Question

a parallel plate capacitor is charged until there are 10^9 electrons on one of its plates. what is the net charge on the two plates?
o -1.6 x 10^-10 c
o +1.6 x 10^-10 c
o +3.2 x 10^-10 c
o -3.2 x 10^-10 c
o 0c
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25.1 capacitance

Explanation:

Step1: Recall the charge of an electron

The charge of a single electron is $e=- 1.6\times10^{-19}\text{ C}$.

Step2: Calculate the charge on one plate

If there are $n = 10^{9}$ electrons on one plate, the charge on that plate $q_1=ne$. Substituting the values, we get $q_1=10^{9}\times(-1.6\times10^{-19}\text{ C})=-1.6\times10^{-10}\text{ C}$.

Step3: Determine the charge on the other plate

In a capacitor, the charges on the two plates are equal in magnitude and opposite in sign. So if the charge on one plate is $q_1=-1.6\times10^{-10}\text{ C}$, the charge on the other plate $q_2 = + 1.6\times10^{-10}\text{ C}$.

Step4: Calculate the net - charge of the two - plate system

The net charge of the two - plate system of a capacitor is $Q=q_1 + q_2=-1.6\times10^{-10}\text{ C}+1.6\times10^{-10}\text{ C}=0\text{ C}$.

Answer:

0 C