Respuesta :
Answer:
The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m
Explanation:
Given that,
First charge = 40 μC
Second charge = 80 μC
Distance between the two particles = 2.0 m
Kinetic energy = 16 J
We need to calculate the distance separating the two particles when the moving particle is momentarily stopped
Using conservation of energy
[tex]K.E+\dfrac{kq_{1}q_{2}}{d}=\dfrac{kq_{1}q_{2}}{x}+K.E[/tex]
Put the value into the formula
[tex]16+\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{2}=\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{x}+0[/tex]
[tex]16+14.4=\dfrac{28.8}{x}[/tex]
[tex]30.4x=28.8[/tex]
[tex]x=\dfrac{28.8}{30.4}[/tex]
[tex]x=0.947\ m[/tex]
Hence, The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m
The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m.
The given parameters;
- charge of the first particle, q₁ = 40 μC
- charge of the second particle, q₂ = 80 μC
- initial distance between the particles, x₁ = 2 m
- initial kinetic energy, K.E₁ = 16 J
The kinetic energy is zero at the instant the moving particle is momentarily stopped.
The distance separating the two particles when the moving particle is momentarily stopped is calculated as follows;
[tex]K.E_1 \ + W_1 = K.E_2 + W_2\\\\K.E_1 + \frac{kq_1q_2}{x_1} = K.E_2 + \frac{kq_1q_2}{x_2} \\\\16 \ + \ \frac{(9\times 10^9)\times (40\times 10^{-6})\times (80\times 10^{-6})}{2} = 0 \ + \ \frac{(9\times 10^9)\times (40\times 10^{-6})\times (80\times 10^{-6})}{x_2} \\\\30.4= \frac{28.8}{x_2} \\\\x_2 = \frac{28.8}{30.4} \\\\x_2 = 0.947 \ m[/tex]
Thus, the distance separating the two particles when the moving particle is momentarily stopped is 0.947 m.
Learn more here:https://brainly.com/question/16252268
