Question

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-sectional area in 10s is

Answer 1

Given :

• total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

`1.6 * 10 {}^( - 19) \: C`

For producing 1 Cuolomb charge we need :

• 
  `\mathrm{\frac{1}{1.6 * 10 {}^( - 19) } }`

• 
  `\frac{10 {}^(19) }{1.6}`

• 
  `\frac{10* 10 {}^(19) }{16}`

• 
  `\frac{100 * 10 {}^(18) }{16}`

• 
  `\mathrm{6.24 * 10 {}^(18) \: \: electrons}`

Now, for producing 0.009 C of charge, the number of electrons required is :

• 
  `0.009 * 6.24 * {10}^(18)`

• 
  `0.05616 * 10 {}^(18)`

• 
  `\mathrm{5.616 * 10 {}^(16) \: \: electons}`

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So, Number of electrons passing through the cross section in 3.6 seconds is :

`\mathrm{5.616 * 10 {}^(16) \: \: electrons}`

Number of electrons passing through it in 1 Second is :

• 
  `\frac{5.616 * {10}^(16) }{3.6}`

• 
  `\mathrm{1.56 * 10 {}^(16) \: \: electrons}`

Now, in 10 seconds the number of electrons passing through it is :

• 
  `10 * \mathrm{1.56 * 10 {}^(16) \: \: }`

• 
  `\mathrm{1.56 * 10 {}^(17) \: \: electrons}`

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`\mathrm{ \#TeeNForeveR}`