# A wire contains a steady current of 2 A. The number of electrons that pass a cross section in 2 s is:A) 2 B) 4 C) 6.3 × 1018D) 1.3 × 1019E) 2.5 × 1019

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### 1 Answer

Ampere(A) is defined as Coulomb per second(C/s). Its formula is:

`I=Q/t`

where I - current, Q-charge, and t-time.

Substitute *I=2 A* and *t=2 s* to determine the charge passing through the wire.

`2=Q/2`

`4=Q`

So, the charge is Q=4C.

To determine the number of electrons, take note of the conversion factor:

(1 electron) `1e=-1.6x10^(-19)C`

Using this yields,

`4C * (1e)/(-1.6x10^(-19)C) = -2.5x10^19 e`

Take the positive value of e.

**Hence, the number of electrons that pass though the wire's cross section is (E) `2.5x10^19 ` .**

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