Download Week 5 - Dielectrica, Resistance and Resistivity

Week 5 - Dielectrica, Resistance and Resistivity
Further, the dignity of the science itself
seems to require that every possible means
be explored for the solution of a problem so
elegant and so celebrated.
Carl Friedrich Gauss
Exercise 5.1: Simple dielectric capacitor
Two conducting plates are placed in parallel to form a capacitor. Each plate has an area of 1 × 10−1 m.
The distance between the plates is 2.00 cm. The potential difference between the plates is 5 kV. There
is vacuum between the plates.
a) What is the capacitance?
Answer:
A
= 44.25 pF
d
(1)
Q0 = C0 V0 = 0.22 µC
(2)
C 0 = ε0
b) What is the charge on each plate?
Answer:
1
c) What is the electric field between the plates?
Answer:
E0 =
V0
V
= 2.5 × 105
d
m
(3)
We disconnect the power supply and assume that all charge stays on the plates, while we insert a dielectric
between the plates. The potential drops to 1 kV.
d) What is the capacitance after the dielectric is inserted?
Answer:
C=
Q
= 221.2 pF
V
(4)
e) What is the dielectric constant of the material?
Answer:
K=
C0
4425 pF
=
= 5.0
C
221.2 pF
(5)
f) What is the permittivity of the dielectric?
Answer:
ε = Kε0 = 4.425 × 10−11 C2 /N · m2
(6)
g) What is the induced charge on each face of the dielectric?
Answer:
Qi = σi A = σA(1 −
1
1
) = Q(1 − ) = 0.177 µC
K
K
(7)
h) What is the electric field?
Answer:
E=
E0
= 5 × 104 V/m
K
(8)
Exercise 5.2: Audio Cables
Copper and silver are the two materials which most often are used for Audio Cables. Silver has a slightly
lower resistivity than copper. This means that for a given resistance, silver cables be made thinner than
copper cables.1 Suppose that you have a copper wire of diameter 0.5 mm and length 5.0 m.
1 This
has implications for audio quality. See ’Audio Wire’ Wikipedia.
2
a) What is the resistance of this wire? (Copper has ρC = 1.72 × 10−8 .)
Answer: R = 0.44 Ω.
b) What would be the diameter of a silver cable with the same length and resistance? (Silver has
ρS = 1.47 × 10−8 .)
Answer: d = 0.46 mm
Exercise 5.3: Resistance and Resistivity Nuts
a) The definition of resistivity (ρ = E/J) implies that an electric field exist inside a conductor. Yet in
Chapter 21 we saw that there can be no electric fields inside a conductor. Is there a contradiction
here?
Answer:
No. One key assumption in the derivation of this fact was that all the charges were static. They settle
into their equilibrium position (where they feel no force) after a very brief period of time and thus
there are no electric fields. However for a closed circuit with an applied field along the length of the
wire, the charges are moving continuously. If you will there are no equilibrium. So the assumption is
broken and therefore we have no contradiction.
b) Suppose we have a closed loop of wire with electrons moving around because of an applied electric
field. It’s a fact that the drift velocity of the electrons is on the order of v ∼ 10−4 m/s, but if there is
an electric field inside the copper wire, how come the electrons does not accelerate?
Answer:
The value v ∼ 10−4 m/s is the average velocity of the electrons in the direction of the wire. They are
continuously accelerating, but they are also continuously crashing into the atoms of the wire, losing
their energy. And since there are so many of them, the net movement is as if they have a constant
velocity of v.
Exercise 5.4: Multiple Dielectrics
Figure 1
3
A parallel-plate capacitor has the space between the plates filled with two slabs of dielectric; one with
constant K1 and another with constant K2 . Each slab has thickness d/2, where d is the plate separation.
Each plate has a charge density σ. See figure 1.
a) Assume that the plates are very large compared to the distance d. Using Gauss law, find the electric
field inside each dielectric.
Answer: E1 =
σ
K 1 ε0
E2 =
σ
K 1 ε0
b) Calculate the potential between the plates.
Answer: V =
Qd K1 +K2
2ε0 A K1 K2 .
c) Find the total capacitance.
Answer:
C=
2ε0 A K1 + K2
.
d
K1 K2
(9)
Exercise 5.5: Leakage of a Dielectric
There is no such thing as a perfect insulator (or conductor). Therefore any dielectric insulator will leak
a little bit of current. Suppose then that you have a parallel plate capacitor with charge Q on each plate
and you have a dielectric material with resistivity ρ and dielectric constant K placed between the two
plates.
a) Find the electric field using that the charge density may be expressed as σ = Q/A.
Answer:
E=
Q
σ
.
=
Kε0
Kε0 A
(10)
b) What is the relation between current density and the electric field?
Answer: J = E/ρ
c) Show that the ’leakage’ current I carried by the dielectric is given by
I=
Q
.
Kε0 ρ
(11)
d) How does the capacitance C vary with time because of this leakage current? Explain.
Exercise 5.6: Resistivity of a Conic conductor
A material of resistivity ρ is formed into a solid, truncated cone of height h and radii r1 and r2 at either
end (see figure 2).
4
Figure 2
a) Calculate the resistance of the cone between the two flat end faces. Hint: Imagine slicing the cone
into very many thin disks, and calculate the resistance of one such disk. You might also need to use
R = ρL/A.
Answer: R =
ρh
πr1 r2
b) Check your result by showing it agrees with the resistance for a cylinder of length L and radius r,
R = ρL/πr2 .
c) The cone has a current density |J| = ar directed along the axis of the cone. Find the difference in
current at the two ends of the cone. Would this conductor remain neutral over time?
Answer:
∆I =
aπ 4
r2 − r14
4
5
(12)