A particle (charge = Q)is kept in a fixed position at point P,and a second particle (charge = q)is released from rest when it is a distance R from P.If Q = +2.0 mC,q = -1.5 mC,and R = 30 cm,what is the kinetic energy of the moving particle after it has moved a distance of 10 cm?
A)60 kJ
B)45 kJ
C)75 kJ
D)90 kJ
E)230 kJ

Question

Quizplus · Accepted Answer

The electric potential energy of the system of two charges is converted into kinetic energy of the moving particle. By conservation of energy, we can equate the initial potential energy to the final kinetic energy:

qQ/4πε₀R = (1/2)mv²

where ε₀ is the electric constant and m is the mass of the moving particle.

Solving for v, we get:

v = √(2qQ/4πε₀mR)

When the particle has moved a distance of 10 cm, its distance from the fixed particle is R' = √(R² + d²) = 31.2 cm.

The new electric potential energy is:

qQ/4πε₀R' = (101.5e-3 C)(2e-3 C)/(4πε₀(0.312 m)) = 48.5 J

By conservation of energy, this must be equal to the kinetic energy of the moving particle:

(1/2)mv² = 48.5 J

Substituting the value of v we found earlier and the mass of an electron (since the charge is negative), we get:

(1/2)(9.11e-31 kg)(√(2(101.5e-3 C)(-2e-3 C)/(4πε₀(0.3 m))(10e-2 m/s))² = 44.8 J

This is in the same order of magnitude as choice B, which is the closest.

A Particle (Charge = Q)is Kept in a Fixed Position