Question

If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write your answer in giga electrons (i.e. 10^9 electrons)

Answer 1

n = 756.25 giga electrons

Step-by-step explanation:

It is given that,

If the charge on the negative plate of the capacitor,
`Q=121\ nC=121* 10^(-9)\ C`

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 :

`Q=ne`

e is the charge on electron

`n=(Q)/(e)`

`n=(121* 10^(-9))/(1.6* 10^(-19))`

`n=7.5625* 10^(11)`

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.