex]\frac{Kq_1q_2}{d}[/tex]
[tex]=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}[/tex]
=(1280-337.5)J
=942.5 J
Total energy of a system remains constant.
Therefore,
E [tex]=\frac{1}{2} mv^2[/tex] + [tex]\frac{Kq_1q_2}{d}[/tex]
[tex]\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}[/tex]
[tex]\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750[/tex]
[tex]\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750[/tex]
[tex]\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}[/tex]
[tex]\Rightarrow v= 1961.19[/tex] m/s
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
