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Two identical particles of mass m carry a charge `Q`, each. Initially one is at rest on a smooth horizontal plane and the other is projected along the plane directly towards first particle from a large distance with speed v. The closest distance of approach be .

Text Solution

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The correct Answer is:
` Q^2 /( m pi in_0 V^2)`
At closest distance both will have equal veleoc ` v`, From conservation linear momentum
`mv + mv= mv rArr v= v/2`
From conservation of energy , loss of `KE =` Gain of ` PE`
`1/2 mv^2 -2 xx 1/2 m (v/2)^2 = (Kq^2 )/r^0 rArr 1/4 mv^2 = (KQ^2)/r_0`
`rArr r_0 = ( 4KQ^2 )/( mv^2) = Q^2 /( m pi in_0 v^2)`.
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