Electrostatic Fields:

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CHAPTER 2
ELECTROMAGNETIC FIELDS THEORY

• Electrostatic Fields
• Electric Field Intensity

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In this chapter you will learn..

• Electrostatic Fields
• -Charge, charge density
• -Coulomb's law
• -Electric field intensity
• - Electric flux and electric flux density
• -Gauss's law
• -Divergence and Divergence Theorem
• -Energy exchange, Potential difference, gradient
• -Ohm's law
• -Conductor, resistance, dielectric and
  capacitance
• - Uniqueness theorem, solution of Laplace and
  Poisson equation

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• Two like charges repel one another, whereas two
  charges of opposite polarity attract
• The force acts along the line joining the charges
• Its strength is proportional to the product of
  the magnitudes of the two charges and inversely
  propotional to the square distance between them

Charles Agustin de Coulomb
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Electric Field Intensity
This proof the existence of electric force field
which radiated from Q1
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Electric Field Intensity

• Consider a charge fixed in a position, Q1.
• Lets say we have another charge, say Qt which is
  a test charge.
• When Qt is moved slowly around Q1, there exist
  everywhere a force on this second charge.
• Thus proof the existence of electric force
  field.

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Electric Field Intensity
Experience E due to Q1

• Electric field intensity, E is the vector force
  per unit charge when placed in the electric field

Q1
Experience the most E due to Q1
Experience less E due to Q1
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Electric Field Intensity

• According to Coulombs law, force on Qt is
• We can also write
• Which is force per unit charge

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Electric Field Intensity

• If we write E as
• Thus we can rewrite the Coulombs law as
• Which gives the electric field intensity for Qt
  at r due to a point charge Q located at r

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Electric Field Intensity

• For N points charges, the electric field
  intensity for at a point r is obtained by
• Thus

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Field Lines
The behavior of the fields can be visualized
using field lines
Field vectors plotted within a regular grid in 2D
space surrounding a point charge.
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Field Lines
Some of these field vectors can easily be joined
by field lines that emanate from the positive
point charge.
The direction of the arrow indicates the
direction of electric fields
The magnitude is given by density of the lines
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Field Lines
The field lines terminated at a negative point
charge
The field lines for a pair of opposite charges
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Example 4.1

• Point charges of 1 mC and -2mC are located at (3,
  2, -1),
• B(-1, -1, 4) respectively in free space.
  Calculate the electric force on a 10 nC charge
  located at (0, 3, 1) and the electric field
  intensity at that point.

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Solution
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Line charge
Imagine charges distributed along a line
We can find the charge density over the line as
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Charge density and charge distribution

• We can find the total charge by applying this
  equation

Charge density
Length of the line
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Surface Charge
Similarly, for surface charge, we can imagine
charges distributed over a surface
We can find the charge density over the surface
as

And the total charge is
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Volume Charge
We can find the volume charge density as

And the total charge is
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Continuous Charge Distribution

• for line charge distribution
• for surface charge distribution
• for volume charge distribution

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E due to Continuous Charge Distributions
Line charge
?S






Surface charge


?v






Volume charge


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What About the Field Lines?
E field due to line charge
E field due to surface charge
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E field due to line charge
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What about volume charge?
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LINE CHARGE
Infinite Length of Line Charge
To derive the electric field intensity at any
point in space resulting from an infinite length
line of charge placed conveniently along the
z-axis

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LINE CHARGE (Contd)
Place an amount of charge in coulombs along the z
axis. The linear charge density is coulombs of
charge per meter length, Choose an arbitrary
point P where we want to find the electric
field intensity.
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LINE CHARGE (Contd)
The electric field intensity is
But, the field is only vary with the radial
distance from the line. There is no segment of
charge dQ anywhere on the z-axis that will give
us . So,
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LINE CHARGE (Contd)
Consider a dQ segment a distance z above radial
axis, which will add the field components for the
second charge element dQ.
The components cancel each other (by
symmetry) , and the adds, will give
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LINE CHARGE (Contd)
Recall for point charge,
For continuous charge distribution, the summation
of vector field for each charges becomes an
integral,
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LINE CHARGE (Contd)
The differential charge,
The vector from source to test point P,
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LINE CHARGE (Contd)
Which has magnitude, and
a unit vector,
So, the equation for integral of continuous
charge distribution becomes
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LINE CHARGE (Contd)
Since there is no component,
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IMPORTANT!!
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IMPORTANT!!
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LINE CHARGE (Contd)
Hence, the electric field intensity at any point
? away from an infinite length is
For any finite length, use the limits on the
integral.
? is the perpendicular distance from the line to
the point of interest a? is a unit vector along
the distance directed from the line charge to the
field point
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Example of Line Charge

• A uniform line charge of 2 µC/m is located on the
  z axis. Find E in cartesian coordinates at P(1,
  2, 3) if the charge extends from
• a) -8 lt z lt 8
• b) -4 z 4

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Example of Line Charge
z
dEz
4
dE
?
dE?
P(1, 2, 3)
r
dl
z
r
y
x
-4
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Solution (a)

• With the infinite line, we know that the field
  will have only a radial component in cylindrical
  coordinates (or x and y components in cartesian).
  
• The field from an infinite line on the z axis is
  generally
• Therefore, at point P

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Solution (b)

• Here we use the general equation
• Where the total charge Q is

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Solution (b)

• Or, we can also write E as
• Where
• and

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Solution (b)

• So E would be
• Using integral tables, we obtain

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What if the line charge is not along the z-axis??
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Line Charge
z
P(x, y, z)
?
R
y
dl
(0, y, 0)
(0, y, 0)
x
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What if the line is not along any axis??
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Example

• A uniform line charge of 16 nC/m is located along
  the line defined by
• y -2, z 5. If e e0.
• Find E at P(1, 2, 3)

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Solution

• Draw the line y -2, z 5.
• Find a?
• Find E

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Solution
z
(1, -2, 5)
5
?
P(1, 2, 3)
-2
y
x
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Solution

• To find a? , we must draw a projection line
  from point P to our line. Then find out what is
  the x-axis value at that projection point
• Then a? can be calculated as

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Then finally, calculate E
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What if it is a circular line of charge??
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EXAMPLE
Use Coulombs Law to find electric field
intensity at (0,0,h) for the ring of charge, of
charge density, centered at the origin in
the x-y plane.
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Solution

• Step 1 derive what are dl and R, R2 and aR
• Step 2 Replace dl, R2 and aR into line charge
  equation

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Solution
By inspection, the ring charges delivers only
and contribution to the field.
component will be cancelled by symmetry.
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Solution
Each term need to be determined
The differential charge,
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Solution
The vector from source to test point,
Which has magnitude,
and a unit vector,
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Solution
The integral of continuous charge distribution
becomes
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Solution
Rearranging,
Easily solved,
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Infinite Surface charge

• Static charges resides on conductor surfaces and
  not in their interior, thus the charges on the
  surface are known as surface charge density or ?s
  
• Consider a sheet of charge in the xy plane
  shown below

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Infinite Surface charge
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Infinite Surface charge
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Infinite Surface charge

• The field does not vary with x and y
• Only Ez is present
• The charge associated with an elemental area dS
  is

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Infinite Surface charge

• The total charge is
• From figure,

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Infinite Surface charge

• From figure,

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Infinite Surface charge

• Replace in
• Thus becomes

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Infinite Surface charge
For an infinite surface, due to symmetry of
charge distribution, for every element 1, there
is a corresponding element 2 which cancels
element 1.
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Infinite Surface charge

• Thus the total of E? equals to zero. E only has
  z-component.
• For infinite surface, 0 lt ? lt 8

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Solution (cont)
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Infinite Surface charge

• Thus in general, for an infinite sheet of charge,
• Where an is a vector normal to the sheet.
• E is normal to the sheet and independent of the
  distance between the sheet and the point of
  observation, P.

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Infinite Surface charge

• In parallel plate capacitor, the electric field
  existing between the two plates having equal and
  opposite charges is given by

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EXAMPLE
An infinite extent sheet of charge
exists at the plane y -2m. Find the
electric field intensity at point P (0, 2m, 1m).
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SOLUTION
Step 1 Sketch the figure
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SOLUTION
Step 2 find an. The unit vector directed away
from the sheet and toward the point P is Thus
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SOLUTION
Step 3 solve the problem!
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Example

• Given the surface charge density, ?s 2µC/m2, in
  the region ? lt 0.2 m, z 0, and is zero
  elsewhere, find E at PA(? 0, z 0.5)

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Solution

• Sketch the figure

(0,0,0.5)
2
2
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Solution

• First, we recognize from symmetry that only a z
  component of E will be present. Considering a
  general point z on the z axis, we have r zaz.
  Then, with r ?a?, we obtain r - r zaz - ?a?.
  The superposition integral for the z component of
  E will be

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Solution

• With z 0.5 m,

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Exercise

• Planes x 2, and y -3, respectively, carry
  charges 10nC/2 and 15nC/m2. if the line x 0, z
  2 carries charge 10p nC/m, calculate E at (1,
  1, -1) due to three charge distributions.

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Volume charge
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Volume charge sphere

• Consider the volume charge distribution with
  uniform charge density, in the figure below

P(0,0,z)P(z,0,0)
a
R
?v
dv at (r,?,F)
Z
r
?
y
F
x
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Volume charge sphere

• The charge dQ is
• The total charge in a sphere with radius a is

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Volume charge sphere

• The electric field dE at point P(0,0,z) due to
  the elementary volume charge is
• Where

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Volume charge sphere

• Due to symmetry of charge distribution, Ex or Ey
  add up top zero.
• Thus

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Volume charge sphere

• For spherical coordinate,
• Then by applying the cosine rule,

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Volume charge sphere

• Or we can express it as
• Differentiate with respect to ?

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Volume charge sphere

• Then dv now becomes
• And Ez becomes

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Volume charge sphere

• And Ez becomes (cont)

(a)
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Volume charge sphere

• But
• So
• Replace in equation (a). So E now becomes

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Volume charge sphere

• For a point P2 at (r,?,F) E is
• Which is identical to the electric field at the
  same point due to a point charge Q at the origin.
  

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EXAMPLE 6
Find the total charge over the volume with volume
charge density,
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SOLUTION
The total charge, with
volume
Thus,
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SOLUTION
To find the electric field intensity resulting
from a volume charge, we use
Since the vector R and will vary over the
volume, this triple integral can be difficult. It
can be much simpler to determine E using Gausss
Law.
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Principle of superposition

• Total E field at a point is the vector sum of the
  fields caused by all the individual charges.
• Consider the figure below. Positions of the
  charges q1, q2, q3, qn, (source points ) be
  denoted by position vectors R1, R2, R3, R4,
  .. Rn. The position of the field point at which
  the electric field intensity is to be calculated
  is denoted by R

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Principle of superposition
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Principle of superposition

• Thus

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Exercise

• Three infinite lines of charge, all parallel to
  the z-axis, are located at the three corners of
  the kite-shaped arrangement shown in the figure
  below. If the two right triangles are symmetrical
  and of equal corresponding sides, show that the
  electric field is zero at the origin.