Question

A positive charge of 5.30 μC is fixed in place. From a distance of 3.50 cm a particle of mass 5.10 g and charge +3.20 μC is fired with an initial speed of 62.0 m/s directly toward the fixed charge. How close to the fixed charge does the particle get before it comes to rest and starts traveling away?

Answer 1

The initial Kinetic Energy (KE) of the fired particle is:

KE = (1/2)*m*v^2 = (1/2)*(.00510)*(62)^2 = 9.8 J

The particle comes to rest when it gains an equal amount of Potential Energy (PE):

Uf - Ui = k*Q*q*( 1/rf - 1/ri ) = KE = 9.8 J

Where Uf - Ui is the change in PE, Q is one charge, q is the other charge, rf is the radius as the fired charge comes to rest, and ri is the initial radius, and k = 1/(4*π*ε):

Solve for rf:

( 1/rf - 1/ri ) = KE/(k*Q*q)

1/rf = KE/(k*Q*q) + 1/ri

rf = 1/[ KE/(k*Q*q) + 1/ri ] = 1/[ 9.8/(9*10^9*(5.3*10^-6)*(3.2*10^-6)) + 1/(.035) ]

rf = 0.010778791 m [or = 1.077879106 cm]