Question

Two stationary positive point charges, charge 1 of magnitude 3.10 nC and charge 2 of magnitude 1.80 nC , are separated by a distance of 42.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

Answer 1

the difference between the potential energy of the electron at the beginning and the end and converting that to kinetic energy (and from there to velocity). Start by finding the potential energy of the electron midway between the two charges (the radius will be 0.42m / 2 = 0.21m):

Vi = kQ1/r + kQ2/r
Vi = k(Q1 + Q2)/r
Vi = k * (3.10nC + 1.80nc)/ 0.21

Convert to coulombs (from nC) and solve:

Vi = 2.333 * 10^-8 * k
Vi = 2.333* 10^-8 * (8.998 * 10^9)
Vi = 209.92 V

Now find the beginning potential energy of the electron:

Ui = q * Vi
Ui = 1.602*10^-19 * 209.92
Ui = 3.362*10^-17

Now do the same thing, but for the position of the electron that this problem is asking for (10.0 cm from charge 1). The radius will be different for each charge this time around:

Vf = kQ1/r1 + kQ2/r2
Vf = k * 3.10nC / 0.1 + k * 1.80nC / 0.32

Convert to coulombs again and solve:

Vf = 329.55 V

Now find the final potential energy of the electron:

Uf = q * Vf
Uf = 1.602*10^-19 * 329.55
Uf = 5.279 * 10^-17

Now solve for kinetic energy (use the absolute value of the difference in potential energies):

KE = abs(Ui – Uf)
KE = abs((3.362*10^-17) – (5.279 * 10^-17))
KE = 1.917 * 10^-17

Convert to velocity (the mass of an electron is 9.1094 * 10^-31 kg):

KE = 1/2mv^2
1.917 * 10^-17 = 1/2mv^2
5.279 * 10^-17 = mv^2
v^2 = 3.834 * 10^13
v = 6191930 m/s (difference due to rounding)

v = 6.19*10^6m/s