a) [tex]3.0\cdot 10^{12}[/tex]
b) [tex]5.61\cdot 10^{-7} C[/tex]
Explanation:
a)
The charge on one electron is negative and equal to the fundamental charge, so the charge on one electron is
[tex]q_e =-e= -1.6\cdot 10^{-19}C[/tex]
Protons instead are positively charged particles, so they have a charge of
[tex]q_p =+e=+1.6\cdot 10^{-19}C[/tex]
This means that the net charge of an object can be written as follows:
[tex]Q=N_p q_p + N_e q_e=e(N_p-N_e)[/tex]
where:
[tex]N_p[/tex] is the number of protons
[tex]N_e[/tex] is the number of electrons
This can also be rewritten as
[tex]N_p - N_e = \frac{Q}{e}[/tex]
This equation gives the difference between the number of protons and the number of electrons in an object.
In this problem, object A has a charge of:
[tex]Q=-4.81\cdot 10^{-7}C[/tex]
Substituting, we find
[tex]N_p-N_e=\frac{-4.81\cdot 10^{-7}}{1.6\cdot 10^{-19}}=-3.0\cdot 10^{12}[/tex]
So, before the two objects come in contact, object A has [tex]3.0\cdot 10^{12}[/tex] electrons more than protons.
b)
When the two objects are touched together, the number of electrons transferred from object A to object B is:
[tex]N_e' = 5.0\cdot 10^{11}[/tex]
So the total charge transferred from object A to object is
[tex]\Delta Q=N_e' (-e)=(5.0\cdot 10^{11})(-1.6\cdot 10^{-19})=-8\cdot 10^{-8} C[/tex]
At the beginning, object B has a charge of:
[tex]Q_B=+6.41\cdot 10^{-7}C[/tex]
So, the final charge on object B after the electrons are transferred will be:
[tex]Q=Q_B+\Delta Q=+6.41\cdot 10^{-7}+(-8\cdot 10^{-8})=5.61\cdot 10^{-7} C[/tex]
