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If  \(10^9\)electrons move out of a body to another body every second, how much time is required to get a total charge of  \(1\;C\)  on the other body?
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If  \(10^9\)electrons move out of a body to another body every second, how much time is required to get a total charge of  \(1\;C\)  on the other body?

Answer

\(198\) years
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Solution

The basic unit of charge, \(e=1.602192\times 10^{-19}\mathit{C}\)
Given that  \(10^9\)electrons move out in one second. Therefore the charge move out in one second can be found as,  \(e\times 10^9\)  .
\({\therefore}\)Rate of flow of charge\(= \ 1.6\times 10^{-19}\times 10^9C/s=1.6\times 10^{-10}\;\mathit{C/s.} \)
The time required to get a total charge of  \(1\;C\)on the other body can be found by dividing the total charge by the charge moving outin one second.
 i.e., \(\frac{1\;C}{ (1.6\times 10^{-10}\;C/s)}=0.625\times 10^{10}=6.25\times 10^9\mathit{s}\)
Time required in years can be estimated as,
 \(6.25\times 10^9s=\frac{(6.25\times 10^9)}{ (365\times 24\times 60\times 60)\mathrm{}}years=198\;\mathrm{years}\)
It reveals that the total time required to collect a charge of one coulomb from a body from which  \(10^9\)
electrons move out every second to another body, we need nearly  \(198\)years of time.
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