Physics 2 (9 CFU)
Name
Surname
Exercise n.1
A.A. 2011-2012
n. matricola
28.2.2011
A small sphere having mass m=5 g and bearing a positive charge q=+10-9 C, is initially kept steady at a
distance d=1m above an ideally infinite plate uniformly charged with a charge density σ. At a certain time the
sphere is thrown vertically downward with initial velocity v=1 m/s. Calculate the minimum charge density σ of
the plate such that the sphere will never reach the plate.
The problem can be solved applying the total energy conservation law and assuming that the sphere
reaches the plate with zero velocity:
1
𝑚𝑣 2 + 𝑚𝑔ℎ = 𝑞∆𝑉
2
Where: 𝑞∆𝑉 = 𝑞ℎ
𝜎
2𝜀0
.
Solving for σ we finally obtain:
𝑚𝜀
𝜎 = 0 (𝑣 2 + 2𝑔ℎ) ≅ 0.9 × 10−4 𝐶/𝑚2
𝑞ℎ
Exercise n. 2
A thin conducting hollow cylinder of outer radius a=5 cm and height h=1 m is surrounded by a coaxial hollow
cylinder having the same height and inner radius b=20 cm. The space in between the two cylinders is
completely filled by a material characterized by a resistivity ρ=10 -1 Ωcm. A difference of electrostatic potential
ΔV= 10 V is applied between the inner and the outer cylinders by an external e.m.f. generator. Calculate the
current flowing between the cylinders.
In order to calculate the resistance between the two cylinders we may think of dividing the material in
between in many thin hollow cylinders of thickness dr and radius r, with r ranging from a to b.
The resistance for each of these cylinders is:
𝑑𝑟
𝑑𝑅 = 𝜌
2𝜋𝑟ℎ
The total resistance will be:
𝑏
𝜌𝑑𝑟
𝜌
𝑏
𝑅=∫
=
𝑙𝑛
2𝜋ℎ 𝑎
𝑎 2𝜋𝑟ℎ
And the total current:
𝐼 =∆V/R≅ 4.5 × 102 𝐴
Exercise n. 3
Two identical homogeneous conducting stripes having
width D=10 cm, indefinite length and negligible thickness,
lay parallel at a distance h= 0.5 cm. A current I=10 A is
flowing along both stripes. The direction of the current in the
two stripes is opposite. Calculate the value and the
direction of the magnetic field in the region between the
two stripes and far from the edges.
The magnetic field can be calculated by the Ampere law, considering i) a rectangular amperian loop perpendicular to
the J vector with both the long side outside of the current double layer and ii) a rectangular amperian loop with one
long side inside and the second outside the double current layer.
The first loop does not enclose current so we can conclude that B out=0 outside.
𝑎
𝐼
From the second loop we see that 𝐵𝑖𝑛 𝑎 = 𝜇0 𝐼 → 𝐵𝑖𝑛 = 𝜇0 ≅ 1.26 × 10−4 𝑇
𝐷
𝐷
The B direction is given by the right hand rule and is indicated in the figure.
Exercise n. 4
An ideal one-dimensional crystal consists of a linear chain of monovalent atoms separated by a distance a=2
Å. Estimate the Fermi energy in eV of this system.
Because the metal is monovalent the first zone is half filled:
𝜋
𝐾𝐹 =
2𝑎
As a consequence:
ћ2 𝜋 2
𝜀𝐹 =
( ) ≅ 2.3 𝑒𝑉
2𝑚 2𝑎