What are the choices ?
Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation. A few of them might be . . .
-- Kevin will have no trouble getting back in time for dinner.
-- Kevin will have no time to enjoy the scenery along the way.
-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.
-- Speed = (distance covered) / (time to cover the distance) .
If time to cover the distance is zero, then speed is huge (infinite).
-- Kinetic energy = (1/2) (mass) (speed)² .
If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.
-- Mass = (Kevin's rest-mass) / √(1 - v²/c²)
-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
-- If he succeeds somehow, his mass becomes imaginary.
-- At that point, he might as well turn around and go home ...
if he ever reached Planet-Y, nobody could see him anyway.
Answer:0.114 C
Explanation:
Given
Total 4.7 C is distributed in two spheres
Let
and
be the charges such that

and Force between charge particles is given by



put the value of 




thus 