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Question

# A point charge +Q having mass m is fixed on horizontal smooth surface. Another point charge having magnitude +2Q and mass 2m is projected horizontal towards the charge +Q from far distance with velocity vo. Acceleration of the particle +2Q when it is closest to fixed particle +Q :

A
Zero
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B
mv4o2kQ2
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C
mv4o4kQ2
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D
mv4okQ2
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Solution

## The correct option is C mv4o4kQ2 At closest approach, +2Q charge will be at instant rest. On applying the law of conservation of mechanical energy between initial and final state, we get: 12×2m×v2o+12×m×02+k(2Q)(Q)∞=0+0+k(2Q)(Q)d ⇒mv2o=2kQ2d ⇒d=2kQ2mv2o ...(i) Now, force of repulsion on the charge +2Q by +Q, F=k(2Q)(Q)d2 So, deceleration, a=F2m=2kQ22md2 ⇒a=2kQ22m×(mv2o2kQ2)2=mv4o4kQ2

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