Download 2. Electrostatics

2. Electrostatics
Dr. Rakhesh Singh Kshetrimayum
1
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
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?
• 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
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.2 Coulomb’s law and electric field
r
And it states that the electric force F 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
inversely proportional to the square of distance r between
them
Mathematically,
ur
ur
q1q 2
q1q 2
ˆ
Fα
r
⇒
F
=
k
rˆ
2
2
r
r
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Electromagnetic Field Theory by R. S. Kshetrimayum
1
k=
≅ 9 ×109
4πε 0
8/11/2014
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
ur
ur
ur F
1 Qq
1 Q
ˆ
F=
r∴ E =
=
rˆ (N/C)
2
2
4πε 0 r
q
4πε 0 r
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
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.3 Electric flux and Gauss’s law
2.3.1 Electric flux:
We can define the flux of the electric field throughuran r
r
area d s to be given by the scalar product dψ = D • d.s
For any arbitrary surface S, the flux is obtained by
integrating over all the surface elements
ur r
ψ = ∫ dψ = ∫ D • d s
S
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S
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
ψ
2.3 Electric flux and Gauss’s law
Gauss’s law
r v
ψ = ∫ D • ds = Qenclosed
S
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
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
ψ
2.3 Electric flux and Gauss’s law
Applying divergence theorem,
r v
r
ψ = ∫ D • ds = ∫ ∇ • D dv = Qenclosed = ∫ ρdv
(
S
)
V
V
Since it is true for any arbitrary volume, we may equate the two
integrands and write,
r
r
ρ
∇•D = ρ ⇒ ∇•E =
ε0
[First law of Maxwell’s Equations]
Next we will discuss
How to find electric field from electric potential?
Easier since electric potential is a scalar quantity
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.4 Electric potential
Suppose we movera potential charge q from point A to B in
an electric field E
r
The work done in displacing the charge by a distance dl
ur r
ur r
dW = - F • dl = -q E • d l
The negative sign shows that the work is done by an external
agent.
B ur
r
∴W = -q ∫ E • dl
A
The potential difference between two points A and B is given
by
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φ AB
B ur
r
W
=
= -∫ E • dl
q
A
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.4 Electric potential
Electric field as negative of gradient of electric
potential:
For 1-D case,
x
φx ( x ) = - ∫ Ex ( x ) dx
∞
Differentiate both sides with respect to the upper limit of
integration, i.e., x
dφx
= - Ex ⇒ dφx = - Exdx
dx
Extending to 3-D case, from fundamental theorem of
gradients,
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.4 Electric potential
⇒ dφ = - E x dx - E y dy - E z dz
∂φ
∂φ
∂φ
dφ =
dx +
dy +
dz
∂x
∂y
∂z
∂φ E = - ∂φ
∴ Ex = y
∂y
∂x
∂φ
Ez = ∂z
E = -∇φ
Electric field intensity is negative of the gradient of
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Electromagnetic Field Theory by R. S. Kshetrimayum
φ
8/11/2014
2.4 Electric potential
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
r r
E • dl =
B
∫
A
Path 1
r r
E • dl
B
∫
A
Path 2
A
r r
r r
Q ∫ E • dl = − ∫ E • dl
B
A
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B
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.4 Electric potential
B
∴
∫
r r
E • dl +
A
Path 1
A
∫
r r
E • dl = 0
B
Path 2
r r
⇒ ∫ E • dl = 0
Applying Stoke’s theorem, we have,
r r
r
r
⇒ ∫ E • dl = ∫ ∇ × E • ds = 0
(
r
∇× E = 0
12
)
[Second law of Maxwell’s
Equations for electrostatics]
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
r
Basically there are three ways of finding electric field E :
First method is using
Coulomb’s law and
Gauss’s law,
when the charge distribution is known
Second method is using
r
E = −∇Φ,
when the electric potential Φ is known
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
Third method
In practical situation,
neither the charge distribution nor the electric potential
is known
Only the electrostatic conditions on charge and potential are
known at some boundaries and
it is required to find them throughout the space
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
In such cases, we may use
Poisson’s or
Laplace’s equations or
method of images
for solving boundary value problems
Poisson’s and Laplace’s equations
r
∇ • D = ρv
r ρv
∇•E =
εo
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
r
Since E = −∇Φ
r
ρ
∇ • E = −∇ • ∇Φ = −∇ 2 Φ = v
Poisson’s equation
εo
ρv
∇ Φ=−
εo
For charge free condition, Laplace’s equation
2
∇ 2Φ = 0
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
Uniqueness theorem:
Solution to
Laplace’s or
Poisson’s equations
can be obtained in a number of ways
For a given set of boundary conditions,
if we can find a solution to
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
Poisson’s or
Laplace’s equation
satisfying those boundary conditions
the solution is unique
regardless of the method used to obtain the solution
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
Procedure for solving Poisson’s or Laplace’s
equation:
Solve the
Laplace’s or
Poisson’s equation
using either direct integration
where Φ
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is a function of one variable
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
or method of separation of variables
if Φ 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
to determine a unique solution for
Once
Φ.
is obtained,
We can find electric field and flux density using
Φ
r
E = −∇Φ
r
r
D = ε oε r E
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
Method of images:
Q
ρL
− ρV
Q
ρL
− ρV
−Q
−ρL
ρV
(a) Point, line and volume charges over a perfectly
conducting plane and its (b) images and equi-potential
surface
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
commonly used to find
electric potential,
field and
flux density
due to charges in presence of conductors
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
States that given a charge configuration above an infinite
grounded perfect conducting plane
may be replaced by the
charge configuration itself,
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
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
In applying image method,
two conditions must always be satisfied:
The image charges must be located within conducting region
and
the image charge must be located such that on conducting
surface S,
the potential is zero or constant
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.5 Boundary value problems for
electrostatic fields
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?
Note that the image method doesn’t give correct potential
inside the conductor
It gives correct values for potential above the conductor only
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
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
them
W = W1 + W2 + W3
=
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Φ 21 × q2 + q3 (Φ 32 + Φ 31 )
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.6 Electrostatic energy
If the charges were placed in the reverse order
W = W3 + W2 + W1
= 0 + q2 (Φ 23 ) + q1 (Φ13 + Φ12 )
Therefore,
2W = q1 (Φ13 + Φ12 ) + q2 (Φ 23 + Φ 21 ) + q3 (Φ 32 × Φ 31 )
⇒ W = 12 (q1Φ1 + q2 Φ 2 + q3Φ 3 )
n
In general, if there are n point charges
W=
1
2
∑q Φ
k
k
k =1
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.6 Electrostatic energy
If instead of point charges,
the region has a continuous charge distribution,
the summation becomes integration
For Line charge
W=
1
2
∫ ρ Φ dl
L
L
For surface charge
W=
1
2
∫ ρ Φ ds
s
S
For volume charge
W=
1
2
∫ ρ Φ dv
v
V
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.6 Electrostatic energy
r
Since ∇ • D = ρv
we have,
W=
1
2
∫(
r
∇ • D Φ dv
)
v
From vector analysis,
r
r
r
∇ • ΦD = D • ∇Φ + Φ∇ • D
(
Hence
)
r
r
r
Φ (∇ • D) = ∇ • ΦD − D • ∇Φ
(
)
Therefore,
r
r
1
W = ∫ ∇ • ΦD dv − 2 ∫ D • ∇Φ dv
(
1
2
V
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Electromagnetic Field Theory by R. S. Kshetrimayum
)
(
)
V
8/11/2014
2.6 Electrostatic energy
Applying Divergence theorem on the 1st integral, we have,
r r 1 r
1
W = ∫ ΦD • ds − ∫ D • ∇Φ dv
2S
2V
(
)
r
r
ΦD remains as 1/r3 while ds remains as 1/r2, therefore
the first integral varies as 1/r,
tend to zero as the surface becomes large and
tends to be infinite
Hence
W =−
1
2
∫(
r
D • ∇Φ dv
)
V
1
2
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r r
2
1
D
•
E
dv
=
ε
E
dv
2 o∫
∫
V
Electromagnetic Field Theory by R. S. Kshetrimayum
V
8/11/2014
2.6 Electrostatic energy
The integral E2 can only increase (the integrand being
positive)
W = 12 ∫ ρv dv
Note that the integral
and is over the region
V
where the charge is located,
so any larger volume would do just as well
The extra space and volume will not contribute to the
integral
Since ρ v = 0 for those regions
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.6 Electrostatic energy
the energy density in electrostatic field is
 D2
dW d  1 r r  d  1
2
w=
=  2 ∫ D •E dv  =  2 ε o ∫ E dv  =
dv dv  V
V
 dv 
 2ε o
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
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
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
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.7 Boundary conditions for
electrostatic fields
∆S
σ
∆S
Fig. 2.8 Boundary for electrostatic fields at the interface of
two media
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.7 Boundary conditions for electrostatic
fields
Note that h0 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
Q
r S r
r
r r
r
Q ∫ E • dl = 0 = ∫ E1 • dl1 + ∫ E2 • dl2 =E1t l − E2t l
C
P
R
⇒ E1t = E2t
The tangential component of electric field vector is continuous
at the interface
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.7 Boundary conditions for
electrostatic fields
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
36
r
vector D is the same
Note that h0 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
Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.7 Boundary conditions for
electrostatic fields
r r
∫ D • ds =
pillbox
r
r
∫ D1 • ds1 +
top surface
r
r
∫ D2 • ds2 = Qenclosed
bottom surface
The normal is in the upward direction in the top surface
and downward direction in the bottom surface
⇒ D2 n ∆S − D1n ∆S =σ ∆S ⇒ D2 n − D1n = σ
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
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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014
2.7 Boundary conditions for
electrostatic fields
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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Electromagnetic Field Theory by R. S. Kshetrimayum
8/11/2014