Two particles with masses of 5 g each and opposite charges of 4.0x10^-5 and -4.0x10^-5 are released from rest with a separation of 1 m. Find the velocities of the particles when the separation is reduced to 50 cm.



The initial potential energy of the system is given by:
Ei = k * q1 * q2 / r, where
k = Coulomb's constant = 9.0x10^9 Nm^2/C^2
q1 = 4.0x10^-5 C (charge of particle 1)
q2 = -4.0x10^-5 C (charge of particle 2)
r = 1 m (initial separation between the particles)

Substituting the values, we get:
Ei = 9.0x10^9 * 4.0x10^-5 * (-4.0x10^-5) / 1
Ei = -1.44 J (negative sign indicates an attractive force)


The final potential energy of the system is given by:
Ef = k * q1 * q2 / r', where
r' = 50 cm = 0.5 m (final separation between the particles)

Substituting the values, we get:
Ef = 9.0x10^9 * 4.0x10^-5 * (-4.0x10^-5) / 0.5
Ef = -28.8 J


The change in potential energy is given by:
ΔE = Ef - Ei

Substituting the values, we get:
ΔE = -28.8 - (-1.44)
ΔE = -27.36 J


According to the law of conservation of energy, the change in potential energy is equal to the kinetic energy gained by the particles. Therefore, the kinetic energy of the particles is given by:
K = ΔE

Substituting the value of ΔE, we get:
K = -27.36 J


The kinetic energy of each particle is given by:
K = 1/2 * m * v^2, where
m = 5 g = 0.005 kg (mass of each particle)
v = velocity of each particle

Substituting the values, we get:
-27.36 = 1/2 * 0.005 * v^2
v^2 = -27.36 / (1/2 * 0.005)
v^2 = -10,944
This is not a valid answer as velocity cannot be negative.

Therefore, there is an error in the calculation. After reviewing the calculations, it is found that the initial potential energy was calculated incorrectly. The correct value is -1.44x10^-7 J.

Substituting this value in the calculations, we get:
ΔE = -27.3593 J