A cylindrical wire of radius 2 mm carries a current of 3.0 A. The potential difference between points on the wire that are 44 m apart is 3.8 V. Required:
a. What is the electric field in the wire?
b. What is the resistivity of the material of which the wire is made?

a. E = 86.36 x 10⁻³ V/m = 86.36 mV/m

b. ρ = 3.6 x 10⁻⁷ Ωm

Explanation:

a.

The electric field in terms of the voltage is given by the following formula:

E = V/d

where,

E = Electric Field in the Wire = ?

V = Potential Difference = 3.8 V

d = distance between the points = 44 m

Therefore,

E = 3.8 V/44 m

E = 86.36 x 10⁻³ V/m = 86.36 mV/m

b.

Now, from Ohm's Law:

V = IR

R = V/I

where,

R = Resistance of wire = ?

I = Current = 3 A

Therefore,

R = 3.8 V/3 A

R = 1.27 Ω

Now, the resistance of a wire can be given as:

R = ρL/A

where,

ρ = resistivity of material = ?

L = Length = 44 m

A = Cross-sectional area = πr² = π(0.002 m)² =  1.25 x 10⁻⁵ m²

Therefore,

1.27 Ω = ρ*44 m/1.25 x 10⁻⁵ m²

(1.27 Ω)(1.25 x 10⁻⁵ m²)/44 m = ρ

ρ = 3.6 x 10⁻⁷ Ωm

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.