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ELECTROSTATICS
COULUMB’S LAW
ELECTRIC FIELD INTENSITY
LINE, SURFACE & VOLUME CHARGES
ELECTRIC FLUX DENSITY
GAUSS’S LAW
ELECTRIC POTENTIAL
BOUNDARY CONDITIONS
CAPACITANCE
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Electrostatic Field Problem –
Example: Parallel Plate Capacitor
Scalar Field: Electrostatic Potential
Vector Field: Electrostatic Field Strength
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Bear in mind!!!!
• Electromagnetics is the study of the effect of
charges at rest and charges in motion.
• Some special cases of electromagnetics:
Electrostatics: charges at rest
Magnetostatics: charges in steady motion
(DC)
Electromagnetic waves: waves excited by
charges in time-varying motion
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Electrostatics
• Electrostatics is the physics term for
static charge.
• Electro means charge, and of course
static means stationary or not
• moving.
Charge
• 2 Types of Charge:
Positive (+) and Negative (-)
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Electrostatics
Attraction and Repulsion
Unlike a gravitational force which always attracts, electrostatic
force may repel or attract depending on the type charge.
Ben's Rule and Paula Abdul - Opposites attract and likes repel.
(+) (-) = attract
(+) (+) = repel
(-) (-) = repel
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Electrostatics:
Microscopic
View
Electrons have a negative charge and protons have a
positive charge.
Neutrons and most materials have a net charge of zero or
neutral.
When certain types of objects are rubbed together,
electrons from one object may be transferred to an object
with a greater affinity for the electrons. When this happens,
the object that gave up the electrons is positive, whereas
the object that collected the electrons is negative.
Key: Only electrons move. They exist in the outer shells of
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the atom.
• Electric charge is inherently quantized
such that the charge on any object is an
integer multiple of the smallest unit of
charge which is the magnitude of the
electron charge e = 1.602  10-19 C.
• On the macroscopic level, we can
assume that charge is “continuous.”
• The following list shows part of the triboelectric sequence. When any two
substances shown in this list are rubbed
together, the top one will become
positively charged while the lower one
will become negatively charged. The
further apart the two substances are in
the list, the greater the electrification.
++
Asbestos
Fur (rabbit)
Glass
Mica
Wool
Quartz
Fur (cat)
Lead
Silk
Human skin, Aluminum
Cotton
Wood
Amber
Copper, Brass
Rubber
Sulfur
Celluloid
India rubber
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Static Charges: Coulomb’s Law
All bodies are able to take a charge of electricity and
this is termed static electricity. The charge on a
body is measure by means of the force between the
charges.
The Coulomb force law, which only applies to
charged points, is stated below..
The force of attraction or repulsion between two charged points
is directly proportional to the charges and inversely proportional
to the square of the distance between them.
In vector form, it is stated thus,
F12 
Q1Q2
40 R 12
2
a12
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Coulomb’s Law
Where;
F = Force between points (N)
Q1, Q2 = Charges on point 1 and point 2
(Coulomb)
R = radial separation on points/distance (m)
a12 = the unit vector in the direction from Q1 to
2
 o =QPermittivity
of the free space (vacuum)
with a value given by:
 0  8.85  10
12
F
10 9 F

m
36 m
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Coulomb’s Law
Charge Q1 exerts
a vector force F12
in Newton's (N)
on charge Q2,
F12 
Q1Q2
40 R 12
2
a 12
Note: a negative force results if the points
have opposite charges and a positive force
results if the points have the same polarity
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Coulomb’s Law
The above equation can be simplified as follows
Q1Q2
F k
R2
The proportional constant, k is:
k
1
40
1
4
,
   r 0
is constant= 8.99x109
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Coulomb’s Law
• If the space between the charges is
another material or air, the law may be
written
F12 
Q1Q2
40 r R 12
2
a12
where
r relative permittivity of material.

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Example 1
Find the force on charge Q1, 20 C,
due to charge Q2, -300 C, where Q1 is
at (0,1,2)m and Q2 at (2,0,0)m
Solution 1
z
F1
Q1
R21
(0,1,2)
y
Q2
x
(2,0,0)
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Solution 1 (cont’d)
• Because 1C is a rather large unit, charges are often
given in microcoulombs (C), nanocoulombs (nC)
and picocoulombs (pC). Referring to figure,
R21=-2ax+ay+2az
R=3
a21=1/3(-2ax+ay+2az)
 using Coulmb' s Law equation; F12 
40 R 12

20 10  300 10    2a


4 8.854 10 3 
6
F21
Q1Q2
6
12
 2a x  a y  2a z 
 N
 6
3


2
x
2
a12
 a y  2a z 

3

The force magnitude is 6N
and the direction is such that
Q1 is attracted to Q2
(unlike charges attract)
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Electric Field Intensity
If Q1 is fixed to be at origin, a second
charge Q2 will have force acting on Q1
and can be calculated using Coulomb’s
Law. We also could calculate the force
vector that would act on Q2 at every
point in space to generate a field of
such predicted force values.
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Electric Field Intensity (Cont’d)
It becomes convenient to define electric field
intensity E1 or force per unit charge as:
F12
E1 
Q2
This field from charge Q1 fixed at origin results
from the force vector F12 for any arbitrarily
chosen value of Q2
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Electric Field Intensity (Cont’d)
Coulomb’s law can be rewritten as
E
Q
40 R
2
aR
to find the electric field intensity at any point in
space resulting from a fixed charge Q.
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Example 2
Find E at (0,3,4) m in cartesian coordinates
due to a point charge Q =0.5C at the origin.
Solution to Example 2
a R  (3a y  4a z ) / 5
R  3a y  4a z
 0.6a y  0.8a z
R 5
0.5 106
0.6a y  0.8a z 
E
2
9
4 10 / 36 5


Thus E  180V / m in the direction a R  0.6a y  0.8a z
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Example 3
Let a point charge Q1 = 25nC be located at P1 (4,-2,7).
If ε = ε0, find electric field intensity at P2 (1,2,3).
Solution to Example 3
By using the electric field intensity,
E
Q
4o R
2
aR
This field will be:
25 10 9
E
aR
4o R
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Solution to Example 3 (Cont’d)
Where,
R12  r2  r1  3a x  4a y  4a z
and
R  41
E

Q
4o R

2
aR 
Q
4o R
25  10 9
4 8.854  10
 ??
12
41
3
3
R
 3a
2
x
 4a y  4a z 
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Electric Field Intensity (Cont’d)
If there are N charges, Q1,Q2...QN located respectively at
point with position vectors r1,r2...rN the electric field
intensity at point r is:
E
Q1
4 0 r  r1
E
2
r  r1   ..
r  r1
1
40
N

k 1
r  rN 
QN
4 0 r  rN
2
r  rN
Qk r  rk 
r  rk
3
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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 (Cont’d)
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 (Cont’d)
The field lines terminated
at a negative point
charge
The field lines for a pair
of opposite charges
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