Question

A particle of charge +9.8 μC is released from rest at the point x = 94 cm on an x axis. The particle begins to move due to the presence of a charge Q that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 21 cm if (a)Q = +96 μC and (b)Q = –96 μC?

Answer 1

here,

q1 = 9.8 uC = 9.8 * 10^-6 C

x1 = 94 cm = 0.94 m

r = 21 cm = 0.21 m

a)

when Q = +96 uC = 96 * 10^-6 C

it moves away from the origin

the kinetic energy gained , Ke = potential energy lost

Ke = K * q1 * Q * ( 1/x1 - 1/(x1 + r))

Ke = 9 * 10^9 * 9.8 * 10^-6 * 96 * 10^-6 * ( 1/0.94 - 1/(0.94 + 0.21))

Ke = 1.64 J

b)

when Q = -96 uC = - 96 * 10^-6 C

it moves towards the origin

the kinetic energy gained , Ke = potential energy lost

Ke = K * q1 * Q * ( 1/(x1) - 1/(x1 - r))

Ke = - 9 * 10^9 * 9.8 * 10^-6 * 96 * 10^-6 * ( 1/(0.94) - 1/(0.94 - 0.21))

Ke = 2.59 J