Question

A particle (charge = 40 μC) moves directly toward a second particle (charge = 80 μC) which is held in a fixed position. At an instant when the distance between the two particles is 2.0 m, the kinetic energy of the moving particle is 16 J. Determine the distance separating the two particles when the moving particle is momentarily stopped.

Answer 1

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

$K.E+\dfrac{kq_{1}q_{2}}{d}=\dfrac{kq_{1}q_{2}}{x}+K.E$

Put the value into the formula

$16+\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{2}=\dfrac{9\times10^{9}\times40\times10^{-6}\times80\times10^{-6}}{x}+0$

$16+14.4=\dfrac{28.8}{x}$

$30.4x=28.8$

$x=\dfrac{28.8}{30.4}$

$x=0.947\ m$

Hence, The distance separating the two particles when the moving particle is momentarily stopped is 0.947 m