If 10^{9} electrons move out of a body to ano | vedclass

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(B) The charge of one electron is $e = 1.6 	imes 10^{-19} \, C$. In one second,$10^{9}$ electrons move out of the body. Therefore,the charge transfer

Vedclass · Accepted Answer

$200 \, 	ext{years}$

Vedclass · Answer

(B) The charge of one electron is $e = 1.6 	imes 10^{-19} \, C$. In one second,$10^{9}$ electrons move out of the body. Therefore,the charge transferred in one second is $q = 10^{9} 	imes 1.6 	imes 10^{-19} \, C = 1.6 	imes 10^{-10} \, C/s$. To accumulate a total charge of $Q = 1 \, C$,the time required $t$ is given by $t = Q / q$. $t = 1 \, C / (1.6 	imes 10^{-10} \, C/s) = 0.625 	imes 10^{10} \, s = 6.25 	imes 10^{9} \, s$. To convert this time into years,we divide by the number of seconds in a year: $365 	imes 24 	imes 3600 \, s \approx 3.1536 	imes 10^{7} \, s$. $t \approx (6.25 	imes 10^{9}) / (3.1536 	imes 10^{7}) \approx 198.18 \, 	ext{years}$. Rounding this value,we get approximately $200 \, 	ext{years}$.

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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