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

1 Answer | Add Yours

mjripalda's profile pic

Posted on

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 ` .

Sources:

We’ve answered 319,967 questions. We can answer yours, too.

Ask a question