A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross

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1. Answer:

   n = 1.56 x 10¹⁷ electrons

   Explanation:

   First of all, we will calculate the current passing through wire:

   [tex]I = \frac{q}{t}[/tex]

   where,

   I = current = ?

   q = charge = 9 mC = 0.009 C

   t = time = 3.6 s

   Therefore,

   [tex]I = \frac{0.009\ C}{3.6\ s}\\\\I = 0.0025\ A = 2.5\ mA[/tex]

   Now, for the same current in 10 s time the charge will be:

   q = It = (0.0025 A)(10 s)

   q = 0.025 C

   Now, the number of electrons can be given as:

   [tex]q = ne\\\\n = \frac{q}{e}\\\\[/tex]

   where,

   n = no. of electrons = ?

   q = charge = 0.025 C

   e = charge on single electron = 1.6 x 10⁻¹⁹ C

   Therefore,

   [tex]n = \frac{0.025\ C}{1.6\ x\ 10^{-19}\ C}[/tex]

   n = 1.56 x 10¹⁷ electrons

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