a. How many conduction electrons are there in a $1.0\mathrm{mm}$diameter gold wire that is $10\mathrm{cm}$long?

b. How far must the sea of electrons in the wire move to deliver $-32\mathrm{nC}$of charge to an electrode?

Gold wire diameter$=1.0mm$

Gold wire length$=10cm$

In our case, the cross-section is a circle, so the volume of the wire is the area $A$times the length$L,$

$V=AL=\frac{\pi D^2}{4}L$

The electron density is the number of electrons per unit volume is,

$n_e=\frac{N}{V}$

Calculate the number of electrons,

$N=n_{e}V=\frac{\pi n_e{D}^{2}L}{4}$

Substitute the values,

role="math" localid="1648878276627" $N=\left(\frac{\pi \times 5.9·{10}^{28}e\times 0.{001}^{2}mm\times 0.1cm}{4}\right)$

$=4.6·{10}^{24}$

Hence, there are $4.6\times {10}^{24}$conduction electrons are there in a $1.0\mathrm{mm}$diameter gold wire that is $10\mathrm{cm}$long.

Gold wire diameter$=1.0\mathrm{mm}$

Gold wire length$=10\mathrm{cm}$

The charge delivered will be given as

$Q=Ne,$

Where $N$is the number of electrons moved, and$e$is the elementary charge.

The number we found is,

$N=\frac{\pi n_e{D}^2}{4}L$

Where $L$is the distance that the charges moved,

$L=\frac{4N}{\pi n_e{D}^2}$

Substituting the number of electrons, $N=\frac{Q}{e}$, we get

$L=\frac{4Q}{\pi e{n}_e{D}^2}$

Substitute the values we get,

role="math" localid="1648878783126" $L=\left(\frac{4\times 3.2·{10}^{-8}J}{\pi \times 1.6·{10}^{-19}c\times 5.9·{10}^{28}e\times 0.{001}^{2}mm}\right)$

$=4.32\mu \mathrm{m}$

Hence, the distance for the sea of electrons in the wire move to deliver $-32nC$of charge to an electrode is$4.2\mu m.$