How much time $10^{20}$ electrons will take to flow through a point in a conductor so that the current is $200$ m $o A$ $[e = 1.6 \times 10^{- 19} C]$ ?

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Solution

Current flowing $I = 200 m A = 0.2 A$
Total charge flowing $Q = n e = 10^{20} \times 1.6 \times 10^{- 19} = 16 C$
Thus time taken $t = \frac{Q}{I} = \frac{16}{0.2} = 80 s$

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How much time 1020 electrons will take to flow through a point in a conductor so that the current is 200m oA[e=1.6×10−19C]?

Current flowing I=200 mA=0.2 A Total charge flowing Q=ne=1020×1.6×10−19=16 CThus time taken t=QI=160.2=80 s

How much time $10^{20}$ electrons will take to flow through a point in a conductor so that the current is $200$ m $o A$ $[e = 1.6 \times 10^{- 19} C]$ ?

Current flowing $I = 200 m A = 0.2 A$
Total charge flowing $Q = n e = 10^{20} \times 1.6 \times 10^{- 19} = 16 C$
Thus time taken $t = \frac{Q}{I} = \frac{16}{0.2} = 80 s$