Answer:
[tex]n=1.56\times 10^{17}[/tex] electrons
Explanation:
Given that,
Total charge = 9 mC = 0.009 C
0.009 C of charge passes through a wire in 3.6 s.
Let q' is the charge that passes through it in 10 s.
So,
[tex]\dfrac{0.009 }{3.6}=\dfrac{q'}{10}\\\\q'=\dfrac{0.009 \times 10}{3.6}\\\\q'=0.025\ C[/tex]
We know that,
q = ne
Where
n is the number of electrons
So,
[tex]n=\dfrac{q}{e}\\\\n=\dfrac{0.025}{1.6\times 10^{-19}}\\\\n=1.56\times 10^{17}[/tex]
So, [tex]1.56\times 10^{17}[/tex]electrons must pass through the cross-sectional area.
