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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?

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Last updated date: 20th Jun 2024
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Answer
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Hint: Remember the law of conservation of energy. When the two particles are released, as the separation increases, the potential energy will get decreased, but the kinetic energy will get increased. That is the potential energy is converted into the kinetic energy of the particles.

Complete step by step answer:
Let’s define the data given in the question.
Charge of the two identical particles, $q = 2.0 \times {10^{ - 4}}C$
Mass of the two identical particles, $m = 10g = 10 \times {10^{ - 3}}kg$
Separation between the two particles, $r = 10cm = 10 \times {10^{ - 2}}m$
Here, we are asked to find the value of the speeds of the particle when the separation becomes large.
It is given that the particles are identical and have the same mass and charge. When it is placed near to each other, there will be a repulsive force and thus a potential energy is generated.
When it is released the particles get apart with the same velocity and the separation will be increased. At this time the potential energy will get decreased, but the kinetic energy will get increased. That is the potential energy is converted into the kinetic energy of the particles.
The potential energy of the two particles is given by,
$U = \dfrac{{k{q_1}{q_2}}}{r} = \dfrac{{kqq}}{r} = \dfrac{{k{q^2}}}{r}$
Where $k$ is a constant with a value equal to $9 \times {10^9}N{m^2}{C^{ - 2}}$
The kinetic energy of these two particles is given by,
$K.E = 2 \times \dfrac{1}{2}m{v^2} = m{v^2}$
Where, $v$ is the velocity of the particles
We know gain kinetic energy = loss potential energy
That is, $m{v^2} = \dfrac{{k{q^2}}}{r}$
$ \Rightarrow v = \sqrt {\dfrac{{k{q^2}}}{{rm}}} $
$ \Rightarrow v = \sqrt {\dfrac{{9 \times {{10}^9} \times {{(2 \times {{10}^{ - 3}})}^2}}}{{10 \times {{10}^{ - 2}} \times 10 \times {{20}^{ - 3}}}}} $
$ \therefore v = 600m{s^{ - 1}}$

That is the speeds of the particle when the separation becomes large, $v = 600m{s^{ - 1}}$.

Note: We use the letter U to denote electric potential energy, which has units of joules. The electrical potential energy will be positive if the two charges are of the same type, that is, either positive or negative. It will be negative if the two charges are of opposite types.