Question

An electron starts from rest 31.0 cm from a fixed point charge with Q = -4.00 nC .

How fast will the electron be moving when it is very far away?

Express your answer with the appropriate units.

v=?

Answer 1

Solution) r = 31 cm = 0.31 m

Q = - 4 nC = - 4×10^(-9) C

Speed , v = ?

Charge of electron , q = - 1.6×10^(-19) C

Mass of electron , m = 9.11×10^(-31) kg

Equating kinetic energy = Potential energy

KE = U

(1/2)(m)(v^2) = ((K)(Q)(q))/(r)

K = 9×10^(9) Nm^2/C^2

(1/2)(9.11×10^(-31))(v^2) = (9×10^(9)×4×10^(-9)×1.6×10^(-19))/(0.31)

(4.55×10^(-31))(v^2) = 1.858×10^(-17)

v^2 = 0.4083×10^(14)

v = (0.4083×10^(14))^(1/2)

v = 6.39×10^(6) m/s