A potential difference of 3.27 nV is set up across a 2.16 cm length of copper wire that has a radius of 2.33 mm. How much charge drifts through a cross section in 3.23 ms? Assume that the resistivity of copper is 1.69 * 10-8 Ω·m.

Charge = 4.9096 x 10⁻⁷ C

Explanation:

First, we find the resistance of the copper wire.

R = ρL/A

where,

R = resistance = ?

ρ = resistivity of copper = 1.69 x 10⁻⁸ Ω. m

L = Length of wire = 2.16 cm = 0.0216 m

A = Cross-sectional area of wire = πr² = π (0.00233 m) ² = 1.7 x 10⁻⁵ m²

Therefore,

R = (1.69 x 10⁻⁸ Ω. m) (0.0216 m) / (1.7 x 10⁻⁵ m²)

R = 2.14 x 10⁻⁵ Ω

Now, we find the current from Ohm's Law:

V = IR

I = V/R

I = 3.27 x 10⁻⁹ V/2.14 x 10⁻⁵ Ω

I = 1.52 x 10⁻⁴ A

Now, for the charge:

I = Charge/Time

Charge = (I) (Time)

Charge = (1.52 x 10⁻⁴ A) (3.23 x 10⁻³ s)

Charge = 4.9096 x 10⁻⁷ C