Week 3
Electric Charge, Force, and Field
Charge Density, Current Density
Continuity Equation
Electric Charge
 Around 600 BC, Thales of
Miletus reported that charge
could be accumulated by
rubbing fur on various
substances such as amber.
 He noted that the charged
amber buttons could attract
light objects such as hair.
 He also noted that if he rubbed
the amber for long enough, he
could even get a spark to jump.
Amber is “electron” in Greek ,
which is spelled ήλεκτρον
. Electric Charge Convention
 The convention of
positive and negative
charges is accredited to
Benjamin Franklin.
 The electron charge
turned out to be negative
under this convention.
 The electron has a
negative charge of 1.6 x
10-19 As.
Electric Charge Convention
 The force of a positive
charge is radially outward.
 The force of a negative
charge is radially inward.
 If the charges are opposite,
the charges will attract
each other.
 If the charges are similar,
the charges will repel each
other.
The Electroscope
 It consists of a vertical metal rod from the
end of which hang two parallel strips of
thin flexible gold leaf.
 A metal ball is attached to the other end
where the charge to be tested is applied.
 When the metal sphere is touched with a
charged object the gold leaves spread
apart in a 'V'.
 Since the leave they receive the same sign
charge they repel each other.
Virtual Demonstration
Electronic Electroscope
 Build the circuit below to detect the electric field E
surrounding an electrified object
Sensor built atop a 9v battery connector
Coulomb’s Law
 The law of electrostatic attraction
and repulsion, developed in the
1780s by French physicist Charles
Augustine de Coulomb, may be
stated in scalar form as follows:
The magnitude of the
electrostatic force between two
point electric charges is directly
proportional to the product of the
magnitudes of each charge and
inversely proportional to the
square of the distance between
the charges.
Inverse-Square Law
 The magnitude of the force exerted by two charges over
each other is given by
F = k Q1 Q2
R2
k=
Newtons (N= kg m s-2)
__1___ = 9 x 109
4πεo
N m2 / (As)2
εo = 8.85 x 10-12 F/m electric constant
or permittivity of free space
Exercise
 Find the magnitude of the force F for Q1 = 5 nAs (or nC
in old convention), Q2 = -3 nAs, and R = 2 m.
Inverse-Square Law
 Lines represent the “flux”
emanating from the source.
 The total number of flux
lines depends on the
strength of a charge and is
constant with increasing
distance.
 A greater density of flux
lines (lines per unit area)
means a stronger force field.
Inverse-Square Law
 The density of flux lines is
inversely proportional to the
square of the distance from
the source because the
surface area of a sphere
increases with the square of
the radius.
 Thus the strength of the field
is inversely proportional to
the square of the distance
from the source.
Electric Force Vector
 The force exerted by Q1 on a positive unit charge located at
a distance R is given by
F12 = k Q1Q2 a12
R2
Newtons ( N = kg m s-2 )
 The direction of the force due to a positive charge Q1 (x1, y1,
z1) at an observation point P (x2, y2, z2) is given by:
a12 =
x (x2 – x1) + y (y2 – y1) + z (z2 – z1)
[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2 ] 1/2
Exercise
 Determine the force F that Q1 = 1 x 10-9 (C) exerts on Q2
= 2 x 10-9 (C). Assume x and y coordinates are given in
meters.
Electric Field
 The space surrounding an electric charge has a
property called an electric field. This electric field
exerts a force on other electrically charged objects.
 The electric field (force per unit charge) at a distance R
is given by
E = k Q1 a12
R2
Newtons/meter (N/m)
Examples of Electric Fields
Exercise
 Find the force exerted on a charge Q1 = + 10 nAs by a
field E = 3 N/m ay.
Electric Dipole
 The simplest example of a
pair of electric charges of
equal magnitude but
opposite sign, separated by
some, usually small,
distance.
 The E field of an electric
dipole behaves as shown
here.
Electric Dipole
 In the far field (R>>L), the electric field due to two point charges
on the z axis separated by a short distance L is given by:
E=
QL
4πεo R2
(2cosθ aR + sinθ aθ )
 Dipoles can be characterized by their dipole moment
p = QL z
 For the simple electric dipole given above, the electric dipole
moment would point from the negative charge towards the
positive charge, and have a magnitude equal to the strength of
each charge times the separation between the charges.
Exercise
 Calculate the divergence of the far field produced by
an electric dipole
E=
QL (2cosθ aR + sinθ aθ )
4πεo R2
Institute of Electrical and Electronics
Engineers
 The logo of the or IEEE (read eye-
triple-e) depicts Franklin’s kite, the
electric dipole symbol, and the
magnetic dipole symbol.
 IEEE is an international non-profit,
professional organization for the
advancement of technology related
to electricity. It has the most
members of any technical
professional organization in the
world, with more than 365,000
members in around 150 countries.
Charge Distributions
 Electric field intensity away from an
infinite line charge distribution ρ along z
axis
E = ____ρ_____ ar
2πεo r
 Electric field intensity away from an
infinite surface charge distribution ρs on
x-y plane
E = ____ρs_____ az
2εo
z >0
Charge Density
 The linear, surface, or volume
charge density is the amount of
electric charge in a line, surface,
or volume.
 It is measured in Coulombs per
meter (As/m), square meter
(C/m²), or cubic meter (C/m³),
respectively.
 Since there are positive as well
as negative charges, the charge
density can take on negative
values.
 Like any density it can depend
on position.
Notation
 Different variables are used to denote the various
dimensions of charge density.
 Commonly used variables:
 λ or ρl for line charge density (C/m)
 σ or ρs for surface charge density(C/m²)
 ρ or ρv for volume charge density (C/m³)
Differentials
 Length
 dx, dy, or dz
 r dφ, dz
 R sinθ dφ
 Surface
 dx dy, dx dz, or dy dz
 r dφ dz
 R2 sinθ dθ dφ
 Volume
 dx dy dz
 r dr dφ dz
 R2 sinθ dR dθ dφ
(angular aperture on x-y plane)
(angular aperture)
(cylindrical surface)
(spherical surface)
(cylindrical volume)
(spherical volume)
Volume Differential Examples
Line Charge Density
 Consider a line density λ on a thing wire along the y
axis as shown here.
Surface Charge Density
 Consider a surface density σ on a thing wire along the
y axis as shown here.
Volume Charge Density
 Consider a volume density ρ in the spherical
configuration shown here.
Getting E from a Charge Distribution
 Use Coulomb’s Law:
 The field differential dE at an observation point P(x, y, z)
is proportional to the charge differential dQ
 The field differential dE is inversely proportional to the
square of the distance R’ between the observation point
P and where dQ is located
 The total field E is the integral of dE
 Take advantage of geometrical symmetry whenever
possible
Field Produced by Line Charge
 The field differential dEz is:
 proportional to the charge
differential dQ = λdr
 Inversely proportional to
R’ = (r2 + z2) ½
 The projection of dE along
the z axis
 Due to symmetry:
 The r component of E will
be zero on the z-axis
 E will point in the zdirection
Field Produced by Ring Charge
 The field differential dEz is:
 proportional to the charge
differential dQ = λ rdφ
 Inversely proportional to R’ = (r2
+ z2) ½
 The projection of dE along the z
axis
 Due to symmetry:
 The r component of E will be
zero
 E will point in the z- direction
Exercise
 Assume that σ is
constant.
 Determine the
differential dE produced
by a ring
 Determine the
projection of dE along
the x axis
Current Density J
 It’s a measure of the density of
flow of the electric charge or
electric current per unit area of
cross section. In SI units, the
electric current density is
measured in amperes per
square meter.
 Current density J defined as a
vector whose magnitude is the
current per cross-sectional
area.
Current Density J
 A common approximation to the
current density assumes the current
simply is proportional to the electric
field, as expressed by
J=σE
where E is the electric field strength
and σ is the electrical conductivity.
 Conductivity σ is the reciprocal
(inverse) of electrical resistivity and
has the SI units of Siemens per meter
(S m-1)
Current I
 The current through a surface S
can be calculated using a surface
integral:
 The current is the net flux of the
current density vector field
flowing through the surface S.
Practical Issues
 In the domain of electrical wiring (isolated copper), maximum current
density can vary from 4A/mm2 for a wire isolated from free air to
6A/mm2 for a wire at free air. If the wire is carrying high frequency
currents (above 100kHz) the skin effect may affect the distribution of
the current across the section by concentrating the current on the
surface of the conductor.
 In the domain of printed board, for top and bottom layers, maximum
current density can be as high as 35A/mm2 with a copper thickness of
35 µm. Inner layers cannot dissipate as much power as outer layers,
thus it is not a good idea to put high power lines in inner layers.
 In the domain of semiconductor, the maximum current density is given
by the manufacturer. But a common average is 1mA/µm (180 nm
technology) (Ampere per width of the line).
Exercise
 The maximum current density of an insulated copper
wire is 4 A/mm2. The conductivity of the metal is
59.6 × 106 S/m. Determine the maximum electric field
that the wire can sustain.
Charge Conservation
Charge Conservation
 It’s the principle that electric charge can neither be created nor
destroyed.
 The quantity of electric charge is always conserved.
 The net flow out of a chosen volume must equal the net change
in charge held inside the volume:
 The surface integral on the left expresses the current outflow
from the volume, and the negatively signed volume integral on
the right expresses the decrease in the total charge inside the
volume.
Continuity Equation
 From the divergence theorem
 Hence
 Because this relation is valid for any volume, no matter
how small, no matter where located, we have
Exercise
 Suppose that a charge of electrons is suddenly injected
in a block of silver. As time goes by, the electrons
would separate due to Coulomb forces.
 Let the charge density be given by
ρ = ρo e –αt
where ρo is the initial charge density and α is a
material constant.
 Determine the initial rate of change of this charge
density.
“Hyperphysics” Website
Homework
 Read book sections 1-2.2, 4-1, 4-2, 4-3
 Solve problems 4.9, 4.11, 4.13, 4.15, 4.1, 4.3, 4.5, 4.7