If the sprinter from the previous problem accelerates at that rate for 20 m, and then maintains that velocity for the remainder
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Answer:
a) 0.15 μC b) 9.4*10¹¹ electrons.
Explanation:
As the total charge must be conserved, the total charge on the spheres, after being brought to contact each other, and then separated, must be equal to the total charge present in the spheres prior to be put in contact:
Q = +8.2μC +9.0 μC +(-7.8 μC) + (-8.8 μC) = +0.6 μC
As the spheres are assumed perfect conductors, as they are identical, once in contact each other, the excess charge spreads evenly on each sphere, so the final charge, on each of them, is just the fourth part of the total charge:
Qs = Qt/4 = 0.6 μC / 4 = 0.15 μC.
b) As the charge has a positive sign, this means that each sphere has a defect of electrons.
In order to know how many electrons are absent in each sphere, we can divide the total charge by the charge of one electron, which is the elementary charge e, as follows: