Question

1) Two stationary positive point charges, charge 1 of magnitude 3.30 nCnC and charge 2 of magnitude 1.60 nCnC , are separated by a distance of 45.0 cmcm . 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 vfinalvfinalv_final of the electron when it is 10.0 cmcm from charge 1?

Answer 1

Solution) q1 = 3.30 nC = 3.30×10^(-9) C

q2 = 1.60 nC = 1.60×10^(-9) C

R = 45 cm = 0.45 m

d = 10 cm = 0.10 m

Vfinal = ?

Work done = Change in Kinetic energy

W = KEf - KEi

KEi = 0 ( at rest velocity = 0 so KE = 0)

KEf = (1/2)(m)(v^2)   

Here Vfinal = v

W = F.d

F = ((K)(q1)(q2))/(R^2)

K = 9×10^(9) Nm^2/C^2

F = (9×10^(9)×3.30×10^(-9)×1.60×10^(-9))/(0.45^2)

F = 234.67×10^(-9) N

W = F.d = 234.67×10^(-9)×0.10 = 23.467×10^(-9) J

W = 23.5×10^(-9) J

KEf = (1/2)(m)(v^2)

Here m is mass of electron

m = 9.11×10^(-31) kg

KEf = (1/2)(9.11×10^(-31))(v^2) = (4.55×10^(-31))(v^2)

W = KEf

23.5×10^(-9) = (4.55×10^(-31))(v^2)

v^2 = (23.5×10^(-9))/(4.55×10^(-31))

v^2 = 5.16×10^(22)

v = (5.16×10^(22))^(1/2)

v = 2.27×10^(11) m/s

Therefore , Vfinal = v = 2.27×10^(11) m/s

Vfinal = 2.27×10^(11) m/s