Electrostatics

Prepared by: Syamsul Adlan
Mahrim
Electrostatic
“Imagination is more important
than knowledge” – Albert Einstein

Mahrim
Charge Density
 Volume charge density definition:
3
0
/lim mC
dv
dq
v
q
v
v 





Where q is the charge contained in v.
v is defined at a given point in space, specified by (x, y, z) in a
Cartesian coordinate system, and at a given time t.
Thus, v = v(x, y, z, t).

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Charge Density
 The variation of v with spatial location is called its
spatial distribution, or simply its distribution.
 The total charge contained in a given volume v is
 CdvQ
v
v 

Mahrim
Charge Density
 Electric charge may be distributed across
the surface of a material, in which case
the relevant quantity of interest is the
surface charge density:
2
0
/lim mC
ds
dq
s
q
s
s 





Where q is the charge present across an elemental surface area s.

Mahrim
Charge Density
 If the charge is distributed along a line,
which need not be straight, we
characterize the distribution in terms of
the line charge density:
mC
dl
dq
l
q
l
l /lim
0







Mahrim
Line Charge Distribution
 Calculate the total charge Q
contained in the cylindrical tube of
charge oriented along the z-axis as
shown in the figure. The line charge
density is l = 2z, where z is the
distance in meters from the bottom
end of the tube. The tube length is
10 cm.

Mahrim

Mahrim
 The total charge is:
Czzdz
dzQ l
21.0
0
2
1.0
0
1.0
0
102 



 

Mahrim
Surface Charge Distribution
 The circular disk of electric charge is
characterized by an azimuthally
symmetric surface charge density
that increases linearly with r from
zero at the center to 6 C/m2 at r = 3
cm. Find the total charge present on
the disk surface.

Mahrim

Mahrim
 Since s is symmetrical with respect to the
azimuth angle , its functional form
depends only on r and is given by
22
2
/102
103
6
mCr
r
s 

 

Where is in meters.

Mahrim
 From the cylindrical coordinates:
 
mC
r
rdrdr
dsQ
r
s
s
31.11
3
1022
102
2
2
103
0
3
2
2
0
103
0
2








 







Mahrim
Tutorial 1
 A square plate in the x-y plane is
situated in the space defined by -3m
 x  3m and -3  y  3m. Find the
total charge on the plate if the
surface charge density is given by s
= 4y2 (C/m2).
 Solution:

Mahrim
Tutorial 2
 A spherical shell centered at the
origin extends between R = 2cm and
R = 3cm. If the volume charge
density is given by v = 3R x 10-4
(C/m), find the total charge
contained in the shell.
 Solution:

Mahrim
Tutorial 3
 Find the total charge on a circular
disk defined by r  a and z = 0 if
s = s0e-r (C/m2) where s0 is a
constant.
 Solution:
 Solutionb:

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Current Density
 Consider the general case where the
charges are flowing through a
surface s whose surface normal n is
not necessarily parallel to u.

Mahrim
Current Density
 The current density in ampere per square
meter, for an arbitrary surface S, the total
current flowing through it is then given
by:
 AsdJI
S

.

Mahrim
Current Density

Mahrim
Current Density
 Convection Current
– The current is generated by the actual
movement of electrically charged
matter, and J is called the convection
current density. eg. – wind-driven
charged cloud.
**Do not obeys Ohm’s
Law.

Mahrim

Mahrim
Current Density
 Conduction Current
– Atom of the conducting medium do not
move.
– Eg. The positive and negative charges in
the metal. Electrons moves only from
atom to atom. The electron emerge
from the wire are not necessarily the
same electrons that entered the wire at
the other end.
– Obeys Ohm’s law.

Mahrim
Coulomb’s Law
 Coulomb’s Law states that
1. An isolated charge q induces an electric field
E at every point in space, and at any specific
point P, E is given by
mV
R
q
RE /
4
ˆ
2



Where R is a unit vector pointing from q to P (in the figure), R is the distance between
them, and  is the electrical permittivity of the medium containing the observation point
P; and

Mahrim
Coulomb’s Law

Mahrim
Coulomb’s Law
2. In the presence of electric field E, at a given
point in space, which may due to a single
charge or a distribution of many charges,
the force acting on a test charge q’, when
the charge is placed at that point, is given
by
)(' NEqF



Mahrim
Coulomb’s Law
 With F measured in newtons (N) and q’ in
coulombs (C), the unit of E is (N/C), which
is same as Volt per meter (V/m).
 For a material with electrical permittivity ,
the electric field quanties D and E are
related by
ED



Mahrim
Coulomb’s Law
   mF
where
with
o
or
/1036/11085.8 912 




Is electrical permittivity of free space, and r = /o is called the relative
permittivity (or dielectricconstant) of the material.
 - constant value independent of both the magnitude and direction of E.

Mahrim

 If  is independent of the magnitude
of E, then the material is said to be
linear because D and E are related
linearly, and if it is independent of
the direction of E, the material is
said to be isotropic.

Mahrim
Electric Field due to a Charge
Distribution

Mahrim
 The differential electric field at point P due
to a differential amount of charge dq =
vdv’ contained in a differential volume
dv’ is
2'
'
2'
'
4
'ˆ
4
ˆ
R
dv
R
R
dq
REd v





Where R’ is the vector from differential volume dv’ to point P.
**- R’ and R’ vary as a function of position over the
integration volume v’

Mahrim
 
' 2'
'
'
'ˆ
4
1
v
v
v R
dv
REdE



 
' 2'
'
'
'ˆ
4
1
S
s
S R
ds
REdE



 
' 2'
'
'
'ˆ
4
1
l
l
l R
dl
REdE



For volume distribution
For surface distribution
For line distribution

Mahrim
Electric field of a Ring of Charge
 A ring of charge with a radius b is
characterized by a uniform line
charge density of positive polarity l.
With the ring in free space and
positioned in the x-y plane as shown
in the figure, determine the electric
field intensity E at a point P(0,0,h)
along the axis of the axis of the ring
at a distance h from its center.

Mahrim
 Segment 1 located at (b,,0).
 The length dl = bd and contain charge
dq = ldl = lbd.
 The distance vector R’1 from segment 1 to
point P(0,0,h) is
hzbrR ˆˆ'
1 


Mahrim
 From which we have
22'
1
'
1'
1
22'
1
'
1
ˆˆˆ
hb
hzbr
R
R
R
hbRR








Mahrim
 The electric field at P(0,0,h) due to the
charge of segment 1 is
 
 




d
hb
hzbrb
R
dl
REd
o
ll
o
2/3222
1
'
11
ˆˆ
4'
ˆ
4
1





Mahrim
 The field dE1 has component dE1r along –r
and component dE1z along z.
 From symmetry considerations, the field
dE2 generated by segment 2 fig(b), which
is located diametrically opposite the
location of segment 1, is identical with dE1
except that the r-component of the sum
cancel and the z-contributions add.

Mahrim
 The sum of contributions is
  2/32221
2
ˆ
hb
dbh
zEdEdEd
o
l





 Since for every ring segment in the semicircle
defined over the range 0     (the right-hand
half of the circular ring) there is a corresponding
segment located diametrically opposite at (+),
we can obtain the total field generated by the
ring by integrating over a semicircle as follows:

Mahrim
 
 
 
Q
hb
h
z
hb
bh
z
d
hb
bh
zE
o
o
l
o
l
2/322
2/322
02/322
4
ˆ
2
ˆ
2
ˆ





 





 
Where Q = 2bl is the total charge contained in the ring.

Mahrim
Electric Field of a Circular Disk of
Charge pg. 75

Mahrim
Gauss’s Law
“It is not knowledge,
but the act of learning;
not possession but the
act of getting there,
which grants the
greatest enjoyment.”
- Carl Friedrich Gauss

Mahrim
Gauss’s Law
 Differential form of Gauss’s Law:
• “differential” refers to the fact that
the divergence operation involves
spatial derivatives.
 LawsGaussD v '


Mahrim
Integral form of Gauss’s Law
 Multiply both sides by dv and take the
volume integral over an arbitrary volume
v. Hence,
 
v
v
v
QdvdvD 


Mahrim
Integral form of Gauss’s Law
 From the Divergence theorem,
 
Sv
sdDdvD

 Thus leads to
 LawsGaussQsdD
S
'


Mahrim
 The integral form of Gauss’s Law is
illustrated diagrammatically:

Mahrim
 For each element ds, D.ds is the
electric field flux flowing outwardly
through ds, and the total flux
through the surface S is equal to the
enclosed charge Q. The surface S is
called Gaussian surface.

Mahrim
 q may be regarded as a point
charge.
 The integral form of Gauss’s Law can
be applied as follows:

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"Throughout the centuries, there were men
who took first steps, down new roads,
armed with nothing but their own vision."
-- Ayn Rand, Novelist

Mahrim
1. Electric Potential
2. Electric Dipole and Flux Lines
3. Energy Density

Mahrim
What is….
 Voltage?...
 Current?...

Mahrim
Electric Scalar Potential
 The voltage V between 2 points in the
circuit represents the amount of work, or
potential energy, required to move a unit
charge between the 2 points.
 It is the existence of electric field between
2 points that gives rise to the voltage
difference between them, such as across a
resistor or a capacitor.

Mahrim
Electric Potential as a function of E
 In electrostatics, the potential
difference between P2 and P1 is the
same irrespective of the path used
for calculating the line integral of the
electric field between them.

Mahrim

Mahrim
 The potential difference between any 2
points P2 and P1 is obtained by
integrating along any path between them:
 
2
1
1212
P
P
ldEVVV

Where V1 and V2 are the electrostatic potentials at points P1 and P2, respectively.
The result of the line integral on the RHS of the above equation should be
independent of the specific integration path taken between points P1 and P2.

Mahrim
 The voltage difference between 2
nodes in an electrical circuit has the
same value regardless of which path
in the circuit we follow between the
nodes.

Mahrim
 In fact, the line
integral of the
electrostatic field E
around any closed
contour C is zero:
  
C
ticElectrostaldE 0


Mahrim
 A vector field whose line integral
along any closed path is zero is
called a conservative or an
irrotational field.
 Hence, the electrostatic field E is
conservative.

Mahrim
 But in the time-varying case, E is no
longer conservative, and its line
integral along a closed path is not
necessarily equal to zero.
 In electric circuit analogue, absolute
voltage at a point in a circuit has no
defined meaning, nor does absolute
electric potential at a point in a
space.

Mahrim
 When we talk of the voltage V of a
point in a circuit, we do so in
reference to the voltage of some
conveniently chosen point to which
we have assigned a reference
voltage of zero which we call
ground.

Mahrim
 The same principle applies to the
electrical potential V.
 Usually the reference potential point is
chosen to be at infinity.
 That is, in equation:
 
2
1
1212
P
P
ldEVVV


Mahrim
 We assume that V1 = 0 when P1 is at
infinity, and therefore the electric
potential V at any point P is given by:


P
ldEV


Mahrim
Electric Potential due to Point
Charges
 For a point charge q located at the origin
of a spherical coordinate system, the
electric field at a distance R is:
2
4
ˆ
R
q
RE




Mahrim
Charges
 The choice of integration path
between the two end points is
arbitrary.

Mahrim
Charges
 Hence, we will conviniently choose
the path to be along the radial
direction R, in which the case dl = R
dR and
 V
R
q
dRR
R
q
RV
R
 44
ˆ
2






 


Mahrim
Charges
 If the charge q is at a location other than
the origin, specified by a source position
vector R1, then V at observation position
vector R becomes
   V
RR
q
RV
14





Where [R – R1] is the distance between the observation point and the location of the
charge q.

Mahrim
Charges
 The principle of superposition that is
applied to the electric field E also
applies to the electric potential V.
 Hence for N discrete point charges
q1, q2,…qN having position vectors
R1, R2,…,RN, the electric potential is
   V
RR
q
RV
N
i i
i
 

14
1




Mahrim
Electric Potential due to
Continuous Distribution
   
   
   ondistributiline'
'4
1
ondistributisurface'
'4
1
ondistributivolume'
'4
1
'
'
'
dl
R
RV
ds
R
RV
dv
R
RV
l
l
s
s
v
v
















Mahrim
Electric Field as a Function of
Electric Potential
 This relationship between V and E in
differential form allows us to determine
E for any charge distribution by first
calculating V using the expression given
previously.
VE 


Mahrim
Eg. 4.12, pg 142 - Sadiku
 Given the potential
 Find the electric flux density D at (2, /2, 0)
 cossin
10
2
R
V 

Mahrim
Ans:





ˆsin
10ˆcoscos
10
ˆcossin
20
ˆ
sin
1ˆ1
ˆ
333
rr
r
r
V
r
V
r
r
r
V
VE


















Table

Mahrim
Electric Field as a Function of
Electric Potential
 Even though the electric potential
approach for finding E is a two step
process, it is computationally simpler
to apply than the direct method
based on Coulomb’s Law.

Mahrim
Electric Field of an Electric Dipole
(eg)
 An electric dipole consists of two
point charges of equal magnitude
and opposite polarity, separated by a
small distance, as shown in the
figure. Determine V and E at any
point P in free space, given that P is
at a distance R>>d, where d is the
spacing between the two charges.

Mahrim
(solution)
 The electric potential due to a single point
charges is given previously by
 V
R
q
dRR
R
q
RV
R
 44
ˆ
2






 

 For the two charges shown in the figure,
application of equation
   V
RR
q
RV
14






Mahrim
(solution)
 gives





 





 

21
12
21 44
1
RR
RRq
R
q
R
q
V
oo 
 Since d<<R, the lines labeled R1 and R2
in the figure are approximately parallel to
each other, in which case the following
approximations apply:
2
2112 ,cos RRRdRR  

Mahrim
(solution)
 Hence,
2
4
cos
R
qd
V
o


 The numerator can be written as the dot
product of qd, where d is the distance
vector from charge –q to charge +q,

Mahrim
(solution)
 and the unit vector R pointing from the
center of the dipole toward the
observation point P ,
RRqqd ˆˆcos  pd
Where p = qd is called the dipole moment of the eletric dipole.
Therefore:
 dipoleelectric
4
ˆp
2
oR
R
V




Mahrim
(solution)
 In spherical coordinates,




















V
R
V
R
V
R
VE
sin
1ˆˆˆ
Upon taking the derivatives of the expression for V, with
respect to R and  and then substituting the results in the
above equation, we have:

Mahrim
(solution)
 Note that the expression for V and E
above equation as well as for the (electric
dipole) are only applicable when R>>d.
   mVR
R
qd
E
o
/sinˆcos2ˆ
4 3





Mahrim
(solution)
 To compute V and E at point in the vicinity
of the 2 charges making up the dipole, it
is necessary to perform the calculation
without resorting to the far-distance
approximations that led to eq:
2
4
cos
R
qd
V
o



Mahrim
(solution)
 Such an exact
calculation for E
leads to the
pattern shown:

Mahrim
Poisson’s and Laplace’s Equation
equationsLaplace'0
equationsPoisson'
2
2


V
V v


 Useful for determining the electrostatic
potential V in regions at whose boundaries
V is known, such as the region between
the plates of a capacitor with a specified
voltage difference across it.

Mahrim
Conductors
 Conductivity is represented by a symbol,
.
 The unit of conductivity is siemen per
meter, S/m.
 The conduction current density is
 
lawsOhm'theofformpoint-
/ 2
mAEJ



Mahrim
Eg 4-8, Conduction Current in a
Copper wire
 A 2-mm-diameter copper wire with
conductivity of 5.8 x 107 S/m is
subjected to electric field of 20
mV/m. Find
a) The current density
b) The current flowing in the wire

Mahrim
Eg 4-8: Ans
 
A
d
JJAIb
mAEJa
64.3
4
104
1016.1
4
)
/1016.11020108.5)
6
6
2
2637





 














Mahrim
Resistance
Linear resistor of cross section A and length l connected to a dc voltage source V.

Mahrim
Resistance
 The current flowing through the cross
section A at x2 is
 AAEsdEdsJI x
AA
  


Mahrim
Resistance
 The resistance of the linear resistor is
 
 71.4











S
l
S
l
sdE
ldE
sdJ
ldE
I
V
R
A
l
R







Mahrim
Resistance
 The reciprocal of R is called conductance
G, and the unit of G is (-1), or siemens
(S). For linear resistor,
 S
l
A
R
G


1

Mahrim
Eg 4-9: Conductance of Coaxial
Cable
 The radii of the inner and outer
conductors of a coaxial cable of
length l are a and b, respectively.
The insulation material has
conductivity . Obtain an expression
for G’, the conductance per unit
length of the insulation layer.

Mahrim
Eg 4-9: Ans
 Let I be the total current flowing from the inner conductor
to the outer conductor through the insulation material. At
any radial distance r from the axis of the center conductor,
the area through which the current flows is A=2rl.

Mahrim
Eg 4-9: Ans
rl
I
rE
EJ
rl
I
r
A
I
rJ
Hence



2
ˆ
fromand
2
ˆˆ
,







Mahrim
Eg 4-9: Ans
 In a resistor, the current flows from
higher electric potential to lower
potential. Hence, if J is in the r-
direction, the inner conductor must
be at a higher potential than the
outer conductor. Accordingly, the
voltage difference between the
conductors is

Mahrim
Eg 4-9: Ans
 
 mS
ablV
I
lRl
G
G
a
b
l
I
r
dr
rr
l
I
ldEV
ab
a
b
a
b
ab
/
ln
211'
'
thenislengthunitpereconductancThe
ln
2
ˆˆ
2

















  


Mahrim
Dielectric

Mahrim
Capacitance
 Any two conducting bodies,
regardless of their shapes and sizes,
When separated by insulating
(dielectric) medium - form a
capacitor.

Mahrim
Capacitance

Mahrim
Capacitance
 If a d-c voltage source is connected to the
conductors (Fig. 4-23) for two arbitrary
conductors, charge of equal and opposite
polarity is transferred to the conductors
surfaces.
 The surface of the conductor connected to
the positive side of the source will
accumulate charge +Q, and charge –Q will
accumulate on the surface of the other
conductor.

Mahrim
Capacitance
 When a conductor has excess
charge, it distributes the charge on
its surface in such a manner as to
maintain a zero electric field
everywhere within the conductor.
 This ensures that a conductor is an
equipotential body, meaning that the
electric potential is the same at
every two point in the conductor.

Mahrim
Capacitance
 Capacitance of a two-conductor capacitor
is defined as
 Where V is the potential (voltage)
difference between the conductor with
charge +Q and –Q. Capacitance is
measured in farads (F) or coulombs per
volt (C/V)
   105.4/ ForVC
V
Q
C 

Mahrim
Capacitance
 The normal component of E at any point
on the surface of either conductor is given
by
 
 surfaceconductorat
106.4ˆ

s
n EnE 

s – surface charge density at that point
n – outward normal unit vector at the same location
 - permitivity of dielectric medium separating the conductor

Mahrim
Capacitance
 The charge Q is equal to the integral of s
over surface S (Fig. 4-23):
 Where use was made of Eq. (4.106). The
voltage V is related to E by Eq. (4.39):
 107.4ˆ  
SSS
s sdEdsEndsQ



Mahrim
Capacitance
 108.4
1
2
12  
P
P
ldEVV

 Where P1 and P2 are any two points on
conductors 1 and 2, respectively.

Mahrim
Capacitance
 Substituting Eqs. (4.107) and (4.108) into
Eq. (4.105) gives
 Where l is the integration path from
conductor 2 to conductor 1.
   109.4F
ldE
sdE
C
l
S




 



Mahrim
Capacitance
 To avoid making sign errors when
applying Eq. (4.109), it is important
to remember that surface S is the
+Q surface and P1 is on S.
 Because E appears on both the
numerator and denominator of Eq.
(4.109), the of C obtained for any
specific capacitor configuration is
always independent of E.

Mahrim
Capacitance
 C depends only on the capacitor
geometry (sizes, shapes and relative
positions of the two conductors) and
the permittivity of the insulating
material.

Mahrim
Capacitance
 If the material between the
conductors is not a perfect dielectric
(i.e., if it has a small conductivity ),
then current can flow through the
material between the conductors,
and the material will exhibit a
resistance R.

Mahrim
Capacitance
 The general expression for R for a resistor of
arbitrary shape is given by
 For a medium with uniform  and , the product
of Eqs. (4.109) and (4.110) gives
 This simple relation allows us to find R if C is
known, or vice versa.
   71.4





S
l
sdE
ldE
R 


 111.4


RC

Reference
 Fawwaz T. Ulaby, “Fundamentals of
Applied Electromagnetics”
 Sadiku, “Elements of
Electromagnetics”
Mahrim

Electrostatics

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