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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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## QuestionPhysicsClass 12

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?

$$198$$ 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.