Question

two stationary positive point charges, charge 1 of magnitude 3.70nC and charge 2 of magnitude 1.50nC , are separated by a distance of 39.0cm . 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 10.0cm from charge 1?

Please show all steps

Answer 1

You can solve it by finding 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 0.39 m / 2 = 0.195m):

V_i = kQ₁/r + kQ₂/r
V_i = k(Q₁ + Q₂)/r
V_i = k * (3.7nC + 1.5nc)/ 0.195

V_i = 240 V

Now find the beginning potential energy of the electron:

U_i = q * V_i
U_i = 1.602*10^-19 * 240
U_i = 3.8448*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:

V_f = kQ₁/r₁ + kQ₂/r₂
V_f = k * 3.7nC / 0.1 + k * 1.5nC / 0.29

V_f = 379.552 V

Now find the final potential energy of the electron:

U_f = q * V_f
U_f = 1.602*10^-19 * 379.552
U_f = 6.0804 * 10^-17

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

KE = abs(U_i