Question

A particle of charge ?7.5 mC is released from rest at the point x ? 60 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 40 cm if (a) Q ? ?20 mC and (b) Q ? ?20 mC?

Answer 1

(a)

If the charge Q is +20 mC, then the charge +7.5 mC will move away from +Q.

Therefore the initial position of the charge +7.5 mC is 60cm = 0.6m from +Q and the final position is 60cm + 40cm = 100cm = 1m from +Q

Then the decrease in potential energy of +7.5 mC charge is

U = k*Q*q[ (1/0.6) - (1) ]

U =(9*10⁹)*(20*10^-3)*(7.5*10^-3)*(1.66-1)

U = 8.91*10⁵ J

Then the kinetic energy at the instant it has moved 40cm is equal to the loss in potential energy.

Therefore kinetic energy K = 8.91*10⁵ J

(b)

If the charge Q is -20 mC, then the charge +7.5 mC will move toward -Q.

Therefore the initial position of the charge +7.5 mC is 60cm = 0.6m from -Q and the final position is 60cm - 40cm = 20cm = 0.2m from -Q

Then the decrease in potential energy of +7.5 mC charge is

U = k*Q*q[(1/0.6) - (1/0.2)]

U = (9*10⁹)*(-20*10^-3)*(7.5*10^-3)*(-3.24)

U = 4.50*10⁶ J

Then the kinetic energy at the instant it has moved 40cm isequal to the loss in potential energy.

Therefore kinetic energy K = 4.50*10⁶ J .