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Q.

A positively charged particle starts at rest 25 cm from a second positively charged particle, which is held stationary throughout the experiment. The first particle is released and accelerates directly away from the second particle. When the first particle has moved 25 cm, it has reached a velocity of 102  ms−1. What is the maximum velocity (in  ×10  ms−1) that the first particle will reach?

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answer is 2.

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Detailed Solution

kq2(1r−12r)=12m(V2)or kq22r=12mV2or kq2r=12m(2 V)2=12m(Vmax2)or Vmax=V2=(102)2=20  ms−1=2×10  ms−1
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A positively charged particle starts at rest 25 cm from a second positively charged particle, which is held stationary throughout the experiment. The first particle is released and accelerates directly away from the second particle. When the first particle has moved 25 cm, it has reached a velocity of 102  ms−1. What is the maximum velocity (in  ×10  ms−1) that the first particle will reach?