Electric current and circuits question?

Question

How many conduction electrons are there in a 5.00 mm diameter gold wire that is 30.0 cm long?

using golds electron density as 5.9 * 10^28 I got 3.48*10^23

but now

How far must the sea of electrons in the wire move to deliver -21.0 nC of charge to an electrode?

and Im stuck... again >.>!
Thanks for the help guys :)

Answer 1

21 nC is also a count of electrons.

There are 6.2414 * 10^18 electrons (charges) in a Coulomb (a physical constant in your textbook inside cover)

21 nC = 21 * 10^-9 * 6.2414 * 10^18 = 1.31 * 10^11 (conduction) electrons

How much volume will that many atoms take up in the gold wire, and then what distance of the cylinder(ical wire) is represented.. that's your answer. I get 113 femtometers


Your first number 3.48 *1-0^23 checks with my math.

.

Answer 2

The numbers of electrons in the aforesaid wire are simply given by multiplying the electron density of gold to its volume, which it self is out by the formula V=AL in which A is its section area equal to ¼πd² (with d the diameter of wire)and L is its length.

Now for the second part, u must find the length of a volume of the wire that contains -21.0 nC of charge. (-21.0 nC equals to -21.0*6.3*10^20 electrons). This way you've found the length that the sea of electrons must move through the wire to deliver such a charge. The mentioned volume is obtained by dividing 21.0 *6.3*10^20 by 5.9 * 10^28. Now dividing the volume to its section area we are up with the answer.

Hope this helped.