Answer:
[tex]d = 0.0306 m[/tex]
Explanation:
Here we know that for the given system of charge we have no loss of energy as there is no friction force on it
So we will have
[tex]U + K = constant[/tex]
[tex]\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2[/tex]
now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.
So we have
[tex]\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0[/tex]
[tex]7.05 + 30.5 = \frac{0.266}{r}[/tex]
[tex]r = 7.08 \times 10^{-3} m[/tex]
so distance moved by the particle is given as
[tex]d = r_1 - r_2[/tex]
[tex]d = 0.0377 - 0.00708[/tex]
[tex]d = 0.0306 m[/tex]
