Answer:
Explanation:
Given
maximum rate of flow is [tex]\frac{\mathrm{d} i}{\mathrm{d} t}=20,000\ C/s[/tex]
Time of flow [tex]t=100\ \mu s[/tex]
Charge flows during this process
[tex]=\frac{\mathrm{d} i}{\mathrm{d} t}\times t[/tex]
[tex]Q=20,000\times 100\times 10^{-6}[/tex]
[tex]Q=2\ C[/tex]
(b)Total no of electron(n)
If charge on one electron is [tex]q=1.6\times 10^{-19}[/tex]
[tex]n=\frac{Q}{q}=\frac{2}{1.6\times 10^{-19}}=1.25\times 10^{19}[/tex]
