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Two particles having equal masses of 10 \mathrm{gm} each and have opposite charges of magnitudes +8 \times 10^{-5} \mathrm{C} and -8 \times 10^{-5} \mathrm{C} respectively are released from the rest position with a separation of 2 \mathrm{~m} between them. Deduce the speed of the particles when the separation is reduced to 1 \mathrm{~m}.

Option: 1

v=53.6 m \mathrm{~s}^{-1}


Option: 2

v=54.6 \mathrm{~ms}^{-1}


Option: 3

v=532.6 m s^{-1}


Option: 4

None of the above.


Answers (1)

best_answer

Mass of both particles =10 \times 10^{-3} \mathrm{~kg}

q_1=+8 \times 10^{-5} \mathrm{C}  and  q_2=-8 \times 10^{-5} \mathrm{C}, r_1=2.0 \mathrm{~m}, r_2=1.0 \mathrm{~m} \text {. }

Increase in kinetic energy = Loss in potential energy 

 = Final potential energy - Initial potential energy 

\begin{gathered} 2 \times \frac{1}{2} m v^2=\frac{\left|q_1\right|\left|q_2\right|}{4 \pi \epsilon_o}\left[\frac{1}{r_2}-\frac{1}{r_1}\right] \\ v^2=\frac{1}{m} \times \frac{\left|q_1\right|\left|q_2\right|}{4 \pi \epsilon_o} \times\left[\frac{1}{r_2}-\frac{1}{r_1}\right] \end{gathered}

After putting all the values, value of \mathrm{v} obtained is v=53.6 \mathrm{~ms}^{-1} \text {. }

Posted by

rishi.raj

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