Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor, [tex]Q=121\ nC=121\times 10^{-9}\ C[/tex]
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
[tex]Q=ne[/tex]
e is the charge on electron
[tex]n=\dfrac{Q}{e}[/tex]
[tex]n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}[/tex]
[tex]n=7.5625\times 10^{11}[/tex]
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
