Minimum velocity from infinity

cupid.callin

<In Pic 1>

(1/2)mv2 = ΔU

The Attempt at a Solution

I thought that if i apply the equations i'll have my answer but i got 2.5m/s while answer is 3m/s

Please refer the solution given by some book --- Pic2, Pic3
<sorry for bad image, my scanner nt working properly >

Please tell me what is wrong i my method?

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Mentor
What method? You haven't shown what you've done, only what the book solution is.

cupid.callin
I mentioned it that i used energy conservation:

decrease in kinetic energy from infinity to 0 results in inc. in potential energy

My work's in pic

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Mentor
Ah. Well, the problem is that the force upon the charge as it moves inwards from infinity is not monotonic. It changes sign (direction) along its path. If you plot the potential along the trajectory, you'll see that there's a significant "hill" in the way that the charge has to climb and then ride down towards the origin.

The energy the particle needs to get over the hill is the energy it needs to have at infinity. Once it has cleared the top of the hill, it can just 'roll' down to the finish line, gaining KE accordingly. If it just barely has enough initial energy to reach the top of the hill (making its velocity approach zero), then the KE it will have at the finish line will be equal to the PE at the top of the hill minus the PE at the finish line.

cupid.callin
But then my answer should have been more than the correct answer, isn't it?

But i got 2.8 but answer is 3

Mentor
No, your answer should be less than the real answer because the hill is higher than the finish line. You calculated the energy required as the difference between the potential at infinity and at the finish line (origin). The hill in between means that more initial energy is required to make it to the finish line.

cupid.callin
OH!

Now i get it ... that was one tricky question