Answer:
1) Approximately [tex]4\,\,\,10^{13}[/tex] excess protons
2) negative [tex]4\,\,\,10^{13}[/tex] ([tex]-4\,\,\,10^{13}[/tex])
Explanation:
1)
Recall that the charge of an electron or proton is approximately: [tex]1.6\,\,\,10^{-19} \,C[/tex]
Therefore, to find the number of protons transferred in 7 micro-Coulombs of charge, we do:
[tex]\frac{7\,\,10^{-6}\,}{1.6\,\,10^{-19}} \approx 4\,\,10^{13}[/tex]
Approximately [tex]4\,\,\,10^{13}[/tex] excess protons
2)
The sign and number of uncanceled elemental charges on plate A is therefore negative [tex]4\,\,\,10^{13}[/tex] , because the same number of positive charges were removed from it, changing its neutrality