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# two identical particles carry a charge Q each. initially one is at rest on a smooth horizontal and the other is projected along the plane directly towards the first particle from a large distance with a speed v. what is the closest distance of approach ?

28 Points
11 years ago

Dear Anand Shukla,

Ans:- As there is no external force present hence the system's total energy will remain constant. So, the initial KE will be totally converted to the system's PE when they are closest to each other. Hence from there we get,

(1/4∏ε)Q²/R=(1/2)mv²

R=Q²/(2∏εv²)

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Regards,

Soumyajit Das IIT Kharagpur

one year ago
Dear student,

As there is no external force present hence the system's total energy will remain constant. So, the initial KE will be totally converted to the system's PE when they are closest to each other. Hence from there we get,
(1/4πε)Q²/R
= (1/2)mv²
R = Q²/(2πεv²)

Thanks and regards,
Kushagra