Question

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Two stationary positive point charges, charge 1 of magnitude 3.50 nCand charge 2 of magnitude 1.95 nC , are separated by a distance of 35.0 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

Given that,

q1 = 3.5 nC

q2 = 1.95 nC

d = 35 cm = 0.35 m

r = 10 cm

Electric potential at the midpoint between the two point charges,

V1 = kq1 / (d/2) + kq2 / (d / 2)

V1 = 9*10^9*3.50*10^(-9) / (0.35 / 2) + 9*10^9*1.95*10^(-9) / (0.35 / 2)

V1 = 280.28 V

Electric potential at a distance r from the charge q1,

V2 = kq1 / r + kq2 / (d - r)

V2 =  (9*10^9*3.5*10^(-9) / 0.10) + 9*10^9*1.95*10^(-9) / (0.35 - 0.10)

V2 = 385.2 V

Work done by potential,

W = q(V2 - V1)

Kinematic energy of electron,

K = (1/2)mvf^2 - (1/2)mvi^2

vi = 0 (released from rest )

From work energy theorem,

q(V2 - V1) = (1/2)mvf^2

1.6*10^(-19) * (385.2 - 280.28) = (1/2)*9.1*10^(-31)*vf^2

vf = 1.92*10^6 m/s