Ajal electrostatics slides

ELECTROSTATICS
By
Prof . AJAL A. J
HEAD OF THE ACADEMICS
Phone : 890 730 5642
e- mail : mentorajal@gmail.com

Electromagnetism
Electromagnetism is one of the fundamental forces
in nature, and the the dominant force in a vast range
of natural and technological phenomena
 The electromagnetic force is solely responsible for the
structure of matter, organic, or inorganic
 Physics, chemistry, biology, materials science
 The operation of most technological devices is based on
electromagnetic forces. From lights, motors, and batteries,
to communication and broadcasting systems, as well as
microelectronic devices.
 Engineering

Electromagnetism
Electricity
Electromagnetism Magnetism
Optics
In this course we are going to discuss the
fundamental concepts of electromagnetism:
charge force field potential current
electric
circuit
magnetic
field
induction alternating
currents
waves
reflection refraction image interference diffraction
Once you master these basic concepts, you will be ready to move forward,
into more advanced subjects in your specific field of interest

System of Units
We will use the SI system – SI  International System of Units
Fundamental Quantities
Length  meter [m]
Mass  kilogram [kg]
Time  second [s]
Other Units
Current  ampere [A]
Derived Quantities
Force  newton 1 N = 1 kg m / s2
Energy  joule 1 J = 1 N m
Charge  coulomb 1 C = 1 A s
Electric Potential  volt 1 V = 1 J / C
Resistance  ohm 1  = 1 V / A

Electric Charge
The Transfer of Charge
SILK
Glass Rod
-
+
+
+
+
+
Glass and silk are insulators:
charges stuck on them stay put.
---
-

Electric Charge
+ +
Two positively charged rods
repel each other.

Electric Charge
History
600 BC Greeks first discover attractive
properties of amber when rubbed.
1600 AD Electric bodies repel as well as attract
1735 AD du Fay: Two distinct types of electricity
1750 AD Franklin: Positive and Negative Charge
1770 AD Coulomb: “Inverse Square Law”
1890 AD J.J. Thompson: Quantization of
electric charge - “Electron”

Electric Charge
Summary of things we know:
– There is a property of matter called electric
charge. (In the SI system its units are Coulombs.)
– Charges can be negative (like electrons) or
positive (like protons).
– In matter, the positive charges are stuck in place in
the nuclei. Matter is negatively charged when
extra electrons are added, and positively charged
when electrons are removed.
– Like charges repel, unlike charges attract.
– Charges travel in conductors, not in insulators
– Force of attraction or repulsion ~ 1 / r2

Charge is Quantized
q = multiple of an elementary charge e:
e = 1.6 x 10-19 Coulombs
Charge Mass Diameter
electron - e 1 0
proton +e 1836 ~10-15m
neutron 0 1839 ~10-15m
positron +e 1 0
(Protons and neutrons are made up of quarks, whose charge is
quantized in multiples of e/3. Quarks can’t be isolated.)

Forces
 By the early 19th century, physicists
had classified the apparent myriad of
forces in nature to just 3 kinds:
 Gravitational force
 Electric force
 Magnetic force

Forces
 By the end of the 19th century, they had
narrowed the list to just 2 forces:
 Gravitational force
 Electromagnetic force

Coulomb’s Law
q1 q2
r12r12
F12
Force on 2 due to 1122
12
21
12
ˆrF
r
qkq

k = (4pe0)-1 = 9.0 x 109 Nm2/C2
e0 = permitivity of free space
= 8.86 x 10-12 C2/Nm2
Coulomb’s law describes the interaction between bodies due to their charges

Gravitational and Electric Forces
in the Hydrogen Atom
+e
-e
M
m
r12
m = 9.1 10-31 kg
M = 1.7 10-27 kg
r12 = 5.3 10-11 m
Gravitational force Electric Force

+e
-e
M
m
r12
m = 9.1 10-31 kg
M = 1.7 10-27 kg
r12 = 5.3 10-11 m
Fg = 3.6 10-47 N

F G
Mm
r
rg 
12
2


+e
-e
M
m
r12
m = 9.1 10-31 kg
M = 1.7 10-27 kg
r12 = 5.3 10-11 m
Fg = 3.6 10-47 N

F G
Mm
r
rg 
12
2


F
Qq
r
re 






1
4 0 12
2pe

Fe = 3.6 10-8N

Coulomb’s Law
The force between two charges depends
on the magnitude of the charges and
the distance between them
F = k
q1
q2
d
2

Coulomb’s Law
•Units of charge = Coulomb (C)
•k is a proportionality constant called the
Coulomb constant
•k  9,000,000,000 N·m2
or 9·10
9 N·m2
F = k
q1
q2
d2
C2 C2

Other Useful Numbers
Charge of an
Electron
qe
= -1.6 · 10
-19
C
Charge of a
Proton
qp
= +1.6 · 10
-19
C

P01 -22
Coulomb's Law
Coulomb’s Law: Force
on q2 due to interaction
between q1 and q2
9 2 2
0
1
8.9875 10 N m /C
4
ek
pe
  
1 2
12 2
ˆe
q q
k
r
F r
r
ˆ :r unit vector from q1 to q2
r
r
r

ˆ 1 2
12 3e
q q
k
r
 F r
r r
:r
r
vector from q1 to q2

Sample Problem
Two electrons are a meter apart.
What is the force between them?
What direction is it in?

Sample Problem
d = 1m
q
p
= -1.6*10
-19
C q
e
= -1.6*10
-19
C
F = k
q1
q2
d2
F = (9*10
9
N*m
2
/C
2
)
(-1.6*10
-19
C)* (-1.6*10
-19
C)
(1 m)
2
F = 2.3*10
-28
N

P01 -25
Coulomb's Law: Example
 31
32 2 2
ˆ ˆ m
1mr
 

r i j
r
a = 1 m
q1 = 6 C
q3 = 3 C
q2 = 3 C
?32 F

32 3 2 3ek q q
r

r
F
rr
32r
 
9
81 10 ˆ ˆ3 N
2

 i j
   
 
 
1
29 2 2
3
ˆ ˆ3 m
9 10 Nm C 3C 3C
1m

 
i j

P01 -26
The Superposition Principle
3 13 23 F F F
r r r
1
N
j ij
i
 F F
r r
Many Charges Present:
Net force on any charge is vector sum of forces
from other individual charges
Example:
In general:

P01 -
Superposition of forces from two charges
Blue charges fixed , negative, equal charge (-q)
What is force on positive red charge +q ?
x
y

P01 -
x
y
Consider effect of each charge separately:

P01 -
x
y
Take each charge in turn:

P01 -
x
y
Create vector sum:

P01 -
x
y
Find resultant:
NET
FORCE

P01 -
Superposition Principle
q3
q1
q2
F31
F21
F F31
F31x
F31y
F21x
F21y
F21
F = (F21x + F31x) x + (F21y + F31y) y
Forces add vectorially

P01 -33
Electric Field (~g)
The electric field at a point P due to a charge
q is the force acting on a test charge q0 at
that point P, divided by the charge q0 :
0
0
( )
qq
q P
q

F
E
For a point charge q:
2
ˆ( )q e
q
P k
r
E r
Units: N/C, also Volts/meter

P01 -34
Superposition Principle
The electric field due to a collection of N point
charges is the vector sum of the individual electric
fields due to each charge
1 2
1
.....
N
total i
i
    E E E E

P01 -35
Gravitational & Electric Fields
2
ˆsM
G
r
 g r
r
Mass Ms Charge qs (±)
2
ˆs
e
q
k
r
E r
r
g mF g
r r
E qF E
r r
This is easiest way to picture field
CREATE:
FEEL:
SOURCE:

Charging
•Items may be charged by friction
•Electrons are moved from one object to
another by being scraped away
•Items may be charged by contact
•Electrons are moved without being
scraped off

Charging by Friction
 If one neutral material has more affinity
for electrons than another (neutral)
material, it will attract electrons from the
other.
 One material becomes negatively
charged, the other positively charged.

Charging by Induction
Induction: The charging of an object
without direct contact

Induction and
Lightning
The bottom of the
negatively charged
cloud induces a
positive charge at
the surface of the
ground below.

Grounding
 Providing a path from a charged object
to the Earth is called grounding it.
 Charges will be attracted from (or
repelled to) the Earth by the charged
object.
 Since the Earth is so large, both the
charged object and the Earth are
neutralized.

In Summary. . .
•By friction, when electrons are rubbed from
one object to another
•By contact, when electrons are transferred
through direct contact without rubbing
•Through induction when electrons are
gathered or dispersed by the presence of a
nearby charge (without physical contact)
Objects are electrically charged in one of
three ways:

Field Lines
•Electric Fields have a magnitude and
direction
•Vector Quantities
•Lines go away from positive
•Lines go toward negative

Drawing Field Lines
•From + to –
•Lines start
perpendicular to
the surface of the
charge
•Field strength is
shown by the
density of the field
lines

P01 -44
Electric Field Lines
1. Join end-to-end infinitesimal vectors representing E…the curve
that results is an electric field line (also known as line of force).
2. By construction then, the direction of the E field at any given
point is tangent to the field line crossing that point.
3. Field lines point away from positive charges and terminate on
negative charges.
4. Field lines never cross each other.
5. The strength of the field is encoded in the density of the field
lines.

P01 -45
Question
The force between
the two charges is:
0%
25%
25%
50% 1. Attractive
2. Repulsive
3. Can’t tell without more information
4. I don’t know

P01 -46
Answer:
One way to tell is to notice that they both must
be sources (or sinks). Hence, as like particles
repel, the force is repulsive.
You can also see this as tension in the field lines
The force between
the two charges is:
2) Repulsive

Electric Shielding
•All charge on a conductor gathers on the
outside
•If a charge is contained inside a conductor
the electric field is zero
•If a charge is outside a conductor the inside
of the container will not be affected by the
outside charge
•Example: Faraday Cage

Charge Polarization
Electrical
Polarization:
When one side of a
molecule is induced
to be slightly more
positive than the
other side

Capacitor
•A capacitor stores
difference in electric
charge
•Usually stored in
parallel plates
•Consists of two
conductors with an
insulator in between

Capacitors
•Can hold this
separation of charge
for large periods of
time
•This is why it is not
smart to take apart
an Old Television
Set

Electric Dipoles
 An object that is electrically neutral
overall, but permanently polarized, is
called an electric dipole.
 Example: H20 molecule

P01 -54
Electric Dipole
Two equal but opposite charges +q and –q,
separated by a distance 2a
charge×displacement
ˆ ˆ×2 2q a qa

 
p
j j
r
points from negative to positive chargep
r
q
-q
2a
Dipole Moment
p

P01 -55
Why Dipoles?
Nature Likes To Make Dipoles!
Animation

P01 -56
Continuous Charge
Distributions

P01 -57
  ?P E
r
V
Continuous Charge Distributions
i
i
Q q 
Break distribution into parts:
2
ˆe
q
k
r

 E r
r
E field at P due to q
Superposition:
 E E
r r
V
dq 
d  E
r
2
ˆe
dq
d k
r
 E r
r

P01 -58
Continuous Sources: Charge Density
L
Q

Q
A
 
Q
V
 
Length L
L
w
L
Area A wL 
R
L
2
Volume V R Lp 
dLdQ 
dAdQ 
dVdQ 

P01 -59
Examples of Continuous Sources:
Line of charge
L
Q

Length L
L
dLdQ 
Link to
applet

P01 -60
Ring of Charge
2
Q
R

p
dLdQ 
Link to
applet

P01 -61
Ring of Charge
2
Q
R

p
dLdQ 
Link to
applet

P01 -62
Example: Ring of Charge
P on axis of ring of charge, x from center
Radius a, charge density .
Find E at P

P01 -63
In-Class: Uniformly Charged Disk
P on axis of disk of charge, x from center
Radius R, charge density .
Find E at P
( 0)x 

P01 -64
Scaling: E for Plane is Constant
1) Dipole: E falls off like 1/r3
2) Point charge: E falls off like 1/r2
3) Line of charge:E falls off like 1/r
4) Plane of charge: E constant

Electric Flux
We define the electric flux ,
of the electric field E,
through the surface A, as:
 = E . A
Where:
A is a vector normal to the surface
(magnitude A, and direction normal to the surface).
 is the angle between E and A
area A
E
A
 = E A cos ()

Electric Flux
Here the flux is
 = E · A
You can think of the flux through some surface as a measure of
the number of field lines which pass through that surface.
Flux depends on the strength of E, on the surface area, and on
the relative orientation of the field and surface.
Normal to surface,
magnitude A
area A
E
A
E
A 

Electric Flux
The flux also depends on orientation
area A

A cos 
The number of field lines through the tilted surface equals the
number through its projection . Hence, the flux through the tilted
surface is simply given by the flux through its projection: E (A cos.
 = E . A = E A cos 
area A

A cos 
E E
A A

Calculate the flux of the electric field E,
through the surface A, in each of the
three cases shown:
a)  =
b)  =
c)  =


dA
E
What if the surface is curved, or the field varies with position ??
1. We divide the surface into small
regions with area dA
2. The flux through dA is
d = E dA cos 
d = E . dA
3. To obtain the total flux we need
to integrate over the surface A
A
 =  d =  E . dA
 = E . A

In the case of a closed surface
     

d E dA
q
inside
e0
The loop means the integral is over a closed surface.

dA
E

Spherical surface with point charge at center
d E dA     
ur uur
Ñ Ñ
Flux of electric field:
2
0
1
cos
4
q dA
E dA
r

pe
   Ñ Ñ
2
but , then:
dA
d
r
 
0 0 0
4
4 4
q q q
d p
pe pe e
    Ñ
0
Gauss's Law
enclosedq
E dA
e
 
ur uur
Ñ

Gauss’s Law
This is always true.
Occasionally, it provides a very easy way
to find the electric field
(for highly symmetric cases).
The electric flux
through any closed surface
equals  enclosed charge / e0
     

d E dA
q
inside
e0

Calculate the electric field produced
by a point charge using Gauss Law
We choose for the gaussian surface a sphere
of radius r, centered on the charge Q.
Then, the electric field E, has the same
magnitude everywhere on the surface
(radial symmetry)
Furthermore, at each point on the surface,
the field E and the surface normal dA are
parallel (both point radially outward).
E . dA = E dA [cos  = 1]

E
Q
Coulomb’s Law !
E 
1
4pe0
q
r 2
 E . dA = Q / e0
 E . dA = E  dA = E A
A = 4 p r2
E A = E 4 p r2 = Q / e0
Electric field produced
by a point charge
E
Q
k = 1 / 4 p e0
e0 = permittivity
e0 = 8.85x10-12 C2/Nm2

Is Gauss’s Law more fundamental than
Coulomb’s Law?
• No! Here we derived Coulomb’s law for a point charge
from Gauss’s law.
• One can instead derive Gauss’s law for a general (even
very nasty) charge distribution from Coulomb’s law. The
two laws are equivalent.
• Gauss’s law gives us an easy way to solve a few very
symmetric problems in electrostatics.
• It also gives us great insight into the electric fields in and
on conductors and within voids inside metals.

 
r
E  d
r
A =
Qenclosed
e0
Gauss’s Law
The total flux within
a closed surface …
… is proportional to
the enclosed charge.
Gauss’s Law is always true, but is only useful for certain
very simple problems with great symmetry.

Applying Gauss’s Law
Gauss’s law is useful only when the electric field
is constant on a given surface
Infinite sheet of charge
1. Select Gauss surface
In this case a cylindrical
pillbox
2. Calculate the flux of the
electric field through the
Gauss surface
 = 2 E A
3. Equate  = qencl/e0
2EA = qencl/e0
4. Solve for E
E = qencl / 2 A e0 =  / 2 e0
(with  = qencl / A)

GAUSS LAW – SPECIAL SYMMETRIES
SPHERICAL
(point or sphere)
CYLINDRICAL
(line or cylinder)
PLANAR
(plane or sheet)
CHARGE
DENSITY
Depends only on
radial distance
from central point
Depends only on
perpendicular distance
from line
Depends only on
perpendicular distance
from plane
GAUSSIAN
SURFACE
Sphere centered at
point of symmetry
Cylinder centered at
axis of symmetry
Pillbox or cylinder
with axis
perpendicular to plane
ELECTRIC
FIELD E
E constant at
surface
E ║A - cos  = 1
E constant at curved
surface and E ║ A
E ┴ A at end surface
cos  = 0
E constant at end
surfaces and E ║ A
E ┴ A at curved surface
cos  = 0
FLUX 

Cylindrical geometry
Planar geometry
Spherical geometry
E

A charge Q is uniformly distributed through a sphere of radius R.
What is the electric field as a function of r?. Find E at r1 and r2.
Problem: Sphere of Charge Q
r2
r1
R

A charge Q is uniformly distributed through a sphere of radius R.
What is the electric field as a function of r?. Find E at r1 and r2.
Use symmetry!
This is spherically symmetric.
That means that E(r) is radially
outward, and that all points, at a
given radius (|r|=r), have the same
magnitude of field.
r2
r1
R
E(r1)
E(r2)

E & dA
r
R
What is the enclosed charge?
First find E(r) at a point outside the charged sphere. Apply Gauss’s
law, using as the Gaussian surface the sphere of radius r pictured.

r
R
E & dA What is the enclosed charge? Q

r
R
What is the flux through this surface?

r
R
 
r
Ed
r
A  EdA
 E dA  EA  E(4p r
2
)

r
R
 
r
Ed
r
A  EdA
 E dA  EA  E(4p r
2
)
Gauss    Q /eo
Q/e0    E(4p r2
)

r
R
 
r
Ed
r
A  EdA
 E dA  EA  E(4p r
2
)
Gauss:   Q /eo
So
r
E(
r
r) 
1
4peo
Q
r2
ˆr
Exactly as though all the
charge were at the origin!
(for r>R)
Q/e0    E(4p r2
)

R
r
E(r
)
Next find E(r) at a point inside the sphere. Apply Gauss’s law,
using a little sphere of radius r as a Gaussian surface.
What is the enclosed charge?
That takes a little effort. The little sphere has
some fraction of the total charge. What fraction?
That’s given by volume ratio: Qenc 
r3
R3 Q
Again the flux is:  = EA = E(4p r 2
)
Setting   Qenc /eo gives E =
(r3
/R3
)Q
4peor2
r
E(
r
r) =
Q
4peoR3 r ˆrFor r<R

Look closer at these results. The electric field at comes
from a sum over the contributions of all the little bits .
It’s obvious that the net E at this point will be horizontal.
But the magnitude from each bit is different; and it’s completely
not obvious that the magnitude E just depends on the distance
from the sphere’s center to the observation point.
Doing this as a volume integral would be HARD.
Gauss’s law is EASY.
R
Q r
r > R


Problem: Infinite charged plane
Consider an infinite plane with a constant surface charge density 
(which is some number of Coulombs per square meter).
What is E at a point located a distance z above the plane?
x
y
z


Consider an infinite plane with a constant surface charge density 
(which is some number of Coulombs per square meter).
What is E at a point located a distance z above the plane?
x
y
z
Use symmetry!
The electric field must point straight away
from the plane (if  > 0). Maybe the
Magnitude of E depends on z, but the direction
is fixed. And E is independent of x and y.
E

E
E
Gaussian “pillbox”

So choose a Gaussian surface that is a “pillbox”, which has its top
above the plane, and its bottom below the plane, each a distance z
from the plane. That way the observation point lies in the top.
z
z

E
E

z
z
Let the area of the top and bottom be A.

E
E

z
z
Total charge enclosed by box =

E
E

z
z
Total charge enclosed by box = A

E
E

z
z
Outward flux through the top:

E
E

z
z
Outward flux through the top: EA

E
E

z
z
Outward flux through the bottom:

E
E

z
z
Outward flux through the bottom: EA

E
E

z
z
Outward flux through the sides:

E
E

z
z
Outward flux through the sides: E x (some area) x cos(900) = 0

E
E

z
z
Outward flux through the sides: E x (some area) x cos(900) = 0
So the total flux is: 2EA

E
E

z
z
Gauss’s law then says that A/e0=2EA so that E=/2e0, outward.
This is constant everywhere in each half-space!
Notice that the area A canceled: this is typical!

Imagine doing this with an integral over the charge distribution:
break the surface into little bits dA …

dE
Doing this as a surface integral would be HARD.
Gauss’s law is EASY.

Consider a long cylindrical charge distribution of radius R,
with charge density  = a – b r (with a and b positive).
Calculate the electric field for:
a) r < R
b) r = R
c) r > R

• A conductor is a material in which charges can move
relatively freely.
• Usually these are metals (Au, Cu, Ag, Al).
• Excess charges (of the same sign) placed on a conductor
will move as far from each other as possible, since they
repel each other.
• For a charged conductor, in a static situation, all the
charge resides at the surface of a conductor.
• For a charged conductor, in a static situation, the
electric field is zero everywhere inside a conductor, and
perpendicular to the surface just outside
Conductors

Conductors
Why is E = 0 inside a conductor?

Conductors
Conductors are full of free electrons, roughly one per
cubic Angstrom. These are free to move. If E is
nonzero in some region, then the electrons there feel
a force -eE and start to move.

Conductors
Conductors are full of free electrons, roughly one per
cubic Angstrom. These are free to move. If E is
nonzero in some region, then the electrons there feel
a force -eE and start to move.
In an electrostatics problem, the electrons adjust their
positions until the force on every electron is zero (or
else it would move!). That means when equilibrium is
reached, E=0 everywhere inside a conductor.

Conductors
Because E = 0 inside, the inside of a conductor is neutral.
Suppose there is an extra charge inside.
Gauss’s law for the little spherical surface
says there would be a nonzero E nearby.
But there can’t be, within a metal!
Consequently the interior of a metal is neutral.
Any excess charge ends up on the surface.

Electric field just outside a charged conductor
0 0
0
enclosed
E dA EA
q A
EA
E

e e

e
   
 


ur uur
Ñ
The electric field just outside a charged conductor
is perpendicular to the surface and has magnitude E = / e0

Properties of Conductors
In a conductor there are large number of electrons free to move.
This fact has several interesting consequences
Excess charge placed on a conductor moves to the exterior
surface of the conductor
The electric field inside a conductor is zero when charges
are at rest
A conductor shields a cavity within it from external electric
fields
Electric field lines contact conductor surfaces at right angles
A conductor can be charged by contact or induction
Connecting a conductor to ground is referred to as grounding
The ground can accept of give up an unlimited number of electrons

a
b
Problem: Charged coaxial cable
This picture is a cross section of an infinitely long line of charge,
surrounded by an infinitely long cylindrical conductor. Find E.
This represents the line of charge.
Say it has a linear charge density of 
(some number of C/m).
This is the cylindrical conductor. It
has inner radius a, and outer radius b.
Use symmetry!
Clearly E points straight out, and its
amplitude depends only on r.

First find E at positions in the space inside the cylinder (r<a).
Choose as a Gaussian surface:
a cylinder of radius r, and length L.
r
L

What is the charge enclosed?  L
What is the flux through the end caps?  zero (cos900)
What is the flux through the curved face?  E x (area) = E(2prL)
Total flux = E(2prL)
Gauss’s law  E(2prL) = L/e0 so E(r) = / 2pre0
First find E at positions in the space inside the cylinder (r<a).
r
L

Now find E at positions within the cylinder (a<r<b).
Make the same kind of cylindrical Gaussian
surface. Now the curved side is entirely
within the conductor, where E=0; hence the
flux is zero.
There’s no work to do: within a conductor E=0.
Still, we can learn something from Gauss’s law.
r+
Thus the total charge
enclosed by this surface
must be zero.

There must be a net charge per unit length -
attracted to the inner surface of the metal, so
that the total charge enclosed by this Gaussian
surface is zero.
r+
-
-
--
-
-

+
+
r+
-
-
--
-
-
+
++
+
There must be a net charge per unit length
– attracted to the inner surface of the
metal so that the total charge enclosed by
this Gaussian surface is zero.
And since the cylinder is neutral, these
negative charges must have come from
the outer surface. So the outer surface
has a charge density per unit length of
+ spread around the outer perimeter.
So what is the field for r>b?

Ajal electrostatics slides

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Ajal electrostatics slides