a) object A has [tex]3.0\cdot 10^{12}[/tex] electrons more than protons
b) [tex]+5.61\cdot 10^{-7} C[/tex]
Explanation:
a)
Electrons are negatively charged particles, so they have a charge of
[tex]q_e =-e= -1.6\cdot 10^{-19}C[/tex]
On the other hand, protons are positively charged particles, so they have a charge of
[tex]q_p =+e=+1.6\cdot 10^{-19}C[/tex]
Therefore, the net charge of an object can be written as
[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
So the equation can be rewritten as
[tex]N_p - N_e = \frac{Q}{e}[/tex]
Which gives the difference between number of protons and number of electrons.
Here, object A has a charge of
[tex]Q=-4.81\cdot 10^{-7}C[/tex]
Therefore,
[tex]N_p-N_e=\frac{-4.81\cdot 10^{-7}}{1.6\cdot 10^{-19}}=-3.0\cdot 10^{12}[/tex]
Which means that object A has [tex]3.0\cdot 10^{12}[/tex] electrons more than protons.
b)
When Erin touches the objects together, the number of electrons transferred from object A to object B is
[tex]N_e' = 5.0\cdot 10^{11}[/tex]
The charge corresponding to these electrons transferred is
[tex]Q'=N_e' (-e)=(5.0\cdot 10^{11})(-1.6\cdot 10^{-19})=-8\cdot 10^{-8} C[/tex]
The initial charge on object B was
[tex]Q_B=+6.41\cdot 10^{-7}C[/tex]
This means that the final charge on object B after the electrons are transferred will be:
[tex]Q=Q_B+Q'=+6.41\cdot 10^{-7}+(-8\cdot 10^{-8})=+5.61\cdot 10^{-7} C[/tex]
