Question

Two stationary positive point charges, charge 1 of magnitude 3.40 nC and charge 2 of magnitude 1.50 nC , are separated by a distance of 33.0 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

here,

the magnitude of charge 1 , q1 = 3.4 nC = 3.4 * 10^-9 C

the magnitude of charge 2 , q2 = 1.5 nC = 1.5 * 10^-9 C

r = 33 cm = 0.33 m

mass of electron , m = 9.1 * 10^-31 kg

charge on electron , q = 1.6 * 10^-19 C

let the speed of electron when it is r1 = 10 cm = 0.1 m from the charge 1 be v

using conservation of energy

kinetic energy gained by electron = change in electrostatic potential energy

0.5 * m * v^2 = K * e * (( q1/r1 + q2/(r - r1) - (q1/(r/2) + q2/(r/2)))

0.5 * 9.1 * 10^-31 * v^2 = 9 * 10^9 * 1.6 * 10^-19 * 10^-9 * ((3.4/0.1 + 1.5/(0.33 - 0.1)) - (3.4 /(0.33/2) + 1.5/(0.33/2)) )

solving for v

v = 5.85 * 10^6 m/s

the speed of electron is 5.85 * 10^6 m/s