1 electromagnetic field theory by r. s. kshetrimayum … · maxwell’s second equation for...

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?

8/11/2014Electromagnetic Field Theory 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

Fr


� 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


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


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


�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,


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


� 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


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 = - Exx x x

dd dx

dx

φφ⇒


= - E - E - Ex y zd dx dy dzφ⇒

= + + d dx dy dzx y z

φ φ φφ

∂ ∂ ∂

∂ ∂ ∂


� Electric field intensity is negative of the gradient of

E = -x

x

φ∂∴

∂E = -

yy

φ∂

∂E = -

zz

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 APath 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 BPath Path

E dl E dl∴ • • =∫ ∫r rr r

∫ =•⇒ 0ldErr


� 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


� when the charge distribution is known

� Second method is using ,

� when the electric potential is knownE = −∇Φr

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

Eρ

ε∇ • =

r



� Since

� Poisson’s equation

E = −∇Φr

2 v

o

Eρ

ε∇ • = −∇ •∇Φ = −∇ Φ =

r

ρ


� For charge free condition, Laplace’s equation

2 v

o

ρ

ε∇ Φ = −

20∇ 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 = −∇Φr

o rD Eε ε=r r


electrostatic fields � Method of images:

Q QLρ

LρV

ρ−V

ρ−


� (a) Point, line and volume charges over a perfectly conducting plane and its (b) images and equi-potential surface

Q− Lρ−V

ρ



� 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

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∑


� 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∫


� For surface charge

� For volume charge

L

12

s

S

W dsρ= Φ∫

12

v

V

W dvρ= Φ∫


� Since

� we have,

� From vector analysis,

vD ρ∇ • =r

( )12

v

W D dv= ∇ • Φ∫r


� 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


� 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


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


� 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= ∫


� 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

221 1

2 2

2o

oV V

dW d d Dw D E dv E dv

dv dv dvε

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 interfaceDr


� 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 1 S S = S

n 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


1 electromagnetic field theory by r. s. kshetrimayum … · maxwell’s second equation for...

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