Download Electromagnetic Fields

University of Tikrit
College of Engineering
Electrical Engineering Department
Electromagnetic Fields
Done By:
Mohammed Kamil Salh
‫م‬1023
‫ هـ‬2333
University of Tikrit
College of Engineering
Electrical Engineering Department
Class : Second
Theory : 2
Hrs. /Wk
Subject: Electromagnetic Fields
Tutorial : 1
Hrs. /Wk
Practical : -
Hrs. /Wk
Code : EEE207
Unit : 4
Article
Vectors Analysis:
Vector components, Unit vector, Vector algebra, Rectangular Coordinate , Cylindrical and Spherical
Coordinates
Coulombs Law and Electric Field Intensity:
Coulombs Law, Electric Field Intensity, Field of n Point Charge, Field Due to a Continuous Volume
Charge Distribution, Field of a Line Charge and a sheet charge , Streamlines and Sketches of Fields
Electric Flux Density, Gauss' Law, and Divergence:
Electric Flux Density, Gauss's Law, Applications of Gauss' Law, Divergence, Maxwell's First
Equation
Energy and Potential:
Energy and Potential in a Moving Point Charge in an Electric Field, Definition of Potential
Difference and Potential, Potential Field of a Point Charge, Potential Field of a System of Charges,
Potential Gradient, Dipole
Conductor , Dielectric and Capacitance:
Current and Current Density, Continuity of Current, Metallic Conductors, Conductor Properties and
Boundary Conditions, The Nature of Dielectric Materials, Boundary Conditions for Perfect
Dielectric Materials, Capacitance, Poisson's and Laplace's Equations, Examples of the Solution of
Laplace's Equation, Example of the Solution of Poisson's Equation
The Steady Magnetic Field:
Biot-Savart Law, Ampere's Circuital Law, Stokes' Theorem, Magnetic Flux and Magnetic Flux
Density, Inductance, Scalar and Vector Magnetic Potentials
Magnetic Forces, Materials and Inductance:
Force on a Moving Charge, Force on a Differential Current Element, Force Between Differential
Current Elements, Force and Torque on a Closed Circuit, Magnetization and Permeability, Magnetic
Boundary Conditions, The Magnetic Circuit
Time-Varying Fields and Maxwell's Equations:
Faraday's Law, Induced voltage, Displacement Current, Maxwell's Equations in Point Form,
Uniform Plane Wave:
Wave Equation, Wave Propagation in Free Space, Wave Propagation in Dielectrics, Wave
Propagation in Good Conductors
Textbook: 1- Engineering Electromagnetics , William H. Hayt, Published By Mcgraw-Hill,
2- Elements of Electromagnetics , Matthew N.O. Sadiku
Useful Web pages:
1- http://www.youtube.com/watch?v=686R2TWCIcI&list=PLEJFWUfvvpOFRKj7inlmgJrL1YW9j089r
2- https://www.facebook.com/groups/265743656900260/
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
1. Scalars and Vector
The term scalar refers to a quantity whose value may be represented by a single (positive
or negative) real number e.g. length, time, voltage etc. A scalar is represented simply by a letter
e.g. A, B and E, or by| | | |
| |.
A vector quantity has both a magnitude and a direction in space. Force, velocity and
electric field are examples of vectors. Each quantity is characterized by both a magnitude and a
direction. A vector is represented by a letter with an arrow on top of it, such as ⃗ ⃗
by a letter in boldface type such as A, B and E
1.1 Vector Addition and Subtraction
Two vectors A and B can be added together to give another vector C: i.e.
The vector addition is carried out component by component. Thus,
if ⃗
⃗
and ⃗
⃗
(
)
, then
(
(
)
)
Vector subtraction is similarly carried out as
⃗
⃗
(
)
(
Example: If ⃗
)
(
and ⃗⃗
)
. Find A+B and A-B?
Solution:
⃗
⃗⃗
(
)
(
)
(
)
⃗
⃗⃗
(
)
(
)
(
)
1
⃗ , or
DATE
/ /2013
Vector Analysis
Lecturer: Mohammed Kamil Salh
2. The Cartesian Coordinate System
In the Cartesian coordinate system we set up three coordinate axes mutually at right angles to
each other, and call them the x, y, and z axes. It is customary to choose a right-handed coordinate
system, in which a rotation (through the smaller angle) of the x axis into the y axis would cause a
right-handed screw to progress in the direction of the z axis. Figure 1.3 shows a right-handed
cartesian coordinate system.
A point is located by giving its x, y, and z coordinates. These are, respectively, the
distances from the origin to the intersection of a perpendicular dropped from the point to the x, y,
and z axes.
Figure 1.3 (a) A right-handed cartesian coordinate system. (b) The location of points P(1, 2, 3) and
Q(2, —2, 1). (c) The differential volume element in Cartesian coordinates
2
DATE
Vector Analysis
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Lecturer: Mohammed Kamil Salh
Also shown in Figure 1.3(c) are differential element in length, area, and volume. Notes
from the figure that in Cartesian coordinate:
1- Differential displacement is given by
2- Differential normal area is given by
3- Differential volume is given by
3. Vector Components and Unit Vectors
A vector ⃗ in Cartesian coordinates may be represented as ⃗
Where
,
and
.
are called the components of ⃗ in x, y, and z directions respectively. A
vector ⃗ has both magnitude and direction. The magnitude of ⃗ is scalar written as A or |⃗ |.
|⃗⃗ |
√
A unit vector
along A is defined as a vector whose magnitude is unity (i.e., 1) and its
direction is along⃗⃗⃗ , i.e.
| |
Notes that |
√
|
The vector between two points P(x0, y0, z0) and Q(x1, y1, z1) is given by:
⃗⃗⃗⃗⃗
(
)
(
)
(
)
3
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
Example: Find the vector between the points P (1, 4, 2) and Q (3, 1, 6)?
Solution: ⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
(
(
)
)
Example: If ⃗
(
(
)
)
(
(
)
)
and ⃗
, find (a) the component of A along
,
(b) the magnitude of A, (c) the magnitude of B, (d) the magnitude of 3A-B, (e) the
unit vector of A
Solution:
( )
| |
( ) ⃗⃗⃗⃗⃗
√
| |
( ) ⃗⃗⃗⃗⃗
√
( )
(
|
|
)
(
)
√
( )
| |
4. The Dot Product
The dot product of two vectors ⃗ and⃗⃗⃗ , written as ⃗ ⃗⃗ , is defined geometrically as the
dot product of the magnitude of ⃗ and the projection of ⃗ onto ⃗ (or vice versa).
Thus:
⃗ ⃗
|⃗ ||⃗ |
4
DATE
Vector Analysis
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where
Lecturer: Mohammed Kamil Salh
is the smaller angle between A and B. the result of A.B is called either scalar
product since it is scalar, or dot product due to sign.
If ⃗
and ⃗⃗
, then
⃗ ⃗⃗
note that
(
)
(
)
Example: The three vertices of a triangle are located at A(6, -1, 2), B(-2, 3, -4) and C(-3, 1, 5).
Find: (a) ⃗⃗
; (b) ⃗⃗
; (c) the angle
at vertex A ?
Solution:
⃗
(
)
(
(
))
(
(
)
(
(
))
(
)
⃗
⃗
)
⃗
|⃗
|
√
|⃗
|
√
⃗
⃗
⃗
⃗
(
|⃗
(
|| ⃗
)
( )
|
)
5
( )
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
5. The Cross Product
The cross product of two vectors A and B, written as A x B, is defined as
⃗
⃗
|⃗ ||⃗ |
Where
is a unit vector normal to the plane containing ⃗ and ⃗ . The direction of
is
taken as the direction of the right thumb when the fingers of the right hand rotate from A to B as
shown in Fig. 1.5(a). Alternatively, the direction of
is taken as that of the advance of a right-
handed screw as A is turned into B as shown in Fig. 1.5(b).
Ax B
Ax B
(a)
(b)
⃗
Figure 1.5: Direction of 𝐀
⃗𝐁
⃗ using (a) right hand rule (b) a right-handed screw
The vector multiplication of equation above is called cross product due to the cross sign;
it is also called vector product since the result is a vector.
If ⃗
and
⃗
⃗
, then
⃗
|
|
(
)
(
)
6
(
)
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
Note that
(
)
While
Example: If three points A(6, -1, 2), B(-2, 3, -4) and C(-3, 1, 5). Find ⃗
⃗
Solution:
⃗
(
)
(
(
))
(
(
)
(
(
))
(
)
⃗
⃗
)
⃗
|⃗
|
√
|⃗
|
√
⃗
⃗
|
⃗
⃗
(
⃗
⃗
|⃗
⃗
|
|
√
)
|
|
(
)
|⃗
|| ⃗
|
7
(
(
)
)
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
6. The Cylindrical Coordinate System
The circular cylindrical coordinate system is very convenient whenever we are dealing with
problems having cylindrical symmetry. A vector ⃗ in cylindrical coordinates can be written as:
⃗
Where
,
and
are unit vectors in the
, and z-directions as illustrated in Figure 1.6.
The magnitude of ⃗ is:
|⃗ |
√
A point P in cylindrical coordinates is represented as (
) and is as shown in Figure
1.6. Observe Figure 1.6 closely and note how we define each space variable:
the cylinder passing through P or the radial distance from the z-axis;
is the radius of
is (called the azimuthal
angle) measured from the x-axis in the xy-plane; and z is the same as in the Cartesian system.
Figure 1.6: Point P, unit vector, and differential element in cylindrical coordinates
Notice that the unit vectors
system is orthogonal;
increasing , and
,
and
are mutually perpendicular since our coordinate
points in the direction of increasing
in the positive z-direction. Thus
8
,
, in the direction of
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
Note Also from Figure 1.6 that in cylindrical coordinate, differential element can be found:
(i) Differential displacement is given by:
𝑑𝐿
𝑑𝜌
𝑑𝐿
𝑑𝑧
𝑧
Ring 𝑑𝐿
𝜌𝑑
𝜌
(ii) Differential normal area is given by:
𝜌𝑑 𝑑𝜌𝐚𝒛
(iii) Differential volume is given by:
𝜌𝑑 𝑑𝑧𝐚𝝆
𝑑𝜌𝑑𝑧 𝐚
The distance between two points in cylindrical coordinate P1(
given by
√
(
)
(
)
9
) and P2(
) is
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
 Cartesian to Cylindrical Coordinate Transformation
The relationships between the variables (x, y, z) of the Cartesian coordinate system and
those of the cylindrical system (
) are easily obtained as
√
In matrix form, we have transformation of vector ⃗⃗⃗
from Cartesian coordinate ⃗⃗⃗
⃗
to cylindrical coordinate
|
|
|
as
| |
|
 Cylindrical to Cartesian Coordinate Transformation
The relationships between the variables (
) of the cylindrical coordinate system and
those of the Cartesian system (x, y, z) are easily obtained as
In matrix form, we have transformation of vector ⃗⃗⃗
From cylindrical coordinate ⃗
⃗⃗⃗
To Cartesian coordinate
|
|
|√
|
√
√
√
as
|
|
|
|
10
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
Example: Given point P(-2, 6, 3) and vector ⃗⃗⃗
(
)
, express P and ⃗⃗⃗
cylindrical coordinate. Evaluate ⃗⃗⃗ at P in Cartesian and cylindrical system?
Solution:
The vector A in Cartesian coordinate at P is:
⃗⃗⃗
(
)
√
| |
The point P in cylindrical coordinate is:
√
√
P(6.324 , 108.430, 3)
The vector A in cylindrical coordinate is:
|
|
|
| |
|
|
|
| |
|
|
|
|
|
| |
|
|
11
|
in
DATE
| |
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
√
7. The Spherical Coordinate System
The Spherical coordinate system is most appropriate when dealing with problems having
spherical symmetry. A vector ⃗ in spherical coordinates can be written as:
⃗
Where
,
and
are unit vectors in the
, and
directions.
The magnitude of ⃗ is:
|⃗ |
√
A point P in spherical coordinates is represented as (
) and is illustrate in Figure
1.7(a). From this Figure, we notice that r is defined as the distance from the origin to the point P
or the radius of a sphere centred at the origin and passing through P;
z-axis and the position vector of P; and
is the angle between the
is measured from the x-axis
(a)
(b)
Figure 1.7: Spherical coordinate system (a) point P and unit vector (b)
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DATE
Vector Analysis
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Notice that the unit vectors
system is orthogonal;
, and
,
and
Lecturer: Mohammed Kamil Salh
are mutually perpendicular since our coordinate
points in the direction of increasing r,
, in the direction of increasing
in the direction of increasing . Thus
From Figure 1.7(b), we note that in spherical coordinate, differential element can be found:
(iv)
Differential displacement is given by:
𝑟𝑑𝜃𝒂𝜽
𝑑𝑟𝐚𝒓
𝑟
(v) Differential normal area is given by:
𝜃𝑑 𝐚
𝑟
𝜃 𝑑𝜃𝑑 𝐚𝒓
𝑟𝑑𝑟 𝑑𝜃𝐚
(vi) Differential volume is given by:
The distance between two points in spherical coordinate P1(
) and P2(
) is given
by
√
(
13
)
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
 Cartesian to Spherical Coordinate Transformation
The relationships between the variables (x, y, z) of the Cartesian coordinate system and
those of the cylindrical system (
) are easily obtained as
√
√
In matrix form, we have transformation of vector ⃗⃗⃗ from Cartesian coordinate
⃗⃗⃗
|
to spherical coordinate ⃗
|
|
| |
as
|
Figure 1.8 shows the relation between space variables
Figure 1.8: The relation between space variables (x, y, z), (𝑟 𝜃
14
) and (𝑟
𝑧)
DATE
Vector Analysis
/ /2013
Lecturer: Mohammed Kamil Salh
 Spherical to Cartesian Coordinate Transformation
The relationships between the variables (
) of the spherical coordinate system and
those of the Cartesian system (x, y, z) are easily obtained as
In matrix form, we have transformation of vector ⃗⃗⃗
⃗
from spherical coordinate
to Cartesian coordinate ⃗⃗⃗
√(
|√
|
|
)(
√(
√
|
as
)
)(
)
√
|
|
√
|
√
√
|
√
Example: Express ⃗
in Cartesian coordinate? Find ⃗ at (-3, 4, 0)?
Solution:
√
|√
|
|
)(
√(
√
)(
√(
|
)
√
√(
)(
√
|
|√
√
|
√
√
√
)
)
15
√
|
DATE
Vector Analysis
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√
)(
√(
)
√
√
(
√
)
16
√
Lecturer: Mohammed Kamil Salh
DATE
/ /2013
Vector Analysis
Lecturer: Mohammed Kamil Salh
Homework
Q1: Find the distance between (5,
, 0) to (5, , 10) in cylindrical coordinate?
√
Ans:
Q2: Obtain the expression for the volume of a sphere of radius a, from the
differential volume?
Q3: Obtain the expression for the volume and surface area of a cylindrical of radius
b and height h, from the differential volume?
Q4: Using spherical coordinates to write the differential surface areas dS1 and dS2
and then integrate to obtain the surface marked 1 and 2 in Figure below?
Ans:
Q5: Express the unit vector directed toward the origin from an arbitrary point on
the line described by x=0, y=3?
Ans:
Q6: Find the angle between ⃗
and ⃗
√
using both dot
Ans: 161.5o
product and cross product?
17
DATE
Electrostatic Fields
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Lecturer: Mohammed Kamil Salh
COULOMB’S LAW AND ELECTRIC FIELD INTENSITY
‫قانىن كىلىم وشدة المجال الكهربائي‬
1. The Experimental Law of Coulomb
Coulomb stated that “The force between two very small objects separated in a vacuum
or free space by a distance which is large compared to their size is proportional to the charge
on each and inversely proportional to the square of the distance between them”.
‫ " القوة بين جسمين صغيرين جدا يفصمهما في الفراغ أو الفضاء الحر مسافة كبيرة بالنسبة لمقاييسها‬:‫قانون كولوم‬
."‫تتناسب طرديا مع الشحنة عمى كل منهما وتتناسب عكسيا مع مربع المسافة بينهما‬
Where:
F: Force in newton (N),
Q1 and Q2 are the positive or negative quantities of charge in Coulomb(C),
R: is the separation in meters (m)
is called the permittivity of free space and has the magnitude, measured in farads per
meter (F/m)
F/m
The coulomb is an extremely large unit of charge, for the smallest known quantity of
charge is that of the electron (negative) or proton (positive), given in mks units as 1.602 x 10 -19
C; hence a negative charge of one coulomb represents about 6 x 1018 electrons.
18
DATE
Electrostatic Fields
/ /2013
Lecturer: Mohammed Kamil Salh
If point charges Q1 and Q2 are located at points having position vector r1 and r2, then the
vector force F12 on Q2 duo to Q1, shown in Figure 2.1, is given by
| |
⃗
⃗
|⃗
|
If Q1 located at (x0, y0, z0) and Q2 at (x1, y1, z1), then
⃗
|⃗
|
(
)
√(
)
⃗
|⃗
(
)
(
(
)
)
|
√(
|
|
(
(
)
(
)
)
)
(
)
(
)
(
)
(
[(
)
[(
[(
(
)
)
)
(
(
)
(
)
)
(
) ] √(
)
(
) ]
(
19
) ]
)
(
)
(
)
(
)
DATE
Electrostatic Fields
/ /2013
Lecturer: Mohammed Kamil Salh
Example: Find the force on Q1 (20μC) duo to charge Q2 (-300μC), where Q1 located at (0, 1, 2)
and Q2 at (2, 0, 0)?
Solution:
(
|⃗
)
(
)
(
)
√
|
(
|
| |
)
|
| |
√
2. The Electric Field Intensity (E)
)‫(شدة المجال الكهربائي‬
If we now consider one charge fixed in position, say Q1, and move a second charge slowly
around, we note that there exists everywhere a force on this second charge; in other words, this
second charge is displaying the existence of a force field. Call this second charge a test charge
Qt. The force on it is given by Coulomb's law,
|
|
Writing this force as a force per unit charge gives
|
|
( )
20
DATE
Electrostatic Fields
/ /2013
Lecturer: Mohammed Kamil Salh
The quantity on the right side of (*) is a function only of Q2 and the directed line segment
from Q2 to the position of the test charge. This describes a vector field and is called the electric
field intensity.
Using a capital letter E for electric field intensity, we have finally
⃗
| |
Where ⃗ is electric field intensity measured in v/m
Let us arbitrarily locate Q1 at the center of a spherical coordinate system. The unit vector
aR then becomes the radial unit vector ar, and R is r. Hence
⃗
| |
Example: Find the electric field intensity (E) at (0, 2, 3) due to a point charge Q (0.4µC) located
at (2, 0, 4)?
Solution:
(
| |
)
)
(
)
√
| |
| |
(
| |
√
21
DATE
Electrostatic Fields
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Lecturer: Mohammed Kamil Salh
3. Field of n Point Charge
Since the coulomb forces are linear, the electric field intensity due to n point charges, Q1 at
r1 , Q2 at r2,and Qn at rn is the sum of the forces on Qt caused by Q1 and Q2 acting alone, or
⃗
⃗
⃗
⃗
|
|
|
|
|
|
Example: A charge of -0.3µC is located at (25, -30, 15) (in cm), and a second charge of 0.5µC is
at (-10, 8, 12) cm. Find E at: (a) the origin; (b) (15, 20, 50) cm
Solution:
( )
|
|
(
)
(
(
)
(
√
(
|
|
)
|
|
22
)
)
DATE
Electrostatic Fields
/ /2013
(
Lecturer: Mohammed Kamil Salh
)
(
)
( )
(
)
(
|
)
)
(
(
))
(
)
)
(
)
|
(
|
(
(
))
(
|
[
(
)
(
[
]
)
]
23
DATE
Electrostatic Fields
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Lecturer: Mohammed Kamil Salh
4. Field Due to a Continuous Volume Charge Distribution
If we now visualize a region of space filled with a great number of charges separated by
minute distances, we see that we can replace this distribution of very small particles with a
smooth continuous distribution described by a volume charge density (
) ⁄
.
‫ فبوىب وستطٍع‬, ‫ارا تصىسوب مىطقت مه انفشاغ ممهىءة بعذد هبئم مه انشذىبث انمىفصهت عه بعضهب بمسبفبث صغٍشة جذا‬
‫ادالل هزا انتىصٌع نجسٍمبث صغٍشة جذا بتىصٌع امهس ٌىصف بكثبفت شذىت دجمٍت‬
The total charge within some finite volume is obtained by integrating throughout that
volume,
∫
⃗
∫
Example: Calculate the total charge within each of the indicate volumes:
(a)
;
(b)
,
(c) Universe
,
;
⁄
;
Solution:
( )
∫
∭
24
DATE
Electrostatic Fields
/ /2013
∬
[
[∫
]
[
]
[ ]
]
[
∬
(
Lecturer: Mohammed Kamil Salh
[∫
)
(
)
][
]
(
)
(
(
)
)
](
)
(b)
∫
(
∫∫∫
(
∫∫∫
)
∫∫[
(
∫∫
[
)
]∫
]
)
(
∫[ ]
(
)
[
(
][
][
)]
(c)
∫
∫ ∫∫
] [
[
[
] [ ][
] [ ]
[
]
25
]
)
[
]
DATE
Electrostatic Fields
/ /2013
Lecturer: Mohammed Kamil Salh
5. Field of a Line Charge
If we now consider a filament like distribution of volume charge density, such as a very
fine, sharp beam in a cathode-ray tube or a charged conductor of very small radius, we find it
convenient to treat the charge as a line charge of density
C/m.
‫ أو‬, ‫ مثم دضمت دقٍقت جذا ومشكضة فً اوبىوت اشعت انكبثىد‬,ً‫ارا اعتبشوب كثبفت شذىت دجمٍت عهى هٍئت تىصٌع فتٍه‬
.
C/m ‫ فبوىب وجذ اوت نمه انمىبسب معبمهت انشذىت كخط شحنة ري كثبفت‬,‫مىصال مشذىوب ارا كبن وصف قطشي شذٌذ انصغش‬
Let us assume a straight line charge extending along the z axis from -∞ to ∞, as shown in
Figure. We desire the electric field intensity E at any and every point resulting from a uniform
line charge density
∫
⃗
(
𝑧)
∫
(𝜌
(
)
√
(
∫
(
[
[∫
(
E
)
)
∫
| |
[∫
)
(
(
(√
) ]
)
⁄
(
∫
(
(
) )
∫
⁄
𝑘)
[
(
(
)
26
) ]
⁄
]
)
)
⁄
]
DATE
∫
Electrostatic Fields
/ /2013
(
)
⁄
(
∫
(
)
∫
∫
⁄
(
)
)
(
)
[
[
)
)
∫
⁄
(
∫
[
(
(
⁄
∫
∫
Lecturer: Mohammed Kamil Salh
]
]
⁄
)
⁄
)
[
))
∫
(
∫(
⁄
(
∫(
]
√
√
)
√
]
⃗
⃗
(
)
27
(
)
(
((
)
)
(
(
)
) )
DATE
Electrostatic Fields
/ /2013
Lecturer: Mohammed Kamil Salh
Example: A uniform line charge of 16nC/m is located along the z-axis: (a) find E at P(1, 2, 3)
(b) Q(1, 2, 7) ?
Solution:
(a)
√
√
√
√
(
( )
)
√
√
(
)
√
√
Example: Two uniform line charges of
nC/m each are parallel to the x axis, one at z = 0,
y = -2 m and the other at z = 0, y = 4m. Find E at (4, 1, 3) m?
Solution:
z
(
((
)
)
(
(
)
) )
(
(
(
))
( )
(
(
E2
)
) )
E1
(4,1,3)
(x,-2,0)
(
((
)
)
(
(
)
) )
-2
x
28
(x,4,0)
4
y
DATE
Electrostatic Fields
/ /2013
Lecturer: Mohammed Kamil Salh
6. Field of a Sheet Charge
Another basic charge configuration is the infinite sheet of charge having a uniform density
of
C/m2. Such a charge distribution may often be used to approximate that found on the
conductors of a strip transmission line or a parallel-plate capacitor. Where static charge resides
on conductor surfaces and not in their interiors; for this reason,
is commonly known as
surface charge density.
‫ ومثم هزا انتىصٌع نهشذىت قذ‬. ‫ هى شكم اسبسً اخش نهشذىبث‬,
C/m2 ‫ رو انكثبفت انمىتظمت‬,‫انهىح االوهبئً نهشذىت‬
‫ دٍث ان انشذىت االستبتٍكٍت تستىطه اسطخ‬. ‫ٌستخم كثٍشا نتقشٌب رنك انمىجىد عهى انمكثف ري انهىدٍه انمتىاصٌٍه‬
.‫تعشف عبمت بكثبفت انشذىت انسطذٍت‬
‫انمىصالث ونٍس فً داخههب ونهزا انسبب فبن‬
Consider an infinite sheet charge in the xy-plane with uniform density
. The charge
associated with elemental area dS is
E
∫
(
⃗
)
∫
𝜌𝑠
| |
∫ ∫
(√
)
√
(𝜌
√
∫ ∫
(
)
(
)
∫
∫ (
)
29
)
DATE
Electrostatic Fields
/ /2013
[
]
√
[
Lecturer: Mohammed Kamil Salh
√
]
⃗
Where aN is a unit vector normal to the sheet
Example: Three infinite uniform sheets of charge are located in free space as follows: 3nC/ m2 at
z = -4, 6nC/m2 at z = 1, and - 8nC/m2 at z = 4. Find E at the point: (a) PA(2, 5, -5); (b)
PB(4, 2, -3); (c) Pc(-1, -5, 2); (d) PD(-2, 4, 5)?
Solution:
z
⃗
4
[
]
𝜌𝑠
𝑛
P c(-1,-5,-3)
1
b[
P D(-2, 4, 5)
𝜌𝑠
𝑛
y
]
P B(4,2, -3)
x
c-
-4
[
]
[
]
𝜌𝑠
𝑛
P A (2,5, -5)
30
DATE
Electrostatic Fields
/ /2013
Example: The finite sheet
Lecturer: Mohammed Kamil Salh
on the z=0 plane has a charge density
(
) nC/m2. Find:
a- The electric field (E) at (0, 0, 5)?
b- The force experienced by a -1 mC charge located at (0, 0, 5)?
Solution:
E
⃗
z
∫
(0,0,5)
| |
1
√
∫∫
(x,y,0)
(
1
)
(√
∫∫
√
)
(
)
(
(
] [
)
)
[ ∫∫
∫∫
]
[
( )
(
)
∫∫
[ [
x
[
∫∫
] [
]
[
]
] [
[
(
)
31
]
]
]
]
y
DATE
Electrostatic Fields
/ /2013
nC/m2 is located at z = 5 m and a uniform line
Example: A uniform sheet charge with
charge with
Lecturer: Mohammed Kamil Salh
nC/m at z =-3 m, y = 3m. Find E at (x, -1, 0)m.?
Solution:
E1 due to a surface charge
1
𝜌𝑠
E2 due to a line charge
3
(
)
(
(
((
(
(
)
)
)
-3
)
(
(
(
((
x
)
) )
)
)
(
(
(
(
))
)) )
)
32
𝜌𝐿
y
DATE
Electrostatic Fields
/ /2013
Lecturer: Mohammed Kamil Salh
Home work
Q1: Let a point charge Q1 = 25 nC be located at P1(4, -2, 7) and a charge Q2 = 60 nC be at
P2(-3, 4, -2). (a) Find E at P(1, 2, 3). (b) At what point on the y-axis is Ex = 0?
Q2: A circular ring of radius a carries a charge Q C and is placed on the xy-plane, find electric
field E at (0, 0, h)
Q3: The circular disk
in the z = 0 plane has a charge density
.Determine the
electric field E at the point (0, 0, h)?
Q4: Two infinite sheets of uniform charge density
C/m2 are located at z = -5 m
and y = - 5 m. Determine the uniform line charge density pL necessary to produce the same
value of E at (4,2,2) m, if the line charge is located at z = 0, y = 0.
Ans. 0.667 nC/m
Q5: A uniform line charge of
√
C/m lies along the x axis and a uniform sheet of
charge is located at y = 5 m. Along the line y = 3 m, z = 3 m the electric field E has only az
component. What is
for the sheet?
Ans. 125 pC/m2
33
DATE
/ /2013
ELECTRIC FLUX DENSITY
Lecturer: Mohammed Kamil Salh
ELECTRIC FLUX DENSITY, GAUSS'S LAW, AND DIVERGENCE
1. Electric Flux Density (D) (‫)كثافة التدفق الكهربائي‬
Michael Faraday had a pair of concentric metallic spheres constructed, the outer one consisting
of two hemispheres that could be firmly clamped together. He also prepared shells of insulating
material (dielectric material) which would occupy the entire volume between the concentric
spheres
His experiment, then, consisted essentially of the following steps:
1. With the equipment dismantled, the inner sphere was given a known positive charge.
2. The hemispheres were then clamped together around the charged sphere with about 2 cm of
dielectric material between them.
3. The outer sphere was discharged by connecting it momentarily to ground.
4. The outer space was separated carefully, using tools made of insulating material in order not
to disturb the induced charge on it, and the negative induced charge on each hemisphere was
measured.
Faraday found that the total charge on the outer sphere was equal in magnitude to the
original charge placed on the inner sphere and that this was true regardless of the dielectric
material separating the two spheres. He concluded that there was some sort of "displacement"
from the inner sphere to the outer which was independent of the medium, and we now refer to
this flux as displacement, displacement flux, or simply electric flux.
Faraday's experiments also showed, of course, that a larger positive charge on the inner
sphere induced a correspondingly larger negative charge on the outer sphere, leading to a direct
proportionality between the electric flux and the charge on the inner sphere
Where
(psi) is electric flux in coulombs C
We can obtain more quantitative information by considering an inner sphere of radius a
and an outer sphere of radius b, with charges of Q and — Q, respectively (Fig. 3.1). The paths of
34
DATE
ELECTRIC FLUX DENSITY
/ /2013
electric flux
Lecturer: Mohammed Kamil Salh
extending from the inner sphere to the outer sphere are indicated by the
symmetrically distributed streamlines drawn radially from one sphere to the other.
Figure 3.1: The electric flux in the region between a pair of charged concentric sphere
At the surface of the inner sphere,
coulombs of electric flux are produced by the charge
Q(= ) coulombs distributed uniformly over a surface having an area of 4πa2 m2. The density of
/4πa2 or Q/4πa2 C/m2, and this is an important new quantity.
the flux at this surface is
Referring again to Fig. 3.1, the electric flux density is in the radial direction and has a
value of
⃗(
)
(
)
⃗(
)
(
)
and at a radial distance r, where a ≤ r ≤ b
⃗
If we now let the inner sphere become smaller and smaller, while still retaining a charge of
Q, it becomes a point charge in the limit, but the electric flux density at a point r meters from the
point charge is still given by
⃗⃗
⃗⃗
⃗
(
)
35
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
Example: Determine D at (4, 0, 3) if there is a point charge -5π mC at (4, 0, 0) and a line charge
3π mC/m along the y-axis?
Solution:
|
|
(
|
)
(
)
(
)
|
(
)
(
)
36
DATE
/ /2013
ELECTRIC FLUX DENSITY
Lecturer: Mohammed Kamil Salh
2. Gauss’s Law
These generalizations of Faraday's experiment lead to the following statement, which is
known as Gauss's law:
“The electric flux passing through any closed surface is equal to the total charge
enclosed by that surface”
"‫"التدفق الكهربي المار خالل اي سطح مغمق يساوي الشحنة الكمية المحتواة بذلك السطح‬: ‫قانون جاوس‬
∮
"‫توضع دائرة صغيرة عمى عالمة التكامل لتشير الى ان التكامل مؤدى عمى سطح مغمق ويسمى هذا السطح ب "سطح جاوس‬

The charge enclosed might be several point charges, in which case
∑

or a line charge,
∫

or a surface charge,
∫

or a volume charge distribution,
∫
The last form is usually used, and we should agree now that it represents any or all of the
other forms. With this understanding Gauss's law may be written in terms of the charge
distribution as
∮
∫
37
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
3. Applications of Gauss's Law
The procedure for applying Gauss's law to calculate the electric field involves first
knowing whether symmetry exists. Once symmetric charge distribution exists, we construct a
mathematical closed surface (known as a Gaussian surface). The surface over which Gauss's law
is applied must be closed, but it can be made up of several surface elements. Thus the defining
conditions of a special Gaussian surface are
a- The surface is closed.
b-
At each point of the surface D is either normal or tangential to the surface, so that Ds. dS
becomes either DsdS or zero, respectively
c- D is sectional constant over that part of the surface where D is normal.
3.1 Symmetrical Charge Distributions:
3.1.1 Point Charge:
Suppose a point charge Q is located at the origin. To determine D at a point P, it is easy to
see that choosing a spherical surface containing P will satisfy symmetry conditions. Thus, a
spherical surface centered at the origin is the Gaussian surface in this case and is shown in
Figure 3.2.
∮
∮
∫ ∫
Figure 3.2
38
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
3.1.2 Infinite Line Charge
Suppose the infinite line of uniform charge
C/m lies along the z-axis. To determine D at
a point P, we choose a cylindrical surface containing P to satisfy symmetry condition as shown
in Figure 3.3. D is constant on and normal to the cylindrical Gaussian surface; i.e., D =
we apply Gauss's law to an arbitrary length L of the line
∮
∮
∫ ∫
∫
∫
39
. If
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
3.1.3 Uniformly Charged Sphere
Consider a sphere of radius a with a uniform charge
C/m3. To determine D everywhere,
we construct Gaussian surfaces for cases r < a, and r > a separately. Since the charge has
spherical symmetry, it is obvious that a spherical surface is an appropriate Gaussian surface.
For r < a, the total charge enclosed by the spherical surface of radius r, as shown in
Figure 3.4 (a), is
∮
∮
∫
∫
∫
∫
∫ ∫
(Gauss’s law)
Figure 3.4-a
For r ≥ a, the Gaussian surface is shown in Figure 4.3(b). The charge enclosed by the surface is
the entire charge in this case, i.e.,
∫
∫
∫ ∫
(Gauss’s law)
(
)
Figure 3.4-b
40
DATE
ELECTRIC FLUX DENSITY
/ /2013
Example: A uniform line charge of
C/m lies along the z axis, and a concentric circular
C/m2. Both distributions are infinite in extent
cylinder of radius 2 m has
with z. Use Gauss's law to find D in all regions?
Solution:
Using Gaussian surface cylinder
∫ ∫
∫
Lecturer: Mohammed Kamil Salh
∫
∫
∫ ∫
41
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
Example: A point charge Q=2 c at the origin , a volume charge density of 4 c/m3 at the region
and a sheet of charge -6 c/m2 has
, Find D at
?
Solution:
4
D at
3
D is due to a point charge
(
)
Q
D at
𝜌𝑠
1
𝜌𝑣
r
Gaussian surface
∫
∫
∫ ∫
* +
( )
D at
* +
∫
∫
∫
( )
( )
42
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
3.2 Differential Volume Element
We are now going to apply the methods of Gauss's law to a slightly different type of
problem, one which does not possess any symmetry at all. At first glance it might seem that our
case is hopeless, for without symmetry a simple Gaussian surface cannot be chosen such that
the normal component of D is constant or zero everywhere on the surface. Without such a
surface, the integral cannot be evaluated. There is only one way to circumvent these difficulties,
and that is to choose such a very small closed surface that D is almost constant over the surface,
and the small change in D may be adequately represented by using the first two terms of the
Taylor's-series expansion for D. The result will become more nearly correct as the volume
enclosed by the Gaussian surface decreases, and we intend eventually to allow this volume to
approach zero.
(
)
The expression is an approximation which becomes better as
becomes smaller.
Example: Find an approximate value for the total charge enclosed in an incremental volume of
10-9 m3 located at the origin, if D = e-x siny ax - e-x cosy ay + 2zaz C/m2.
Solution:
,
,
At the origin, the first two expressions are zero, and the last is 2. Thus, we find that the charge
enclosed in a small volume element there must be approximately 2
.If
is 10-9m3, then we
have enclosed about 2nC.
4. Divergence (div)
There are two main indicators of the manner in which a vector field changes from point to
point throughout space. The first of these is divergence, which will be examined here. It is a
scalar and bears a similarity to the derivative of a function. The second is curl.
43
DATE
/ /2013
ELECTRIC FLUX DENSITY
Lecturer: Mohammed Kamil Salh
When the divergence of a vector field is nonzero, that region is said to contain sources or
sinks, sources when the divergence is positive, sinks when negative. In static electric fields there
is a correspondence between positive divergence, sources, and positive electric charge Q.
Electric flux  by definition originates on positive charge. Thus, a region which contains
positive charges contains the sources of . The divergence of the electric flux density D will be
positive in this region. A similar correspondence exists between negative divergence, sinks, and
negative electric charge
∮
(
(
(
)
(
)
)
)
5. Maxwell's First Equation (Electrostatics)
We now wish to consolidate the gains of the last two sections and to provide an
interpretation of the divergence operation as it relates to electric flux density. The expressions
developed there may be written as
∮
44
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
This is the first of Maxwell's four equations as they apply to electrostatics and steady
magnetic fields, and it states that the electric flux per unit volume leaving a vanishingly small
volume unit is exactly equal to the volume charge density there. This equation is called the
point form of Gauss's law. Gauss's law relates the flux leaving any closed surface to the charge
enclosed,
⃗⃗⃗
Example: in the region
and for
⃗⃗⃗
for
⃗⃗⃗
(
(
)
,
)
,
find
in all three regions?
Solution:
(
)
(
(
))
(
(
(
)
))
45
(
)
(
)
(
)
DATE
ELECTRIC FLUX DENSITY
/ /2013
Example: Let⃗⃗⃗
find
Lecturer: Mohammed Kamil Salh
m and ⃗⃗⃗
, for
at r =0.06m, (b) find
, for
at r =0.1m, (c) what surface charge density could be
located at r =0.08m to caused D=0 for
?
Solution:
( )
⃗⃗⃗
(
(
)
)
(
)
( )
⃗⃗⃗
(
)
( )
∫
∫
∫
∫
∫ ∫
∫ ∫
* +
∫
m, (a)
(
∫
46
)
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
6. The Divergence Theorem
Gauss' law states that the closed surface integral of D. dS is equal to the charge enclosed. If
the charge density function v is known throughout the volume, then the charge enclosed may be
obtained from an integration of v throughout the volume. Thus,
∮
∫
But
and so
∮
∫ (
)
This is the divergence theorem, also known as Gauss' divergence theorem. Of course, the
volume v is that which is enclosed by the surface s.
(
Example: Given that
)
, (C/m2) in cylindrical coordinates, evaluate both sides
of the divergence theorem for the volume enclosed by = 1m, = 2m, z = 0 and z = 10
Solution:
The left side of divergence theorem is:
∮
∫ ∫
(
)
The right side is:
∫ (
)
∫ (
∫ (
)
(
∫ (
))
*
(
∫ (
)
+
47
))
∫ ∫
∫
DATE
ELECTRIC FLUX DENSITY
/ /2013
Example: Given that⃗⃗⃗
Lecturer: Mohammed Kamil Salh
, evaluate both sides of the divergence
theorem for the volume enclosed by the shell r = 2?
Solution:
The left side of divergence theorem is:
∮
∮(
∮
∫
)
( )
∫
[
]
[
∫ (
]
The right side is:
∫ (
)
(
∫ (
(
∫ (
)
(
))
∫ (
)
∫ (
[
))
)
∫
∫ ∫ (
∫
∫ ∫
[
] [
][
)
∫
] [ ]
][
∫ ∫
[ ] [
] [ ]
]
48
)
DATE
ELECTRIC FLUX DENSITY
/ /2013
Lecturer: Mohammed Kamil Salh
Home work
(
Q1: In cylindrical coordinates, let
)
. Find D everywhere?
Q2: Spherical surfaces at r = 2, 4, and 6 m carry uniform surface charge densities of 20 nC/m2,
−4 nC/m2, and
respectively. (a) Find D at r = 1,3, and 5 m. (b) Determine
such that
D = 0 at r = 7 m.
Q3: Let a 50-cm length of coaxial cable having an inner radius of 1 mm and an outer radius of 4
mm. The total charge on the inner conductor is 30 nC. Find the charge density on each
conductor, and the E and D fields?
Q4: The volume in cylindrical coordinates between
= 2 m and = 4 m contains a uniform
charge density p (C/m3). Use Gauss' law to find D in all regions.
Q5: Given the flux density
(
)
, use two different methods to find the total
charge within the region 1 < r < 2 m, 1 < < 2 rad, 1 < < 2 rad.
Q6: A point charge Q is at the origin, show that
Q7: Show that
everywhere?
for the field of infinite sheet charge?
49
DATE
/ /2013
ENERGY AND POTENTIAL
Lecturer: Mohammed Kamil Salh
ENERGY AND POTENTIAL
‫الطاقة والجهد‬
4.1 Energy Expended In Moving a Point Charge in an Electric Field
The electric field intensity was defined as the force on a unit test charge at that point at
which we wish to find the value of this vector field. If we attempt to move the test charge against
the electric field, we have to exert a force equal and opposite to that exerted by the field, and this
requires us to expend energy, or do work. If we wish to move the charge in the direction of the
field, our energy expenditure turns out to be negative; we do not do the work, the field does
Suppose we wish to move a charge Q a distance dL in an electric field E. The force on Q
due to the electric field is
where the subscript reminds us that this force is due to the field. The component of this force in
the direction dL which we must overcome is
where
= a unit vector in the direction of dL
The force which we must apply is equal and opposite to the force due to the field
and our expenditure of energy is the product of the force and distance. That is,
Differential work done by external source moving Q
Returning to the charge in the electric field, the work required to move the charge a finite
distance must be determined by integrating
∫
where the path must be specified before the integral can be evaluated. The charge is assumed to
be at rest at both its initial and final positions.
50
DATE
ENERGY AND POTENTIAL
/ /2013
Lecturer: Mohammed Kamil Salh
Example: An electrostatic field is given by E = (x/2 + 2y) ax + 2x ay (V/m). Find the work done
in moving a point charge Q = -20 C (a) from the origin to (4, 0, 0) m, and (b) from
(4, 0, 0) m to (4, 2, 0) m?
Solution:
(a)
The first path is along x-axis, so that the
∫
(b)
∫ (
)
*
+

The second path is in the ay direction, so that the

[ ]
∫
Example: Find the work done in moving a point charge Q = 5 C from the origin to
(2 m, π/4, π/2) spherical coordinates, in the field
Solution:
∫
∫
⁄
∫

(
⁄
(
|
⁄
)
∫
⁄
|
)


⁄
| |
⁄

51
⁄
⁄
?
DATE
/ /2013
ENERGY AND POTENTIAL
Lecturer: Mohammed Kamil Salh
Example: uniform line charge lie along z-axis, determine the work expended in carrying Q
from a to b along:
(a)
Circular path?
(b)
Radial path?
Figure: (a) A circular path and (b) a radial path along which a charge of Q is carried in the field
of an infinite line charge.
Solution:
The electric field of a line charge is ⃗
(a)
for circular path
∫
(b)
for radial path
∫
∫
∫
52
DATE
ENERGY AND POTENTIAL
/ /2013
Lecturer: Mohammed Kamil Salh
4.2 The Line Integral
A line integral is like many other integrals which appear in advanced analysis, including
the surface integral appearing in Gauss's law, in that it is essentially descriptive. It tells us to
choose a path, break it up into a large number of very small segments, multiply the component of
the field along each segment by the length of the segment, and then add the results for all the
segments. This is a summation, of course, and the integral is obtained exactly only when the
number of segments becomes infinite.
For this special case of uniform electric field intensity, we should note that the work
involved in moving the charge depends only on Q, E, and LBA, a vector drawn from the initial to
the final point of the path chosen. It does not depend on the particular path we have selected
along which to carry the charge. We may proceed from B to A on a straight line or via the Old
Chisholm Trail; the answer is the same.
Example: Given the nonuniform field
, determine the work expended in
carrying 2C from B(l, 0, 1) to A(0.8, 0.6,1) along:
(a)
The shorter arc of the circle x2 + y2 = 1 z = 1?
(b)
The straight line path from B to A?
Solution:
∫ (
∫
(a)
∫
* √
∫
∫
+
)(
∫ √
* √
∫ √
+
53
)
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ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
(b) We start by determining the equations of the straight line
At point B
∫
∫
∫
∫
∫
(
)
4.3 Definition of Potential Difference and Potential
We are now ready to define a new concept from the expression for the work done by an
external source in moving a charge Q from one point to another in an electric field E,
∫
In much the same way as we defined the electric field intensity as the force on a unit test
charge, we now define potential difference V as the work done (by an external source) in
moving a unit positive charge from one point to another in an electric field,
∫
Potential difference is measured in joules per coulomb, for which the volt is defined as a
more common unit, abbreviated as V. Hence the potential difference between points A and B is
∫
VAB is positive if work is done in carrying the positive charge from B to A.
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ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
4.4 The Potential Field of a Point Charge
The potential difference between points located at r = rA and r = rB in the field of a point
charge Q placed at the origin
⃗
and
∫
∫
(
∫
)
The potential difference between two points in the field of a point charge depends only on
the distance of each point from the charge and does not depend on the particular path used to
carry our unit charge from one point to the other
How might we conveniently define a zero reference for potential? The simplest possibility
is to let V = 0 at infinity. If we let the point at r = rB recede to infinity the potential at rA becomes
(
)
Since there is no reason to identify this point with the A subscript,
This expression defines the potential at any point distant r from a point charge Q at the origin,
the potential at infinite radius being taken as the zero reference.
A convenient method to express the potential without selecting a specific zero reference entails
identifying rA as r once again and letting Q/4πorB be a constant. Then
C1 may be selected so that V = 0 at any desired value of r. We could also select the zero
reference indirectly by electing to let V be Vo at r = ro.
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ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
Example: Find the potential at rA = 5 m with respect to rB =15 m due to a point charge
Q =500 pC at the origin and zero reference at infinity?
Solution:
(
)
(
)
The zero reference at infinite must be used to find V5 and V15
(
)
(
( )
)
(
)
Example: A 15-nC point charge is at the origin in free space. Calculate V1 if point P1 is located
at P1 (-2, 3, -1) and: (a) V = 0 at (6, 5, 4); (b) V =0 at infinity; (c) V = 5 V at (2, 0, 4)?
Solution:
a-
√
√
56
DATE
ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
b- V=0 at infinity
√
c-
√
4.5 The Potential Field of a Line Charge
The potential difference between points located at = a and = b in the field of a point
charge Q placed at the origin
⃗
and
∫
∫
Example: For a line charge
∫
⁄ C/m on the z-axis, find VAB, where A is (2m, π/2, 0)
and B s (4m, π, 5m) ?
Solution:
57
DATE
ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
4.6 The Potential Field of a System of Charges
The potential field of a single point charge, which we shall identify as Q1 and locate at r1,
involves only the distance |r - r1| from Q1 to the point at r. For a zero reference at infinity, we
have
|
|
The potential due to two charges, Q1 at r1 and Q2 at r2, is a function only of |
|
|, the distances from Q1 and Q2 to the field point, respectively.
and|
|
|
|
|
If the charge distribution takes the form of a line charge, a surface charge, a volume charge
the integration is along the line or over the surface or volume:
∫
∫
∫
| |
| |
| |
Example: Five equal point charges, Q=2 nC, are located at x=2, 3, 4, 5, 6 m. Find the potential at
the origin?
Solution:
(
)
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DATE
ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
Example: Find the potential V at (0, 0, K) for cylindrical surface charge
and
?
z
𝑘
Solution:
∫
| |
,
| |
𝑎
𝜌𝑠
√
∫ ∫
y
√
∫
x
√
∫
√
[
(
)
(
[
)]
(
(
[
(
)
( )]
√
)
)
√(
(
(
)
( )
)
√
√
(
)
59
√( )
]
)
𝑧
DATE
ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
4.7 Gradient
The vector field
4.8
(also written grad V) is called the gradient of the scalar function V
Relationship Between E and V
The electric field intensity E may be obtained when the potential function V is known by
simply taking the negative of the gradient of V. The gradient was found to be a vector normal to
the equipotential surfaces, directed to a positive change in V. With the negative sign here, the E
field is found to be directed from higher to lower levels of potential V
Example: Given the potential field, V = 2x2 y − 5z, and a point P(-4, 3, 6), find at point P: the
potential V, the electric field intensity E, the direction of E, the electric flux density D,
and the volume charge density
?
Solution:
The potential at P is:
The electric field intensity E is :
(
)
60
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ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
E at the point P is:
(
)
Example: Given the potential field in cylindrical coordinates
m,
, and point P
, z=2 m, in free space find at point P: the potential V, the electric
field intensity E, the direction of E, the electric flux density D, and the volume charge
density
?
Solution:
The potential field V at the point P is:
(
(
(
)
)
(
)
61
(
)
)
DATE
ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
(
)
(
(
)
)
(
(
)
)
(
(
)
)
4.9 The Electric Dipole
An electric dipole, or simply a dipole, is the name given to two point charges of equal
magnitude and opposite sign, separated by a distance that is small compared to the distance to
the point P at which we want to know the electric and potential fields.
The dipole is shown in Figure 4.9a. The distant point P is described by the spherical
’s r, θ, and
, in view of the azimuthal symmetry. The positive and negative
point charges have separation d and rectangular coordinates (0, 0, ½ d) and (0, 0, -½ d),
respectively.
Figure 4.9: The geometry of an electric dipole
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ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
The distances from Q and −Q to P be R1 and R2, respectively, and write the total potential as
(
)
The final result is then
Using the gradient relationship in spherical coordinates
(
)
(
)
The potential field of the dipole may be simplified by making use of the dipole moment. We first
identify the vector length directed from −Q to +Q as d and then define the dipole moment as Qd
and assign it the symbol p. Thus
Because
, we then have
This result may be generalized as
̀
|
̀|
where r locates the field point P, and ̀ determines the dipole center
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ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
Example: A dipole 1nC at ( 0 , 0 , 0.1) and -1nC at ( 0, 0 , -0.1) . (a) Calculate V and E at
(0.3, 0, 0.4)?
Solution:
√
√
√
Example: An electric dipole located at the origin in free space has a moment p = 3ax − 2ay + az
nC.m , find V at PA(2, 3, 4)?
Solution:
̀
|
̀|
̀
(
) (
|
)
|
(√
64
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ENERGY AND POTENTIAL
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Lecturer: Mohammed Kamil Salh
Example: A dipole of moment p = 6az nC.m is located at the origin in free space. (a) Find V at
P(r = 4,
= 20◦, = 0◦). (b) Find E at P ?
Solution:
Home work
Q1: Find V on the z axis for a uniform line charge
in the form of a ring,
= a, in the z = 0
plane?
Q2: Let a uniform surface charge density of 5 nC/m2 be present at the z = 0 plane, a uniform line
charge density of 8 nC/m be located at x = 0, z = 4, and a point charge of 2 μC be present at
P(2, 0, 0). If V = 0 at (0, 0, 5), find V at (1, 2, 3)?
Q3: In spherical coordinates,
Ans: 1.98 KV
V/m. Find the potential at any point, using the
reference (a)V = 0 at infinity; (b) V = 0 at r = 0; (c)V = 100 V at r = a.?
Q4: Given
V, assuming free space conditions, find. (a) E; (b) the volume charge
density at = 0.5 m; (c) the total charge lying within the closed surface = 0.6, 0 < z < 1.
Ans:
, -673 C/m3 , -1.96 nC
Q5: A dipole having a moment p = 3ax − 5ay + 10az nC . m is located at Q(1, 2,-4) in free space.
Find V at P(2, 3, 4)?
Ans: V=1.31
Q6: The dipole lie on z-axis and located at the origin, find the difference in potential between
points at
and
, each point having the same r and
65
coordinates?
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Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
CONDUCTORS AND DIELECTRICS
5.1
‫الموصالت والعوازل‬
Current and Current Density
Electric charges in motion constitute a current. The unit of current is the ampere (A),
defined as a rate of movement of charge passing a given reference point (or crossing a given
reference plane) of one coulomb per second. Current is symbolized by I, and therefore
We find the concept of current density, measured in amperes per square meter (A/m2),
more useful. Current density is a vector represented by J.
Total current is obtained by integrating,
∫
Current density may be related to the velocity of volume charge density at a point.
Where
is velocity and
is volume charge density
This last result shows clearly that charge in motion constitutes a current. We call this type of
current convection current
mA/m2: determine the total
Example: Given the current density
current flowing outward through the circular band
Solution:
∫
∫
∫ ∫ (
∫
)
( )
[ ]
66
= 3, 0 <
< 2π,
2 < z < 2.8.?
DATE
Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
Example: Find the current in the circular wire shown in figure below, if the current density
(
)
A/m2: the radius of the wire is 2mm?
Solution:
∫
∫
∫
∫
∫
(
)
∫
∫
Example: Find the total current outward directed from a 1m cube with one corner at the origin
A/m2?
and edge parallel to the coordinate axes if
Solution:
1
∫
x=1
1
∬(
) (
∬
(
)
∬
)
(
)
(
∫ ∫
[ ] [ ]
∬
)
(
)
(
)
∫ ∫
[
(
∫ ∫
] [ ]
67
)
∫ ∫
1
DATE
5.2
Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
Continuity of Current
The continuity equation follows when we consider any region bounded by a closed surface.
The current through the closed surface is
∫
and this outward flow of positive charge must be balanced by a decrease of positive charge (or
perhaps an increase of negative charge) within the closed surface. If the charge inside the closed
surface is denoted by Qi , then the rate of decrease is −dQi /dt and the principle of conservation
of charge requires
∫
∫
∫ (
)
We next represent the enclosed charge Qi by the volume integral of the charge density
∫ (
)
∫
If we agree to keep the surface constant, the derivative becomes a partial derivative and may
appear within the integral
∫ (
)
∫
from which we have our point form of the continuity equation,
Remembering the physical interpretation of divergence, this equation indicates that the
current, or charge per second, diverging from a small volume per unit volume is equal to the time
rate of decrease of charge per unit volume at every point.
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5.3
Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
Metallic Conductors
Let us first consider the conductor. Here the valence electrons, or conduction, or free,
electrons, move under the influence of an electric field. With a field E, an electron having a
charge Q = −e will experience a force
In free space, the electron would accelerate and continuously increase its velocity (and
energy); in the crystalline material, the progress of the electron is impeded by continual
collisions with the thermally excited crystalline lattice structure, and a constant average velocity
is soon attained. This velocity
is termed the drift velocity, and it is linearly related to the
electric field intensity by the mobility of the electron in the given material.
where
is the mobility of an electron, mobility is measured in the units of square meters per
volt-second; typical values are 0.0012 for aluminum, 0.0032 for copper, and 0.0056 for silver.
where
is the free-electron charge density
The relationship between J and E for a metallic conductor, however, is also specified by
the conductivity
where
(sigma),
is measured is Siemens per meter (S/m).
Conductivity may be expressed in terms of the charge density and the electron mobility,
If a conductor of uniform cross-sectional area S and length L, as shown in Figure below,
has a voltage difference V between its ends, then
Assume that J and E are uniform
∫
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Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
∫
The ratio of the potential difference between the two ends of the cylinder to the current
entering the more positive end,
 When the fields are nonuniform,
∫
∫
Example: Find the resistance between the inner and outer curved surfaces of the block shown in
Fig. below, where the material is silver for which
Solution:
∫
∫
∫
70
= 6.17 x 107 S/m. if
⁄
?
DATE
5.4
Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
Conductor Properties and Boundary Conditions
For electrostatics, no charge and no electric field may exist at any point within a
conducting material. Charge may, however, appear on the surface as a surface charge density,
and our next investigation concerns the fields external to the conductor.
To summarize the principles which apply to conductors in electrostatic fields, we may
state that
1. The static electric field intensity inside a conductor is zero.
2. The static electric field intensity at the surface of a conductor is everywhere directed
normal to that surface.
3. The conductor surface is an equipotential surface.
5.4.1 Conductor-Dielectric Boundary Conditions
Under static conditions all net charge will be on the outer surfaces of a conductor and both
E and D are therefore zero within the conductor. Because the electric field is a conservative field,
the line integral of E.dL is zero for any closed path. A rectangular path with corners 1, 2, 3, 4 is
shown in Figure below.
∮
around the small closed path abcda. The integral must be broken up into four parts
∫
∫
∫
∫
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Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
If the path lengths b to c and d to a are now permitted to approach zero, keeping the
interface between them, then the second and fourth integrals are zero.
The path from c to d is within the conductor where E must be zero. This leaves
∫
∫
where Et is the tangential component of E at the surface of the dielectric. Since the interval
a to b can be chosen arbitrarily, at each point of the surface.
To discover the conditions on the normal components, a small, closed, right circular
cylinder is placed across the interface; Gauss' law applied to this surface gives
∮
∫
∫
∫
∫
The third integral is zero since, as just determined, Dt = 0 on either side of the interface.
The second integral is also zero, since the bottom of the cylinder is within the conductor, where
D and E are zero. Then,
∫
∫
∫
The electric flux leaves the conductor in a direction normal to the surface, and the value of
the electric flux density is numerically equal to the surface charge density.
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Example: A solid conductor has a surface described by x + y=3m and extends toward the origin.
At the surface the electric field intensity is 0.35 V/m. Express E and D at the surface
and find
?
Solution:
The unit vector normal to the surface is :
√
√
(
(
|
5.5
)
)
|
The Nature of Dielectric Materials
A dielectric in an electric field can be viewed as a free-space arrangement of microscopic
electric dipoles, each of which is composed of a positive and a negative charge whose centers do
not quite coincide. These are not free charges, and they cannot contribute to the conduction
process. Rather, they are bound in place by atomic and molecular forces and can only shift
positions slightly in response to external fields. They are called bound charges.
The characteristic that all dielectric materials have in common, whether they are solid,
liquid, or gas, and whether or not they are crystalline in nature, is their ability to store electric
energy. This storage takes place by means of a shift in the relative positions of the internal,
bound positive and negative charges against the normal molecular and atomic forces.
The dipole may be described by its dipole moment p
where Q is the positive one of the two bound charges composing the dipole, and d is the vector
from the negative to the positive charge. We note again that the units of p are coulomb-meters.
There is thus an added term to D that appears when polarizable material is present
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Conductors and Dielectrics
Lecturer: Mohammed Kamil Salh
The linear relationship between P and E is
where
(
(chi) is a dimensionless quantity called the electric susceptibility of the material.
)
The expression within the parentheses is now defined as
This is another dimensionless quantity, and it is known as the relative permittivity, or dielectric
constant of the material. Thus
where is the permittivity
Example Two point charges in a dielectric medium where
interact with a force of
8.6 10-3 N. What force could be expected if the charges were in free space??
Solution:
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DATE
5.6
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Conductors and Dielectrics
Lecturer: Mohammed Kamil Salh
Boundary Conditions for Perfect Dielectric Materials
Let us first consider the interface between two dielectrics having permittivities
and
and occupying regions 1 and 2, as shown in Figure below. We first examine the tangential
components by using
∮
Around the small closed path on the left, obtaining
The small contribution to the line integral by the normal component of E along the sections
of length
h becomes negligible as
h decreases and the closed path crowds the surface.
Immediately, then,
The boundary conditions on the normal components are found by applying Gauss’s law to
the small “pillbox” shown at the right in the Figure. The sides are again very short, and the flux
leaving the top and bottom surfaces is the difference
for no free charge is available in the perfect dielectrics, we may assume
75
is zero on the interface and
DATE
Conductors and Dielectrics
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Let D1 (and E1) make an angle
Lecturer: Mohammed Kamil Salh
with a normal to the surface (Figure below) . Because the
normal components of D are continuous,
The ratio of the tangential components is given by
and the division of these equations gives
Example The surface x = 0 separates two perfect dielectrics. For x > 0, let
, while
where x < 0. If E1 = 80ax − 60ay − 30az V/m, find (a) EN1; (b) ET1 ; (c) E1; (d)
the angle
between E1 and a normal to the surface; (e) DN2; ( f ) DT2; (g) D2; (h) P2;
(i ) the angle
between E2 and a normal to the surface.?
Solution:
(a)
𝜀𝑟
(b)
𝜃
E1
X=0
(c) |
|
√
𝜀𝑟
(d)
76
E2
𝜃
DATE
Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
(e)
(f)
(
(h)
(
)
)
(
)
(i)
77
DATE
Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
Example Region 1, z<0 m, is free space where D = 5ay + 7az C/m2. Region 2, 0< z <1 m,
has
. And region 3, z > 1 m, has
Find E2, P2 , and
?
Solution:
(
)
𝜃
E3
𝜀𝑟
Z=1
E2
√
𝜃
𝜀𝑟
Z=0
𝜀𝑟
E1
(
)
√
78
𝜃
Free space
DATE
Conductors and Dielectrics
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Lecturer: Mohammed Kamil Salh
Home work
Q1: Current density is given in cylindrical coordinates as J =
; for
A/m2 in the region
, J = 0. Find the total current crossing the surface z = 0.1 m?
Ans. −39.7μA
Q1: Given J = −10−4(yax + xay)A/m2, find the current crossing the y = 0 plane in the -ay direction
between z = 0 and 1, and x = 0 and 2?
Ans: 2*10-4 A
Q2: Let Region 1 (z < 0) be composed of a uniform dielectric material for which
Region 2 (z > 0) is characterized by
(a) DN1; (b) Dt1 (c) |
= 3.2, while
= 2. Let D1 = -30ax +50ay +70az nC/m2 and find:
|; (d) D1; (e) ; ( f ) P1 (g) DN2; (h) Dt2; (i) D2; (k) P2;(l)
.?
Ans: 70 nC/m2; −30ax + 50ay nC/m2; 58.3 nC/m2; 91.1 nC/m2; 39.8o; −20.6ax +34.4ay + 48.1az
nC/m2, 70az nC/m2; −18.75ax + 31.25ay nC/m2; −18.75ax + 31.25ay + 70az nC/m2;
−9.38ax + 15.63ay + 35az nC/m2; 27.5o
Q2: Two perfect dielectrics have relative permittivity
= 2 and
= 8. The planar interface
between them is the surface x − y + 2z = 5. The origin lies in region 1. If
E1 = 100ax + 200ay − 50az V/m, find E2?
Ans: 125ax + 175ay V/m
Q2: Figure below shows a planar dielectric slab with free space on either side. Assuming a
constant field E2 within the slab, show that E3 = E1?
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DATE
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Capacitance
Lecturer: Mohammed Kamil Salh
CAPACITANCE
‫السعــــــــــــــة‬
A capacitor is a device that stores energy; energy thus stored can either be associated
with accumulated charge or it can be related to the stored electric field.
6.1 Capacitance Defined
Consider two conductors embedded in a homogeneous dielectric. Conductor M2 carries a
total positive charge Q, and M1 carries an equal negative charge. There are no other charges
present, and the total charge of the system is zero.
Let us designate the potential difference between M2 and M1 as V0. We may now define the
capacitance of this two-conductor system as the ratio of the magnitude of the total charge on
either conductor to the magnitude of the potential difference between conductors
In general terms, we determine Q by a surface integral over the positive conductors, and
we find V0 by carrying a unit positive charge from the negative to the positive surface,
∮
∫
The capacitance is independent of the potential and total charge, for their ratio is constant.
If the charge density is increased by a factor of N, Gauss’s law indicates that the electric flux
density or electric field intensity also increases by N, as does the potential difference. The
capacitance is a function only of the physical dimensions of the system of conductors and of the
permittivity of the homogeneous dielectric
Capacitance is measured in farads (F), where a farad is defined as one coulomb per volt
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Capacitance
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Lecturer: Mohammed Kamil Salh
6.2 Parallel-Plate Capacitor
We can apply the definition of capacitance to a simple two-conductor system in which the
conductors are identical, infinite parallel planes with separation d (Figure below). Choosing the
lower conducting plane at z = 0 and the upper one at z = d, a uniform sheet of surface charge
on each conductor leads to the uniform field
where the permittivity of the dielectric is ,
The potential difference between lower and upper planes is
∫
∫
Since the total charge on either plane is infinite, the capacitance is infinite. A more practical
answer is obtained by considering planes, each of area S, whose linear dimensions are much
greater than their separation d.
Example: Calculate the capacitance of a parallel-plate capacitor having a mica dielectric,
a plate area of 10 cm2, and a separation of 0.01 cm.?
Solution:
81
= 6,
DATE
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Capacitance
Lecturer: Mohammed Kamil Salh
6.3 Multiple-Dielectric Capacitors
When two dielectrics are present in a capacitor with the interface normal to E and D, as
shown in Figure below, the equivalent capacitance can be obtained by treating the arrangement
as two capacitors in series
When two dielectrics are present in a capacitor with the interface parallel to E and D, as
shown in Figure below, the equivalent capacitance can be obtained by treating the arrangement
as two capacitors in parallel
S1
S2
𝜀𝑟
82
𝜀𝑟
𝑑
DATE
/ /2013
Capacitance
Lecturer: Mohammed Kamil Salh
Example: A parallel-plate capacitor with area 0.30 m2 and separation 5.5 mm contains three
dielectrics with interfaces normal to E and D, as follows:
d2 = 2mm;
, d3 = 2.5 mm. Find the capacitance?
Solution:
83
, dl = 1mm;
,
DATE
6.4
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
Capacitance of coaxial cable or coaxial capacitor
We choose a coaxial cable or coaxial capacitor of inner radius a, outer radius b, and length L
∫
and the voltage difference between the conductors is
The total charge on the inner conductor is
(
)
( ⁄ )
Example: Find the capacitance between the inner and outer curved conductor surfaces shown in
Figure below?
Solution:
( ⁄ )
( ⁄ )
( ⁄ )
(
⁄
84
)
DATE
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
6.5 Spherical Capacitor
Consider a spherical capacitor formed of two concentric spherical conducting shells of
radius a and b, b > a. The expression for the electric field was obtained previously by Gauss’s
law
where the region between the spheres is a dielectric with permittivity . The expression for
potential difference was found from this by the line integral.
𝜀𝑟
Thus,
(
)
Here Q represents the total charge on the inner sphere, and the capacitance becomes
If we allow the outer sphere to become infinitely large, we obtain the capacitance of an isolated
spherical conductor
Example: Find the capacitance between the inner (r=2) and outer (r=3) sphere conductor
surfaces if
?
Solution:
85
DATE
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
6.6 Energy Stored in a Capacitor
The total energy stored in the capacitor is
∫
∫∫
Example: Find the relative permittivity of the dielectric material present in a parallel-plate
capacitor if: (a) S = 0.12m2, d = 80 m, V0 = 12 V, and the capacitor contains 1 J of
energy; (b) the stored energy density is 100 J/m3, V0 = 200 V, and d = 45 m;
(c) E = 200 kV/m and
= 20 C/m2?
Solution:
( )
(
)
( )
(
)
( )
86
DATE
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
6.7 Poisson’s and Laplace’s Equations
An alternate approach would be to start with known potentials on each conductor, and
then work backward to find the charge in terms of the known potential difference. The
capacitance in either case is found by the ratio Q/V.
Obtaining Poisson’s equation is exceedingly simple, for from the point form of Gauss’s law
by substitution we have
(
)
(
)
Equation above is Poisson’s equation,
If
= 0, indicating zero volume charge density, but allowing point charges, line charge,
and surface charge density to exist at singular locations as sources of the field, then
which is Laplace’s equation. The
2
operation is called the Laplacian of V.
(
(
(
)
(
)
(
)
(
)
87
)
)
(
)
DATE
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
6.8 Examples of The Solution of Laplace’s Equation
Several methods have been developed for solving Laplace’s equation. The simplest method
is that of direct integration. We will use this technique to work several examples involving onedimensional potential variation in various coordinate systems in this section.
The necessary steps are these, after the choice of boundary conditions has been made:
1. Given V, use E = − V to find E.
2. Use D = E to find D.
3. Evaluate D at either capacitor plate, D = DS = DN aN.
4. Recognize that
= DN .
5. Find Q by a surface integration over the capacitor plate,

∫
V is a function only of x:
Laplace’s equation reduces to
and the partial derivative may be replaced by an ordinary derivative, since V is not a
function of y or z,
We integrate twice, obtaining
where A and B are constants of integration
Example: Find the capacitance of a parallel-plate capacitor of plate area S, plate separation d,
and potential difference V0 between plates?
Solution:
88
DATE
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
( )
∫
∫
 V as a function of
Laplace’s equation becomes
(
)
Noting the ρ in the denominator, we exclude
= 0 from our solution and then multiply by
and integrate
where a total derivative replaces the partial derivative because V varies only with . Next,
rearrange, and integrate again,
89
DATE
/ /2013
Capacitance
Lecturer: Mohammed Kamil Salh
Example: Find the capacitance of the coaxial capacitor or coaxial transmission line?
Solution:
conductor
∫
∫ ∫
90
𝜀𝑟
DATE
Capacitance
/ /2013
 V as a function of
Lecturer: Mohammed Kamil Salh
in cylindrical coordinates
Laplace’s equation is now
(
)
We exclude
= 0 and have
The solution is
Example: In cylindrical coordinates two
= const, planes are insulated along the z axis, as
shown in Fig. 8-9. Neglect fringing and find the expression for E between the planes,
assuming a potential of 100 V for
= a, and a zero reference at
Solution:
( )
(
)
91
= 0?
DATE
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
Example: In the capacitor shown in Figure below, the region between the plates is filled with a
dielectric having
= 4.5. Find the capacitance?
Solution:
From example above
∫
∫
∫
 V as a function of in Spherical coordinates
Laplace’s equation is now
(
)
We exclude r = 0 and θ = 0 or π and have
The second integral is then
∫
(
)
92
DATE
Capacitance
/ /2013
Lecturer: Mohammed Kamil Salh
Example: Solve Laplace's equation for the region between coaxial cones, as shown in Figure
below. A potential V, is assumed at , and V = 0 at
at r = 0? (b) let
angle
= 10°,
. The cone vertices are insulated
=30°, and V1 = 100V. Find the voltage at
is the voltage 50 V?
Solution:
(
)
(
)
(
)
(
(
(
)
(
)
)
)
(
(
)
[ (
)
)
(
)
(
)
(
)
(
)
93
(
)]
= 20°, At what
DATE
/ /2013
Capacitance
Lecturer: Mohammed Kamil Salh
Home work
Q1: In the cylindrical capacitor shown in Fig. below each dielectric occupies one-half the
volume. Find the capacitance?
Q2: The coaxial cable in Fig. below has an inner conductor radius of 0.5 mm and an outer
conductor radius of 5 mm. find the capacitance per unit length with spacers as shown.
Ans: 45.9 PF/m
Q3: A parallel-plate capacitor has its dielectric changed from
= 2.0 to
= 6.0. It is noted
that the stored energy remain fixed: W2 = W1. Examine the changes, if any, in V, C, D, E, Q,
and
?
Q4: Find the capacitance between the two cones of Fig. below. Assume free space?
Ans: 12.28
Q5: A parallel-plate capacitor of 8nF has an area 1.51 m2 and separation 10 mm. What separation
would be required to obtain the same capacitance with free space between the plates?
Ans: 1.67 mm
94