Question

Two particles of equal masses (m = 5.5x10-15 kg) are released from rest with a distance between them is equal to 1 m. If particle A has a charge of 12 μC and particle B has a charge of 60 μC, what is the speed of particle B at the instant when the particles are 3m apart?

Answer 1

To solve this problem it is necessary to resort to the energy conservation equations, both kinetic and electrical.

By Coulomb's law, electrical energy is defined as

$EE = \frac{kq_1q_2}{d}$

Where,

EE = Electrostatic potential energy

q= charge

d = distance between the charged particles

k = Coulomb's law constant

While kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m= mass

v = velocity

There by conservation of energy we have that

EE= KE

There is not Initial kinetic energy, then

$\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = 2*\frac{1}{2}mv_f^2$

$\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} = mv_f^2$

$v_f^2= \frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'} }{m}$

$v_f = \sqrt{\frac{\frac{kq_1q_2}{d}-\frac{kq_1q_2}{d'}}{m}}$

Replacing with our values we have,

$v_f = \sqrt{\frac{\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{1}-\frac{(9*10^9)(12*10^{-6})(60*10^{-6})}{3}}{5.50*10^{-15}}}$

$v_f = 2.802*10^7m/s$

Therefore the speed of particle B at the instat when the particles are 3m apart is $2.802*10^7m/s$