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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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