two identical particles of mass m carry charge Q each.initially one of its rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large speed v.the closed disance of approach will be

Question

Vikas TU · Answer

Dear Student,|
When the mass comes the KE slowly decreases because the charge is same. It decreases till the potential energy is maximum and the KE falls to zero.

KE=mv²/2
PE=keq1q2/r
By principle of conservation of energy,
KE=PE
=>r=(q1q2/4πe)/(mv²/2)
=>r=2q²/4πemv²

Cheers!!

Regards,

Vikas (B. Tech. 4^th year
Thapar University)

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two identical particles of mass m carry charge Q each.initially one of its rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large speed v.the closed disance of approach will be

two identical particles of mass m carry charge Q each.initially one of its rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large speed v.the closed disance of approach will be

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two identical particles of mass m carry charge Q each.initially one of its rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large speed v.the closed disance of approach will be

two identical particles of mass m carry charge Q each.initially one of its rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large speed v.the closed disance of approach will be

1 Answers

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