Question

Part A

How many conduction electrons are there in a 1.00 mm diameter gold wire that is 50.0 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.

Answer 1

Given that :

Diameter of gold wire, D = 1 x 10^-3 m

Radius of gold wire, r = 0.5 x 10^-3 m

Length of wire, L = 0.5 m

We know that, volume of wire is given by -

V = A L = (r²) L

V = (3.14) (0.5 x 10^-3 m)² (0.5 m)

V = 3.925 x 10^-7 m³

Part A : The number of conduction electrons required which will be given as -

using a formula, we have

N = n V

where, n = number of free electrons in gold wire = 5.9 x 10²⁸ m^-3

THEN, we get

N = (5.9 x 10²⁸ m^-3) (3.925 x 10^-7 m³)

N = 23.1 x 10²¹

N = 2.31 x 10²² conduction electrons

Part B : How far must the sea of electrons in the wire move to deliver -24 nC of charge to an electrode?

we know that, n = q / e = (24 x 10^-9 C) / (1.6 x 10^-19 C/electron)

n = 1.5 x 10¹¹

using a formula, we have

L = (1.5 x 10¹¹) / n (r²)

L = (1.5 x 10¹¹) / (5.9 x 10²⁸ m^-3) [(3.14) (0.5 x 10^-3 m)²]

L = (1.5 x 10¹¹) / (4.6315 x 10²² m^-1)

L = 3.23 x 10^-12 m