Dr. Rakhesh Singh Kshetrimayum
2. Electrostatics
Dr. Rakhesh Singh Kshetrimayum
8/11/20141 Electromagnetic Field Theory 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?
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• 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
Fr
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� 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 qˆ ˆ = kF r F r
r rα ⇒
ur ur 9
0
1 9 104
kπε
= ≅ ×
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 20 0
1 Qq 1 Qˆ ˆ = = = (N/C)
4 4
FF r E r
r q rπ πε ε∴
urur ur
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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
d sr
= d D d sψ •ur r
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integrating over all the surface elements
= S S
d D d sψ ψ = •∫ ∫ur r
2.3 Electric flux and Gauss’s law
enclosed
S
QsdD =•= ∫vr
ψ
ψ
�Total electrical flux coming out of a closed surface S is equal to
Gauss’s law
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�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
2.3 Electric flux and Gauss’s law
( ) dvQdvDsdDV
enclosed
VS
∫∫∫ ==•∇=•= ρψrvr
ψ
�Since it is true for any arbitrary volume, we may equate the two integrands and write,
�Applying divergence theorem,
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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 Eρ
ρε
∇ • ⇒∇ •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
Er
dlr
= - = -q dW F dl E dl• •ur r ur r
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� The negative sign shows that the work is done by an external agent.
� The potential difference between two points A and B is given by
= -q
B
A
W E dl∴ •∫ur r
= = -
B
AB
A
WE dl
qφ •∫
ur r
2.4 Electric potential
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∞
∫
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� 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 = - Exx x x
dd dx
dx
φφ⇒
2.4 Electric potential
= - E - E - Ex y zd dx dy dzφ⇒
= + + d dx dy dzx y z
φ φ φφ
∂ ∂ ∂
∂ ∂ ∂
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� Electric field intensity is negative of the gradient of
E = -x
x
φ∂∴
∂E = -
yy
φ∂
∂E = -
zz
φ∂
∂
= -E φ∇
φ
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
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� is independent of the path connecting the two points
1 2
B B
A APath Path
E dl E dl• = •∫ ∫r rr r
=
B A
A B
E dl E dl• − •∫ ∫r rr r
Q
2.4 Electric potential
1 2
+ 0
B A
A BPath Path
E dl E dl∴ • • =∫ ∫r rr r
∫ =•⇒ 0ldErr
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� Applying Stoke’s theorem, we have,
∫
( )∫ ∫ =•×∇=•⇒ 0sdEldErrrr
0=×∇ Er [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
Er
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� when the charge distribution is known
� Second method is using ,
� when the electric potential is knownE = −∇Φr
Φ
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
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� Only the electrostatic conditions on charge and potential are known at some boundaries and � it is required to find them throughout the space
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
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� for solving boundary value problems
� Poisson’s and Laplace’s equations
vD ρ∇ • =r
v
o
Eρ
ε∇ • =
r
2.5 Boundary value problems for
electrostatic fields
� Since
� Poisson’s equation
E = −∇Φr
2 v
o
Eρ
ε∇ • = −∇ •∇Φ = −∇ Φ =
r
ρ
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� For charge free condition, Laplace’s equation
2 v
o
ρ
ε∇ Φ = −
20∇ Φ =
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
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� can be obtained in a number of ways
� For a given set of boundary conditions,
� if we can find a solution to
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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� regardless of the method used to obtain the solution
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
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� Poisson’s equation
� using either direct integration
� where is a function of one variableΦ
2.5 Boundary value problems for
electrostatic fields
� 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
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� 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 = −∇Φr
o rD Eε ε=r r
2.5 Boundary value problems for
electrostatic fields � Method of images:
Q QLρ
LρV
ρ−V
ρ−
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� (a) Point, line and volume charges over a perfectly conducting plane and its (b) images and equi-potential surface
Q− Lρ−V
ρ
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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� due to charges in presence of conductors
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
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� 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
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
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and
� the image charge must be located such that on conducting surface S, �the potential is zero or constant
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?
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�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
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� 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=
2.6 Electrostatic energy
� 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=
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� Therefore,
� In general, if there are n point charges
11 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
12
1
n
k k
k
W q=
= Φ∑
2.6 Electrostatic energy
� If instead of point charges, � the region has a continuous charge distribution,
� the summation becomes integration
� For Line charge 12
L
L
W dlρ= Φ∫
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� For surface charge
� For volume charge
L
12
s
S
W dsρ= Φ∫
12
v
V
W dvρ= Φ∫
2.6 Electrostatic energy
� Since
� we have,
� From vector analysis,
vD ρ∇ • =r
( )12
v
W D dv= ∇ • Φ∫r
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� Hence
� Therefore,
( )D D D∇ • Φ = •∇Φ + Φ∇ •r r r
( )( )D D DΦ ∇ • = ∇ • Φ − •∇Φr r r
( ) ( )1 12 2
V V
W D dv D dv= ∇ • Φ − •∇Φ∫ ∫r r
2.6 Electrostatic energy
� 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,
( )dvDsdDWVS
∫∫ Φ∇•−•Φ=rrr
2
1
2
1
Dr
Φ sdr
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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 12 2
o
V V
D E dv E dvε• =∫ ∫r r
2.6 Electrostatic energy
� 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
12 v
V
W dvρ= ∫
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� 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
2.6 Electrostatic energy
� the energy density in electrostatic field is
221 1
2 2
2o
oV V
dW d d Dw D E dv E dv
dv dv dvε
ε
= = • = =
∫ ∫r r
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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
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� 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
electrostatic fields
S∆
S∆
σ
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� 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
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� 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 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
� vector is the same
Note that h�0 at the boundary interfaceDr
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� 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
2.7 Boundary conditions for
electrostatic fields
� The normal is in the upward direction in the top surface � and downward direction in the bottom surface
∫∫∫ =•+•=•surfacebottom
enclosed
surfacetoppillbox
QsdDsdDsdD 2211
rrrrrr
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� 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 1 S S = S
n n n nD D D Dσ σ⇒ ∆ − ∆ ∆ ⇒ − =
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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