Question

Two identical particles, each having a charge of $ 2.0 \times {10^{ - 4}}C $ and mass of $ 10g $ , are kept at a separation of $ 10cm $ and then released. What would be the speeds of the particles when the separation becomes large?

Answer 1

Hint: We use conservation of energy to solve this problem. At an infinite distance between the particles, the potential energy experienced is zero so all the potential energy is converted to kinetic energy as the distance increases. The change in initial potential energy (before the separation is infinite) is equal to the final kinetic energy.
Formula used: coulombs potential energy $ PE = K\dfrac{{{q_1}{q_2}}}{r} $
Kinetic energy $ KE = \dfrac{1}{2}m{v^2} $
Here,
Mass is represented by $m$,
Charge of particle one is represented by ${q_1}$,
Charge of particle two is represented by ${q_2}$,
Potential energy is represented by $PE$,
Kinetic energy is represented by $KE$,
Coulomb's constant is represented by $K$,
Distance between particles is represented by $r$.

Complete step by step answer:
The formula for change in potential energy between two charges is given by
The particles are identical and have the same charge
$ \Rightarrow \Delta PE = K\dfrac{{{q_1}{q_2}}}{r} $
$ \because {q_1} = {q_2} $
$ \Rightarrow \Delta PE = K\dfrac{{{q^2}}}{r} $
The kinetic energy of one particle is
$ KE = \dfrac{1}{2}m{v^2} $
Since the repulsion is between two particles the total change in kinetic energy is
$ \Delta KE = 2 \times \dfrac{1}{2}m{v^2} $
According to law of conservation of energy the change in potential energy is equal to the change in kinetic energy,
From the question, converting the units to SI units,
$ q = 2 \times {10^{ - 4}} $
$ m = 10g = {10^{ - 2}}kg $
$ K = 9 \times {10^9} $
$ r = 10cm = 10 \times {10^{ - 2}}m $
Substituting the above values,
$ \Rightarrow \Delta KE = \Delta PE $
$ \Rightarrow K\dfrac{{{q^2}}}{r} = 2 \times \dfrac{1}{2}m{v^2} $
$ \Rightarrow 9 \times {10^9}\dfrac{{{{(2 \times {{10}^{ - 4}})}^2}}}{{10 \times {{10}^{ - 2}}}} = {10^{ - 2}} \times {v^2} $
$ \Rightarrow v = \sqrt {\dfrac{{9 \times {{10}^9} \times {{(2 \times {{10}^{ - 4}})}^2}}}{{10 \times {{10}^{ - 2}} \times {{10}^{ - 2}}}}} $
Solving for velocity, $v = 600m/s$

Hence, the velocity of the particle is $ v = 600m/s $.

Additional Information:
The electron’s potential energy is a result of the attractive force between the negatively charged electron and the positively charged nucleus. When unlike charges (one negative and the other positive) attract each other, or like charges (both positive and both negative) repel each other, Coulomb’s law governs the force between them.

Note: The separation increases because the charges are of the same sign. The force felt between them is a force of repulsion. At the initial condition, the potential energy is maximum and kinetic energy is zero. At the final condition, the distance between the particles is infinite and the potential energy is zero and the kinetic energy is maximum. Hence the change is nothing but potential energy at the starting and kinetic energy at the end.