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Question: If \(10^{9}\)electrons move out of a body to another body every second, then the time required to ge...

If 10910^{9}electrons move out of a body to another body every second, then the time required to get a total charge of 1 C on the other body is.

A

250 years

B

100 years

C

198 years

D

150 years

Answer

198 years

Explanation

Solution

: The charge given out in one second

=1.6×10−19C×109=1.6×10−10C1.6 \times 10^{- 19}C \times 10^{9} = 1.6 \times 10^{- 10}C

Time required to accumulate a charge of 1 C

=1C1.6×10−10Cs−1=6.25×109s= \frac{1C}{1.6 \times 10^{- 10}Cs^{- 1}} = 6.25 \times 10^{9}s

=6.25×109365×24×3600=198years= \frac{6.25 \times 10^{9}}{365 \times 24 \times 3600} = 198years