Q.

Two small particles, each of mass m carrying positive charge q each are attached to the ends of a non-conducting light thread of length 2l. A third particle of mass 2m is attached at mid-point of the thread. The whole system is placed on a smooth horizontal floor and the particle of mass 2m is given a velocity v as shown in Fig. minimum distance between the two charged particles during the process of motion is expressed as (a.  qblcd0mV2l+qe) find a+b+c+d+e to nearest integer

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answer is 13.

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Detailed Solution

At the instant of minimum separation all three move with same velocity  =VCH=V2
 14πε0q22l+12.2mV2=14πε0q2d+12.4mV24                     q24πε02l+mV22=14πε0q2d
Solving we get 

d=2q2l(4πε0mv2l+q2)

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