Download Optics I - Department of Applied Physics

Hong Kong Polytechnic University
Electrical Charge
Some Concepts:
Electrical charge (电荷); Electric field (电场); Electric current (电流); Conductor (导体);
Semiconductor (半导体); Insulator (绝缘体); Superconductor (超导体);
Two Fundamental Properties of Electrical Charge:
Charge is quantized: e  1.60 1019 C ;
Charge is conserved:
Electrostatic Force:
12
2
2
k  8.99 10 N  m /C
permittivity constant (介电常数):  0  8.85 10 C /N  m
Note that this satisfies Newton's third law because it implies that exactly the same
magnitude of force acts on q2 . Coulomb's law is a vector equation and includes the
fact that the force acts along the line joining the charges. Like charges repel and
unlike charges attract. Coulomb's law describes a force of infinite range which obeys
the inverse square law, and is of the same form as the gravity force.
9
2
2
Shell Theorem:
A shell of uniform charge attracts or repels a charged particle that is outside the shell
as if all the shell’s charge were concentrated at its center. A shell of uniform charge
exerts no electrostatic force on a charged particle that is located inside the shell.
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electrical Charge
Spherical Conductors:
If excess charge is placed on a spherical shell that is made of conducting material,
the excess charge spreads uniformly over the external surface.
Example: The figure shows two particles fixed in a place: a particle of charge q1=+8q at
the origin of an x axis and a particle of charge q2=-2q at x=L. At what point (other than
infinitely far away) can a proton be placed so that it is in equilibrium? Is that equilibrium
stable or unstable?
2qq p
1 8qq p
1
Solution:

x  2L
40 x
40 x  L 
If x  2 L  x
8qq p
2qq p
qq p x
1
1
F


2
2
40 2 L  x  40 L  x 
20 L3
2
2
Example: The arrangement of six fixed charged
particles, where =30, is shown. All six particles
have the same magnitude of charge q; their
electrical signs are as indicated. What is the net
electrostatic force acting on q1 due to the other
charges?
Solution: 0
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electrical Charge
Example: Two identical, electrically isolated conducting spheres A and B are separated by
a (center-to-center) distance a that is large compared to the spheres. Sphere A has a positive
charge of +Q; sphere B is electrically neutral; and initially, there is no electrostatic force
between the spheres. (a) Suppose the spheres are connected for a moment by a conducting
wire. The wire is thin enough so that any net charge on it is negligible. What is the
electrostatic force between the spheres after the wire is removed? (b) Next, suppose sphere A
is grounded momentarily, and then the ground connection is removed. What now is the
electrostatic force between the spheres?
Solution:
2
(a) : q1  q2  Q 2
(b) : q1  Q 2
F
q2  0
q1q2
1 Q

 
2
40 a
160  a 
1
F 0
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electrical Charge
Homework:
1. The figure shows two charges, q1 and q2, held in a
fixed distance d apart. (a)What is the magnitude of the
electrostatic force that acts on q1? Assume that
q1=q2=20.0C and d=1.50m. (b)A third charge
q3=20.0 C is brought in and placed as shown in the
figure. What now is the magnitude of the electrostatic
force on q1?
2. In the basic CsCl (cesium chloride) crystal structure,
Cs+ ions form the corners of a cube and a Cl ion is at
the cube’s center. The edge length of the cube is 0.40
nm. The Cs+ ions are each deficient by one electron
(and thus each has a charge of +e), and the Cl ion has
one excess electron (and thus has a charge of –e). (a)
What is the magnitude of the net electrostatic force
exerted on the Cl ion by the eight Cs+ ions at the
corners of the cube? (b)If one of the Cs+ ions is
missing, the crystal is said to have a defect; what is the
magnitude of the net electrostatic force exerted on the
Cl ion by the seven remaining Cs+ ions?
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Electric Fields
Electric Field
Electric field is defined as the electric force per unit charge. The direction of the field
is taken to be the direction of the force it would exert on a positive test charge. The
electric field is radially outward from a positive charge and radially in toward a
negative point charge.
F
E
q0
Conversely, the electrostatic force acting on a particle with a charge q is F  qE
Electric Field of Point Charge:
E
1 Qsource
40 r 2
A positive number is taken to be an
outward field; the field of a negative
charge is toward it. The electric field
from any number of point charges can
be obtained from a vector sum of the
individual fields.
n
F0   F0i
i 1
n
E 0   E 0i
College Physics----by Dr.H.Huang, Department of Applied Physics
i 1
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Electric Fields
Example: The figure shows three particles with charges
q1=+2Q, q2=2Q, and q3=4Q, each a distance d from the
origin. What net electric field E is produced at the origin?
Solution:
E  E1  E2  E3
Ex  E1 cos 30  E2 cos 30  E3 cos 30

1  2Q 2Q 4Q 
3Q



cos
30



40  d 2 d 2 d 2 
0 d 2
E x   E1 sin 30  E2 sin 30  E3 sin 30  0
E
3Q
i
0 d 2
Example: The nucleus of a uranium atom has a radius R of 6.8 fm. Assuming that the
positive charge of the nucleus is distributed uniformly, determine the electric field at a point
on the surface of the nucleus due to that charge.
Solution: The atomic number Z=92
E
1
Q
1 Ze
21


2
.
9

10
N/C
40 R 2 40 R 2
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Fields
Electric Field Lines (电力线):
Electric field lines provide a means for visualizing the direction and magnitude of
electric fields. The electric field vector at any point is tangent to a field line through
that point. The density of field lines in any region is proportional to the magnitude of
the electric field in that region. Field lines originate on positive charges and
terminate on negative charges. A few examples are shown below:
Electric Dipole (电偶极子) Field:
An electric dipole consists of two particles with charges of equal magnitude q but
opposite sign, separated by a small distance d. It can be proved that the electric
field generated by the dipole along the dipole axis is
1 2p
E// 
40 r 3
where r is the distance to the center of the dipole r>>d.
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Fields
Dipole Moment (偶极矩):
p=qd is called the electric dipole moment. It is a vector with a direction pointing from
the negative to the positive charge of the dipole.
The electric field in a plane perpendicular to the dipole axis that cuts through the
1 p
dipole by half is
E 
40 r 3
Dipole in an Electric Field:
The electric field exerts a torque on the dipole
d
d
sin   F sin    Fd sin 
2
2
 p
 qE   sin    pE sin 
q
  F
τ  pE
The dipole has a potential energy U  p  E
where we choose the potential energy to be zero when the angle  is 90.
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Fields
Electric Field Due to a Charged Ring: (Optional)
As shown in the figure,  is the charge per unit length. The total
charge on the ring is q.
E   dE cos   

z
40 r
dq
1 z
cos


ds
2
3

40 r
40 r
  ds 
3
1
z
40 r
 2R  
3
qz
40 r 3

qz
40 z 2  R 2 
32
Electric Field Due to a Charged Disk: (Optional)
For a uniformly charged disk with a radius R, we can cut the disk
into small pieces of rings. Suppose the charge per unit area is
, then for each piece of ring, it carries a charge of
dq  dA   2rdr 
z 2rdr 
Each ring contributes an electric field dE 
The total field is E   dE 
z
4 0
 z
R
0
2
 r2
40 z 2  r 2 
32

3 2
dr 

z
2rdr
4 0 z 2  r 2 3 2
 
z
1 
2 0 
z 2  R2




College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Fields
Homework:
1.
2.
A clock face has negative point charges –q, 2q, 3q, …, 12q fixed at the positions of
the corresponding numerals. The clock hands do not perturb the net field due to the
point charge. At what time does the hour hand point in the same direction as the electric
field vector at the center of the dial? (Hint: Use symmetry).
In the figure a uniform, upward-pointing electric field E of magnitude 2.00103 N/C has
been set up between two horizontal plates by charging the lower plate positively and the
upper plate negatively. The plates have length L=10.0 cm and separation d=2.00 cm.
An electron is then shot between the plates from the left edge of the lower plate. The
initial velocity v0 of the electron makes an angle =45.0 with the lower plate and has a
magnitude of 6.00106 m/s. (a) Will the electron strike one of the plate? (b) If so,
which plate and how far horizontally from the left edge?
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Potential
Electrical Potential Energy:
The electrostatic force is a conservative force. Thus we can define a change of the
electric potential energy U of a charged particle in an electric field to be,
U  U f  U i  W where W is the work done by the electrostatic force and is path
independent.
If the potential energy is defined to be zero at infinity, the electric potential energy U
of a point charge is U=W , where W is the work done by the electric field on the
point charge as the charge moves from infinity to the particular point.
Concept:
In the figure, a proton moves from point i to point f in a uniform electric field directed as
shown. (a) Does the electric field do positive or negative work on the proton? (b) Does the
electric potential energy of the proton increase or decrease?
E
f
Electrical Potential:
+
i
The electrical potential energy per unit charge is defined as the electric potential
difference (电势差):
U
W
V  V f  Vi 
q

The electrical potential at a point is V  
q
U  q  V
W
; The unit is volt=joule per coulomb.
q
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Potential
Equipotential Surfaces (等势面):
Adjacent points that have the same electric potential form an equipotential surface,
which can be either an imaginary surface or a real, physical one.
(i) No work is done by the
electric field when a charge
is moved on that surface.
(ii) The electric field E is
always directed
perpendicularly to the
equipotential surface.
Potential Calculated from the Field:
f
V f  Vi  W q
W  q  E  ds
i
f
f
V   E  ds
V f  Vi   E  ds
i
i
In a uniform electric field E, the potential
f
difference is,
V f  Vi   Eds   Ed
i
f
V f  Vc   E cos 45 ds   Ed
c
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Electric Potential
Potential due to a Point Charge:
r


qdr

V    E  dr   E  dr  

r
r 4 r 2
0
n
V  Vi 
i 1
V
n
qi

40 i 1 ri
1
V
1
q
40 r
1
dq
1

40  r
40

dV
r
Example: (a) What is the electric potential V at a distance r=2.1210-10 m from the nucleus
of a hydrogen atom (the nucleus consists of a single proton)? (b) What is the electric
potential energy U in electron-volts of an electron at the given distance from the nucleus? (c)
If the electron moves closer to the proton, does the electric potential energy increase or
decrease?
1 e
Solution: (a ) : V 
 6.78 V
(b) : U  qV  eV  6.78 eV
40 r
(c): V increases and U decreases.
Example: What is the potential at point P, located at the center of the
square of point charges shown in the figure? Assume that d=1.3 m and
that the charges are q1=+12 nC, q2=24 nC, q3=+31 nC, q4=+17 nC.
Solution:
4
1 q q q q
V  Vi 
i 1
40
1
2
3
d
2
4
 350 V
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Potential
Example: (a) In figure a, 12 electrons are equally spaced and fixed around a circle of
radius R. Relative to V=0 at infinity, what are the electric potential and electric field at the
center C of the circle due to these electrons? (b) If the electrons are moved along the circle
until they are nonuniformly spaced over a 120 are (figure b), what then is the potential at C?
Solution:
1 12e
40 R
1 12e
(b) : V  
40 R
(a) : V  
E0
Potential of Line Charge: (Optional)
It can be found by superposing the point charge
potentials of infinitesmal charge elements. It is an
example of a continuous charge distribution.
V
1
40

b
a
 b  b2  d 2 


ln 

2
2
2
2
40   a  a  d 
x d
dx
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Potential
Potential for Ring of Charge: (Optional)
It can be found by superposing the point charge potentials of
infinitesmal charge elements. The ring potential can then be used
as a charge element to calculate the potential of a charged disc.
1 Q R

40 r 2 0 r
V
Potential for Disc of Charge: (Optional)
It can be found by superposing the point charge potentials of
infinitesmal charge elements. The evaluation of the potential can
be facilitated by summing the potentials of charged rings.
V
dq
1
 2RdR 




2
2
40 r
40
2 0
z  R
1



R
0
RdR
z 2  R 2

z2  R2  z
2 0
Calculating E from V:

W  V r  dr   V r   V  dr  -E  dr
E  V x, y, z   
V
V
V
i
j
k
x
y
z
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Potential
Electric Potential Energy of a System of Point Charges:
The electric potential energy of a system of fixed point charges is equal to the work
that must be done by an external agent to assemble the system, bringing each
charge in from an infinite distance.
1 n n qq
i j
q1q2
U

For two point charges: U 12 
For many point charges:
80 i 1 j 1 rij
40 r12
i j
Example: Starting with the expression for the potential at any point on the axis of a
1
charged disk, derive the expression for the electric field at any point on the axis of the disk.
Solution:

V
 d
 
z
Ez  
z

2 0

z
dz
2

 R2  z 
1 



2
2
2 0 
z R 
Example: The figure shows three charges held in fixed positions by forces that are not
shown. What is the electric potential energy of this system of charges? Assume
that d=12 cm and that q1=+q, q2=4q, and q3=+2q, in which q=150 nC.
Solution:
U
1  q1q2 q2 q3 q3q1 



  17 mJ
40  d
d
d 
College Physics----by Dr.H.Huang, Department of Applied Physics
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Electric Potential
Homework:
1.
2.
In a given lightning flash, the potential difference between a cloud and the ground is
1.0109 V and the quantity of charge transferred is 30 C. (a) What is the change in
energy of that transferred charge? (b) If all the energy released by the transfer could be
used to accelerate a 1000 kg automobile from rest, what would be the automobile’s final
speed? (c) If the energy could be used to melt ice, how much ice would it melt at 0C?
The heat of fusion of ice is 3.33105 J/kg.
For the charge configuration of the figure, show that V(r) for points such as P on the
axis, assuming r>>d, is given by
1 q  2d 
V
1 

40 r 
r 
(Note: the charges should be treated as point charges and their size being neglected.)
r

q
d
+
+q
d
+
+q
P
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Capacitance
Capacitor:
A capacitor is a device that stores electric potential energy by storing separated
positive and negative charges. It consists of two conductors separated by either
vacuum or an insulating material. The simplest case is a parallel plate capacitor.
Capacitance is defined in terms of charge storage:
C
Q
V
(Farad, F)
where, Q = magnitude of charge stored on each plate.
V = voltage applied to the plates.
E

Q

0 0 A
V  Ed 
Qd
0 A
C
0 A
d
Cylindrical Capacitor: (Optional)
The charge resides on the outer surface of the inner conductor
and the inner wall of the outer conductor. Assume the length of
the cylinder L>>b.
b
b
Q
Q
b
V   Edr  
a
C  20
a
 0 2rL 
dr 
20 L
ln  
a
L
ln b a 
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Spherical Capacitor: (Optional)
b
b
Q
Q  1 1
V   Edr  
dr

  
a
a  4r 2 
40  a b 
0
Capacitance
C  40
ab
ba
An Isolated Sphere: (Optional)
By taking the limits: aR and b,
C  40 R
Concept: For capacitors charged by the same battery, does the charge stored
by the capacitor increase, decrease, or remain the same in each of the
following situations? (a) The plate separation of a parallel capacitor is
increased. (b) The radius of the inner cylinder of a cylindrical capacitor is
increased. (c) The radius of the outer spherical shell of a spherical capacitor is
increased.
Example: The plates of a parallel-plate capacitor are separated by a distance d=1.0 mm.
What must be the plate area if the capacitance is to be 1.0 F?
Solution:
A
Cd
0
 1.1108 m 2
Example: A storage capacitor on a random memory (RAM) chip has a capacitance of 55fF.
If the capacitor is charged to 5.3 V, how many excess electrons are on its negative plate?
Q CV
Solution:
6
n
e

e
 1.8 10
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Capacitance
Capacitors in Parallel and Series:
Equivalent capacitance: Ceq 
n
C
i 1
i
(parallel)
n
1
1
  (series)
Ceq i 1 Ci
Example: (a) Find the equivalent capacitance of the combination as shown. Assume
C1=12.0 F, C2=5.30 F, C3=4.50 F. (b) A potential difference V=12.5 V is applied to the
input terminals. What is the charge on C1?
Solution: (a): C12  C1  C2
C  17.3 F
1
1
1


C123 C12 C3
(b): q123  C123V  44.6 C
12
C123  3.57 F
V12 
q12 q123

 2.58 V
C12 C12
Example: A 3.55 F capacitor C1 is charged to a potential difference V0=6.30 V, using a
6.30 V battery. The battery is then removed and the capacitor is connected as in the figure to
an uncharged 8.95 F capacitor C2. When switch S is closed, charge flows from C1 to C2
until the capacitors have the same potential difference V. What is the common potential
difference?
Solution: q0  q1  q2
C1
C1V0  C1V  C2V
V  V0
C1  C2
 1.79 V
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Capacitance
Potential Energy and Energy Density:
The electric potential energy of a capacitor is the energy stored in the electric field
between the two plates (electrodes). It is the work required to charge the capacitor.
Q
Q
0
0
W   dW   Vdq  
q
Q2
dq 
C
2C
Q2 1
U
 CV 2
2C 2
The energy density is the potential energy per unit volume.
U
CV 2 1  V 
u

 0 
Ad 2 Ad 2  d 
2
1
u  0E 2
2
The above results hold generally for types of capacitors.
Example: An isolated conducting sphere whose radius R is 6.85 cm has a charge q=1.25
nC. (a) How much potential energy is stored in the electric field of this charged conductor?
(b) What is the energy density at the surface of the sphere? (c) What is the radius R0 of an
imaginary spherical surface such that half of the stored potential energy lies within it?
Solution:
2
2
2
a  :
U
c  : R
R0
q
q

 103 nJ
2C 80 R


b  :
2
u 4r 2 dr  
R0
R
1
1  1 q 
3

u   0 E 2   0 

25
.
4

J/m
2
2  40 R 2 
1  1 q
q2  1 1  U
2
 4r dr 
   
 0 
2 
2  40 r 
80  R R0  2


College Physics----by Dr.H.Huang, Department of Applied Physics
R0  2 R
21
Hong Kong Polytechnic University
Capacitance
Dielectrics:
Dielectric material contains polar molecules (or being polarized under an electric
field), they will generally be in random orientations when no electric field is applied.
An applied electric field will polarize the material by orienting the dipole moments of
polar molecules. This decreases the effective electric field between the plates and
will increase the capacitance of the parallel plate structure. The dielectric must be a
good electric insulator so as to minimize any DC leakage current through a
capacitor.
Eeff  E  E p 

0
C
0 A
d
 is called the dielectric constant of a material.
In a region completely filled by a material of
dielectric constant , all electrostatic equation
containing the permittivity constant 0 are to be
replaced by 0.
The Guass’ law need to be generalized to,
0  E  dA  q
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Hong Kong Polytechnic University
Example: A parallel-plate capacitor whose capacitance C
is 13.5 pF is charged to a potential difference V=12.5 V
between its plates. The charging battery is now
disconnected and a porcelain slab is slipped between the
plates. What is the potential energy of the device, both
before and after the slab is introduced?
Solution: U  1 CV 2  1100 pJ
i
2
U
q2
q2
Uf 

 i  160 pJ
2C f 2C 
Capacitance
Room temperature dielectric
constants for some materials
Material

Air (1atm)
1.00054
Polystyrene
2.6
Paper
3.5
Transformer oil
4.5
Pyrex
4.7
Ruby mica
5.4
Porcelain
6.5
Silicon
12
Germanium
16
Ethanol
25
Water (20°C)
80.4
Water (25°C)
78.5
Titania ceramic
130
Strontium titanate
310
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Capacitance
Example: The figure shows a parallel-plate capacitor of plate area A and plate separation d.
A potential difference V0 is applied between the plates. The battery is then disconnected, and
a dielectric slab of thickness b and dielectric constant  is placed between the plates as
shown. Assume, A=115cm2, d=1.24cm, V0=85.5V, b=0.780cm, =2.61. (a) What is the
capacitance C0 before the dielectric slab is inserted? (b) What free charge appears on the
plates? (c) What is the electric field E0 in the gaps between the plate and the dielectric slab?
(d) What is the electric field E1 in the dielectric slab? (e) What is the potential difference V
between the plates after the slab has been introduced? (f) What is the capacitance with the
slab in place?
 A
Solution: (a ) : C0  0  8.21 pF
d
(b) : q  C0V0  720 pC
( c ) :  0  E  dA   0 E 0 A  q
E0 
q
 6.90 kV/m
0 A
(d ) : 0  E  dA  0 E1 A  q
(e) : V  E0 d  b   E1b  52.3 V
E1 
q
0 A
 2.64 kV/m
( f ): C 
q
 13.4 pF
V
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Capacitance
Homework (continued):
1. The figure shows a variable “air gap” capacitor. Alternate
plates are connected together; one group is fixed in position
and the other group is capable of rotation. Consider a pile of
n plates of alternate polarity, each having an area A and
separated from adjacent plates by a distance d. Show that this
capacitor has a maximum capacitance of C=(n-1)0A/d.
2. In the figure, battery B supplies 12 V. (a) Find the charge on
each capacitor first when only switch S1 is closed and (b)
later when switch S2 is also closed. Take C1=1.0 F, C2=2.0
F, C3=3.0 F, and C4=4.0 F.
3. A parallel-plate capacitor of plate area A is filled with two
dielectrics as shown in the figure. Show that the capacitance
is
 A   2
C 0 1
d
2
4. A parallel-plate capacitor of plate area A is filled with two
dielectrics of the same thickness as shown in the figure.
Show that the capacitance is
2 A  1 2
C 0
d 1   2
1
2
2
1
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Current and Resistance
Current:
dq
i

An electric current in a conductor is
dt (ampere, A)
A current direction is the one in which positive charge carriers would move. The
negative charges move in the opposite direction.
Current Density:
i   J  dA
Drift of Charge Carriers:
When an electric field E is established in a conductor, the charge carriers (assumed
positive) acquire a drift speed vd in the direction of E, J  nev d
where n is the number of charge carriers per unit volume.
Example: (a) The current density in a cylindrical wire of radius R=2.0 mm is uniform
across a cross section of the wire and is given by J=2.0105 A/m2. What is the current
through the outer portion of the wire between radial distance R/2 and R? (b) Suppose, instead,
that the current density through a cross section varies with radial distance r as J=ar2, in
which a=3.01011 A/m4 and r is in meters. What now is the current through the same outer
portion of the wire?
Solution: (a) : i  JA  J R 2   R 2 4  1.9 A


(b) : i   JdA   ar 2 2r dr  7.1 A
R
R2
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Current and Resistance
Example: One end of an aluminum wire whose diameter is 2.5 mm is welded to one end of
copper wire whose diameter is 1.8 mm. The composite wire carries a steady current i of 17
mA. (a) What is the current density in each wire? (b) What is the drift speed of the
conduction electrons in the copper wire? Assume that, on the average, each copper atom
contributes one conduction electron.
J Cu  i ACu  6.7 103 A/m 2
Solution: (a) : J Al  i AAl  3.5 103 A/m 2
(b) : vd 
J Cu J Cu M

 4.9 10 7 m/s
ne
e NA
Example: Consider a strip of silicon that has a rectangular cross section with width w=3.2
mm and height h=250 m, and through which there is a uniform current i of 5.2 mA. The
silicon has a number of charge carriers (electrons) per unit volume n=1.51023 m-3. (a) What
is the current density in the strip? (b) What is the drift speed?
Solution:
i
i
J
2
(a) : J 
A

wh
 6500 A/m
(b) : vd 
ne
 27 cm/s
Concept:
The figure shows conduction electrons moving leftward through a wire. Are the following
leftward or rightward: (a) the current i, (b) the current density J, (c) the electric field E in the
wire?
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Resistance:
V
R
i
(ohm, )
Current and Resistance
Resistivity of some materials at
room temperature
Material
 (m)
 (K-1)
Silver
1.6210-8
4.110-3
Copper
1.6910-8
4.310-3
Aluminium
2.7510-8
4.410-3
Tungsten
5.2510-8
4.510-3
Iron
9.6810-8
6.510-3
Platinum
10.610-8
3.910-3
The resistivity for most materials changes with
temperature   0  0 T  T0 
Manganin
48.210-8
-7010-3
Pure Silicon
2.5103
 is the mean temperature coefficient of resistivity.
n-type Silicon
8.710-4
Ohm’s Law:
p-type Silicon
2.810-3
Resistivity:
E  J
Conductivity:   1 
unit of : m
(-1m-1)
J  E
The resistivity R of a conducting wire of length L
and uniform cross-section A is
L
R
A
10
14
Glass
10 -10
The current through a device is always directly
Fused quartz
1016
proportional to the potential difference applied to
the device.
A conducting material or device obeys Ohm’s law when the resistivity is independent
of the magnitude and direction of the applied electric field.
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Current and Resistance
Example: A rectangular block of iron has dimension 1.2 cm1.2 cm15 cm. (a) What is the
resistance of the block measured between the two square ends? (b) What is the resistance
between two opposite rectangular faces?
Solution:
L
L
(a) : R  
A
 100 A
(b) : R  
A
 0.65 A
Power: P  iV  Ri 2  V 2 R
Example: A wire has a resistance R of 72 . At what rate is energy dissipated in each of
the following situations? (1) A potential difference of 120 V is applied across the full length
of the wire. (2) The wire is is cut in half, and a potential difference of 120 V is applied across
the length of each half.
Solution: (1) : P  V 2 R  200 W
(2) : P  2V 2 R 2  800 W
Example: A wire of length L=2.35 m and diameter d=1.63 mm carries a current i of 1.24
A. The wire dissipates electrical energy at the rate P=48.5 mW. Of what is the wire made?
Solution:
i 2 L 4i 2 L
d 2 P
P  Ri 2 
A

d
2

2
4i L
 2.80 10 8   m
It’s aluminum.
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Current and Resistance
Semiconductor (optional):
Semiconductors are materials with intermediate conductivity
between conductors and insulators.
From quantum mechanics, the electrons of an atom may
occupy quantized energy levels. When a large amount of
atoms form a solid, the discrete energy levels may be merged
to form energy bands. The electrons are only allowed to sit
within the energy bands but they cannot have an energy value
within the gaps separating the gaps.
When excited (by thermal activation, for example), electrons
can jump from the valence band to the conduction band and
leave holes in the valence band. The conductivity of the
material is enhanced.
Superconductors (optional):
Superconductivity is the lost of any conductivity at low
temperatures. It was first discovered in mercury by K.Onnes in
1911. Since 1986, high temperature (~90K) superconductor in
ceramics have been discovered and developed.
College Physics----by Dr.H.Huang, Department of Applied Physics
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Hong Kong Polytechnic University
Current and Resistance
Homework:
1.
2.
A charged belt, 50 cm wide, travels at 30 m/s between a source of charge and a sphere.
The belt carries charge into the sphere at a rate corresponding to 100 A. Compute the
surface charge density on the belt.
In the figure, a resistance coil, wired to an external battery, is placed inside a thermally
insulated cylinder fitted with a frictionless piston and containing an ideal gas. A current
i=240 mA exists in the coil, which has a resistance R=550 . At what speed v must the
piston, of mass m=12 kg, move upward to keep the temperature of the gas unchanged?
College Physics----by Dr.H.Huang, Department of Applied Physics
31