In: Physics
Part A
How many conduction electrons are there in a 4.50 mm diameter gold
wire that is 30.0 cm long?
Part B
How far must the sea of electrons in the wire move to deliver -24.0
nC of charge to an electrode?
Express your answer with the appropriate units.
SOLUTION:
A)
Givens/ conversions:
Diameter ≡ d = 4.50 mm = 0.0045 m
Length ≡ l = 30.0 cm = 0.300 m.
neGOLD = 5.9×1028
m-3
And we need to determine Ne...
Ne = ne⋅V
We can determine volume by V = A⋅l = (π/4)d2⋅l
∴ Ne = ne⋅(π/4)d2⋅l
= (5.9×1028 m-3)(π/4)⋅(0.0045 m)2⋅(0.300 m)
⇒Ne = 2.50101×1026 electrons
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PART B
SOLUTION:
So we know from part A that in 0.3 m, there are
2.50101×1026 electrons ...
....
the desired charge = -24×10-9 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 Qtot = Qe⋅Ne
The charge per unit length can be determined by dividing the total
charge by the length from part A.
Q/l = Qe⋅Ne/l = (-1.602×10-19
C)(2.50101×1026)/(0.3 m)
→ Q/l = -1.33553934×108 C/m
lets call this QperLength
The total length (L) required to contain a total charge (Qf) may then be solved for by ... L = Qf/QperLength = l⋅Qf/(Qe⋅Ne)
→ L = (0.2 m)(-24×10-9 C)/[(-1.602×10-19
C)(2.50101×1026)]
⇒ L = 1.19951×10-16 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 = 1.19951×10-14 cm