2 slides

Dr. Rakhesh Singh Kshetrimayum
2. Electrostatics
Dr. Rakhesh Singh Kshetrimayum
8/11/20141 Electromagnetic FieldTheory by R. S. Kshetrimayum

2.1 Introduction
• In this chapter, we will study
• how to find the electrostatic fields for various cases?
• for symmetric known charge distribution
• for un-symmetric known charge distribution
• when electric potential, etc.
• what is the energy density of electrostatic fields?
8/11/2014Electromagnetic FieldTheory by R. S. Kshetrimayum2
• what is the energy density of electrostatic fields?
• how does electrostatic fields behave at a media interface?
• We will start with Coulomb’s law and discuss how to find
electric fields?
What is Coulomb’s law?
It is an experimental law

2.2 Coulomb’s law and electric field
And it states that the electric force between two
point charges q1 and q2 is
along the line joining them (repulsive for same charges and
attractive for opposite charges)
directly proportional to the product q1 and q2
F
r
directly proportional to the product q1 and q2
inversely proportional to the square of distance r between
them
Mathematically,
1 2 1 2
2 2
q q q q
ˆ ˆ= kF r F r
r r
α ⇒
ur ur 9
0
1
9 10
4
k
πε
= ≅ ×

2.2 Coulomb’s law and electric field
Electric field is defined as the force experienced by a unit positive
charge q kept at that point
Principle of Superposition:
2 2
0 0
1 Qq 1 Q
ˆ ˆ= = = (N/C)
4 4
F
F r E r
r q rπ πε ε
∴
ur
ur ur
Principle of Superposition:
The resultant force on a charge due to collection of charges is
equal to the vector sum of forces
due to each charge on that charge
Next we will discuss
How to find electric field from Gauss’s law?
Convenient for symmetric charge distribution

2.3 Electric flux and Gauss’s law
2.3.1 Electric flux:
We can define the flux of the electric field through an
area to be given by the scalar product .
For any arbitrary surface S, the flux is obtained by
integrating over all the surface elements
ds
r
=d D dsψ •
ur r
integrating over all the surface elements
=
S S
d D dsψ ψ = •∫ ∫
ur r

enclosed
S
QsdD =•= ∫
vr
ψ
ψ
Total electrical flux coming out of a closed surface S is equal to
Gauss’s law
Total electrical flux coming out of a closed surface S is equal to
charge enclosed by the volume defined by the closed
surface S
irrespective of the shape and size of the closed surface

( ) dvQdvDsdD
V
enclosed
VS
∫∫∫ ==•∇=•= ρψ
rvr
ψ
Since it is true for any arbitrary volume, we may equate the two
integrands and write,
Applying divergence theorem,
integrands and write,
Next we will discuss
How to find electric field from electric potential?
Easier since electric potential is a scalar quantity
0
= =D E
ρ
ρ
ε
∇ • ⇒ ∇ •
r r
[First law of Maxwell’s Equations]

2.4 Electric potential
Suppose we move a potential charge q from point A to B in
an electric field
The work done in displacing the charge by a distance
The negative sign shows that the work is done by an external
E
r
dl
r
= - = -qdW F dl E dl• •
ur r ur r
The negative sign shows that the work is done by an external
agent.
The potential difference between two pointsA and B is given
by
= -q
B
A
W E dl∴ •∫
ur r
= = -
B
AB
A
W
E dl
q
φ •∫
ur r

Electric field as negative of gradient of electric
potential:
For 1-D case,
Differentiate both sides with respect to the upper limit of
( ) ( )= - dx
x
x xx E xφ
∞
∫
Differentiate both sides with respect to the upper limit of
integration, i.e., x
Extending to 3-D case, from fundamental theorem of
gradients,
=- E =- Ex
x x x
d
d dx
dx
φ
φ⇒

= - E - E - Ex y zd dx dy dzφ⇒
= + +d dx dy dz
x y z
φ φ φ
φ
∂ ∂ ∂
∂ ∂ ∂
Electric field intensity is negative of the gradient of
E = -x
x
φ∂
∴
∂
E = -y
y
φ∂
∂
E = -z
z
φ∂
∂
= -E φ∇
φ

Maxwell’s second equation for electrostatics:
Electrostatic force is a conservative force,
i.e., the work done by the force in moving a unit charge from
one point to another point
is independent of the path connecting the two points
is independent of the path connecting the two points
1 2
B B
A A
Path Path
E dl E dl• = •∫ ∫
r rr r
=
B A
A B
E dl E dl• − •∫ ∫
r rr r
Q

1 2
+ 0
B A
A B
Path Path
E dl E dl∴ • • =∫ ∫
r rr r
∫ =•⇒ 0ldE
rr
Applying Stoke’s theorem, we have,
∫
( )∫ ∫ =•×∇=•⇒ 0sdEldE
rrrr
0=×∇ E
r [Second law of Maxwell’s
Equations for electrostatics]

2.5 Boundary value problems for
electrostatic fields
Basically there are three ways of finding electric field :
First method is using
Coulomb’s law and
Gauss’s law,
when the charge distribution is known
E
r
when the charge distribution is known
Second method is using ,
when the electric potential is known
E = −∇Φ
r
Φ

Third method
In practical situation,
neither the charge distribution nor the electric potential
is known
Only the electrostatic conditions on charge and potential are
Only the electrostatic conditions on charge and potential are
known at some boundaries and
it is required to find them throughout the space

In such cases, we may use
Poisson’s or
Laplace’s equations or
method of images
for solving boundary value problems
for solving boundary value problems
Poisson’s and Laplace’s equations
vD ρ∇• =
r
v
o
E
ρ
ε
∇• =
r

Since
Poisson’s equation
E = −∇Φ
r
2 v
o
E
ρ
ε
∇ • = −∇ •∇Φ = −∇ Φ =
r
ρ
For charge free condition, Laplace’s equation
2 v
o
ρ
ε
∇ Φ = −
2
0∇ Φ =

Uniqueness theorem:
Solution to
Laplace’s or
Poisson’s equations
can be obtained in a number of ways
can be obtained in a number of ways
For a given set of boundary conditions,
if we can find a solution to

Poisson’s or
Laplace’s equation
satisfying those boundary conditions
the solution is unique
regardless of the method used to obtain the solution
regardless of the method used to obtain the solution

Procedure for solving Poisson’s or Laplace’s
equation:
Solve the
Laplace’s or
using either direct integration
where is a function of one variableΦ

or method of separation of variables
if is a function of more than one variable
Note that this is not unique
since it contains the unknown integration constants
Then, apply boundary conditions
Φ
Then, apply boundary conditions
to determine a unique solution for .
Once is obtained,
We can find electric field and flux density using
Φ
Φ
E = −∇Φ
r
o rD Eε ε=
r r

Method of images:
Q QLρ LρVρ− Vρ−
(a) Point, line and volume charges over a perfectly
conducting plane and its (b) images and equi-potential
surface
Q− Lρ− Vρ

commonly used to find
electric potential,
field and
flux density
due to charges in presence of conductors
due to charges in presence of conductors

States that given a charge configuration above an infinite
grounded perfect conducting plane
may be replaced by the
charge configuration itself,
its image and
its image and
an equipotential surface
A surface in which potential is same is known as
equipotential surface
For a point charge the equipotential surfaces are spheres

In applying image method,
two conditions must always be satisfied:
The image charges must be located within conducting region
and
and
the image charge must be located such that on conducting
surface S,
the potential is zero or constant

For instance,
Suppose a point charge q is held at a distance d above an
infinite ground plane
What is the potential above the plane?
What is the potential above the plane?
Note that the image method doesn’t give correct potential
inside the conductor
It gives correct values for potential above the conductor only

2.6 Electrostatic energy
Assume all charges were at infinity initially,
then, we bring them one by one and fix them in different
positions
To find the energy present in an assembly of charges,
we must first find the amount of work necessary to assemble
we must first find the amount of work necessary to assemble
them
1 2 3W W W W= + +
21 2 3 32 31( )q qΦ × + Φ + Φ=

If the charges were placed in the reverse order
Therefore,
3 2 1W W W W= + +
2 23 1 13 120 ( ) ( )q q+ Φ + Φ + Φ=
Therefore,
In general, if there are n point charges
1
1 1 2 2 3 32 ( )W q q q⇒ = Φ + Φ + Φ
1 13 12 2 23 21 3 32 312 ( ) ( ) ( )W q q q= Φ + Φ + Φ + Φ + Φ ×Φ
1
2
1
n
k k
k
W q
=
= Φ∑

If instead of point charges,
the region has a continuous charge distribution,
the summation becomes integration
For Line charge 1
2 L
L
W dlρ= Φ∫
For surface charge
For volume charge
L
1
2 s
S
W dsρ= Φ∫
1
2 v
V
W dvρ= Φ∫

Since
we have,
From vector analysis,
vD ρ∇• =
r
( )1
2
v
W D dv= ∇• Φ∫
r
Hence
Therefore,
( )D D D∇• Φ = •∇Φ + Φ∇•
r r r
( )( )D D DΦ ∇• = ∇• Φ − •∇Φ
r r r
( ) ( )1 1
2 2
V V
W D dv D dv= ∇• Φ − •∇Φ∫ ∫
r r

Applying Divergence theorem on the 1st integral, we have,
remains as 1/r3 while remains as 1/r2, therefore
the first integral varies as 1/r,
( )dvDsdDW
VS
∫∫ Φ∇•−•Φ=
rrr
2
1
2
1
D
r
Φ sd
r
the first integral varies as 1/r,
tend to zero as the surface becomes large and
tends to be infinite
Hence
( )1
2
V
W D dv= − •∇Φ∫
r
21 1
2 2 o
V V
D E dv E dvε• =∫ ∫
r r

The integral E2 can only increase (the integrand being
positive)
Note that the integral and is over the region
where the charge is located,
so any larger volume would do just as well
1
2 v
V
W dvρ= ∫
so any larger volume would do just as well
The extra space and volume will not contribute to the
integral
Since for those regions0=vρ

the energy density in electrostatic field is
2
21 1
2 2
2
o
oV V
dW d d D
w D E dv E dv
dv dv dv
ε
ε
   
= = • = =   
   
∫ ∫
r r

2.7 Boundary conditions for electrostatic
fields
Two theorems or
Maxwell’s first and
second equations in integral form
are sufficient to find the boundary conditions
2.7.1 Boundary conditions for electric field
2.7.1 Boundary conditions for electric field
Let us consider the small rectangular contour PQRSP (see
Fig. 2.8
l is chosen such that E1t and E2t are constant along this
length

2.7 Boundary conditions for
S∆
S∆
σ
Fig. 2.8 Boundary for electrostatic fields at the interface of
two media

2.7 Boundary conditions for electrostatic
fields
Note that h 0 at the boundary interface and
therefore there is no contribution from QR and SP in the above line
integral
Also note that the direction of the line integral along PQ and RS are
in the opposite direction
The tangential component of electric field vector is continuous
at the interface
tt
tt
C
S
R
Q
P
EE
lElEldEldEldE
21
2122110
=⇒
−=•+•==•∫ ∫∫
rrrrrr
Q

2.7.2 Boundary conditions for electric flux density
Let us consider a small cylinder at the interface
Cross section of the cylinder must be such that
vector is the same
Note that h 0 at the boundary interface
D
r
Note that h 0 at the boundary interface
therefore, there are no contribution from the curved surface
of the pillbox in the above surface integral
So only the top and bottom surfaces remains in the surface
integral

The normal is in the upward direction in the top surface
and downward direction in the bottom surface
∫∫∫ =•+•=•
surfacebottom
enclosed
surfacetoppillbox
QsdDsdDsdD 2211
rrrrrr
and downward direction in the bottom surface
the normal component of electric flux density can only
change at the interface
if there is charge on the interface, i.e., surface charge is
present
2 1 2 1S S = Sn n n nD D D Dσ σ⇒ ∆ − ∆ ∆ ⇒ − =

If medium 2 is dielectric and medium 1 is conductor
Then in conductor D1=0 and hence D2n=σ
or in general case, Dn=σ

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