30: Problem 30.6

PART A:
How many conduction electrons are there in a 4.50 mm diameter gold wire that is 20.0 cm long?

SOLUTION:
signature first step...

Givens/ conversions:

Diameter ≡ d = 4.50 mm = 0.0045 m
Length ≡ l = 20.0 cm = 0.200 m

Using Table 30.1...

n_eGOLD = 5.9×10²⁸ m^-3

And we need to determine N_e... 

Equation (30.2) states that N_e = n_e⋅V
We can determine volume by V = A⋅l = (π/4)d²⋅l

∴ N_e = n_e⋅(π/4)d²⋅l

= (5.9×10²⁸ m^-3)(π/4)⋅(0.0045 m)²⋅(0.200 m)

⇒N_e = 1.88×10²³ electrons

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PART B:
How far must the sea of electrons in the wire move to deliver -38.0 nC of charge to an electrode?
Express your answer with the appropriate units.

SOLUTION:
So we know from part A that in 0.2 m, there are 1.877×10²³ electrons ...

each electron has a charge of -1 e, or -1.602×10^-19 C
and that 1 nC = 1×10^-9 C, or 1×10^-9 C/nC = 1×10⁹ C/nC....

this means that the desired charge = -38×10^-9 C

The total charge created by that many electrons is simply equal to the number of electrons multiplied by the charge or each individual electron, or Q_tot = Q_e⋅N_e

The charge per unit length can be determined by dividing the total charge by the length from part A.

Q/l = Q_e⋅N_e/l = (-1.602×10^-19 C)(1.877×10²³)/(0.2 m)
→ Q/l = -1.504×10⁵ C/m
lets call this Q_perLength
​

The total length (L) required to contain a total charge (Q_f) may then be solved for by ... L = Q_f/Q_perLength = l⋅Q_f/(Q_e⋅N_e) 

→ L = (0.2 m)(-38×10^-9 C)/[(-1.602×10^-19 C)(1.877×10²³)]
⇒ L = 2.528×10^-13 m

Which is a really small number... which brings us to the last aspect. Units... I think that the units (which are units of length, of course) are most likely going to be desired in the same units that the length in part A was expressed, so centimeters... 

10² cm/m... so

⇒ L = 2.53×10^-11 cm