The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle e
nds up at the center of the original square and is momentarily at rest. If the mass of this particle is m, what was its initial speed v?Express your answer in terms of q, d, m, and appropriate constants. Use k instead of 14πϵ0. The numeric coefficient should be a decimal with three significant figures.
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Answer:
Explanation:
Expression for escape velocity
ve =
ve² R / 2 = GM
M is mass of the planet , R is radius of the planet .
At distance r >> R , potential energy of object
=
Since the object is at rest at that point , kinetic energy will be zero .
Total mechanical energy = + 0 =
Putting the value of GM = ve² R / 2
Total mechanical energy = ve² Rm / 2 r
This mechanical energy will be conserved while falling down on the earth due to law of conservation of mechanical energy . So at surface of the earth , total mechanical energy
= ve² Rm / 2 r