EMT Electromagnetic Theory MODULE II

MODULE II

ELECTROSTATICS

Compiled by MKP for CEC S5 Batch July 2008

Syllabus – Module IIElectrostatics:Electrostatic fields – Coulomb’s law and field intensity. Electric fields due to continuous charge distributions, Electric flux density, Gauss’s law, Applications of Gauss’s law, Electric potential, Relationship between E and V, Electric dipole, Energy density in electrostatic fields. Electric fields in material space – Properties of materials, Convection and conduction currents, Conductors, Polarization in dielectrics, Dielectric constant and strength, Linear, isotropic and homogeneous dielectrics, Continuity equation, Relaxation time, Boundary conditions. Electrostatic boundary value problems – Poisson’s and Laplace’s Equations, Uniqueness theorem, Resistance and capacitance – Parallel plate, coaxial, spherical capacitors.


ReferencesText Books:

1. Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford University Press

2. Jordan and Balmain, Electromagnetic waves and radiating systems,Pearson Education PHI Ltd.References:

1. Kraus Fleisch, Electromagnetics with applications, McGraw Hill2. William.H.Hayt, Engineering Electromagnetics, Tata McGraw Hill3. N.Narayana Rao, Elements of Engineering Electromagnetics, Pearson

Education PHI Ltd. 4. D.Ganesh Rao, Engineering Electromagnetics, Sanguine Technical

Publishers.5. Joseph.A.Edminister, Electromagnetics, Schaum series-McGraw Hill


ReferencesReferences:

1.K.D.Prasad, Electromagnetic fields and waves, Sathya Prakashan2.Syed Nazar, 2000 solved problems in Electromagnetics, McGraw Hill


Coulombs Law and field intensity

1Q 2Q

R

ˆ Ra

1 22

04ˆ R

QF aQRπε

=


Coulombs Law and field intensityCoulombs law states that the force F between two point charges Q1

and Q2 separated in a vacuum or free space by a distance which is large when compared to their size is:

Along the line joining Q1 and Q2

Directly proportional to the product Q1 Q2 of the charges. Inversely proportional to the square of the distance between them.

1 22

Q QF kR

=

1 2Q ,Q Quantity of positive or negative charges ⇒

1 2 R Separation between charges Q and Q⇒

k Proportionality constant ⇒

coulombs meWhen Q is in and R is in the constant k is found toters be

0

14

kπε

=



Incorporating these values

The force acts along the line joining Q1 and Q2. In order to incorporate this information we may write

0

14

kπε

=

120 8.854 10 /the permitivity of free space which s F miε −×⇒

9

0

1 9 10 /4

k m Fπε

= = ×

1 22

04Q QF

Rπ ε=

1 22

04ˆ R

QF aQRπε

=



ˆ RWhere is a unit vector in the direction of the fa orce

1 22

04ˆ R

QF aQRπε

=

12 2 1 The force F on Q due to Q is given by

1 2

1 21 2 2

0

ˆ4 R

Q QFR

aπ ε

=

12 1 2ˆ RWhere is a unit vector directed from Q to Qa

1 2 If r and r are the position vectors of the points where

1 2 Q and Q is situated

1 2 12 2 1R Vector joining Q and Q r ris = −



Origin

1r

2r

12R

12F

21F1Q

2Q

12

12

1

1

2

2ˆRRaR

RR

= =


Coulombs Law and field intensity12R R= 12

12

1

1

2

2ˆRRaR

RR

= =

1 2 1 21 2 2

0

4

Q Q RN o w FR Rπ ε

⎛ ⎞= ⎜ ⎟

⎝ ⎠1 2

1 2304

Q Q RRπ ε

=

1 21 2 1 23

04Q QF R

Rπ ε=

2121 21 2 1 12 ˆ RForce F on Q due to Q is given b Fy F a=

( )2 1 1 2

1 2 1 22 1 2 2

0 0

ˆ ˆ4 4R R

Q Q Q QFR R

a aπ ε π ε

= −=

( )1 2 2 11 2 3

0 2 14Q Q r r

Fr rπ ε

−=

−


Coulombs Law and field intensityIf there are N point charges Q1,Q2,……….QN located at points with position vectors the resultant force on a charge Q located at point with position vector is the vector sum of the forces exerted on Q by each of the charges Q1,Q2,……….QN

1, 2,...... Nr r r Fr

1 1 2 23 3 3

0 1 0 2 0

( ) ( ) ( )4 4 4

N N

N

QQ r r QQ r r QQ r rFr r r r r rπε πε πε− − −

= + + ⋅⋅ ⋅ ⋅ ⋅ +− − −

310

1 ( )4

Nk k

k k

QQ r rFr rπε =

−=

−∑


Example 1A 2mC positive charge is located at P1(3,-2,-4) in vacuum and a 5 μC negative charge is located at P2(1,-4,2). Find the force on the negative charge.

31 2 10Q C−= + × 6

2 5 10Q C−= − ×

1 ˆ ˆ ˆ3 2 4x y zr a a a= − − 2 ˆ ˆ ˆ4 2x y zr a a a= − +

12 2 1 ˆ ˆ ˆ2 2 6x y zR r r a a a= − = − − +

12 12 4 4 36 44R R= = + + =

:Solution


Example 1

( ) ( ) ( )3 6

1 2 1 2

2 1 0 5 1 0ˆ ˆ ˆ0 .3 0 .3 0 .9

4 8 .8 5 4 1 0 4 4 x y zF a a aπ

− −

−

× − ×= − − +

× × ×

( )ˆ ˆ ˆ2 .0 4 3 0 .3 0 .3 0 .9x y za a a= − − − +

1 2 ˆ ˆ ˆ0 .6 1 0 .6 1 1 .8 4x y zF a a a= + −

1 2

1 21 2 2

0 1 24ˆ RaQ QF

Rπ ε=

12

12

12

ˆ ˆ ˆ2 2 6ˆ

44x y z

R

a a aRaR

− − += = ˆ ˆ ˆ0.3 0.3 0.9x y za a a= − − +


Electric field intensityElectric field intensity at a point is defined as the force on a unit positive test charge placed at that point.

The direction of electric field intensity is the same as that of the force and is measured in newtons/coulombThe electric field intensity at a point with p.v due to a point charge at a point with p.v is obtained as

FEQ

=

r1r

204

ˆ RQ aE

Rπ ε= 1

30 1

( )4

Q r rr rπ ε−

=−

E


Electric field intensityIf there are N point charges Q1,Q2,……….QN located at points with position vectors the electric field intensity at point with position vector is the vector sum of the electric field intensities produced by charges Q1,Q2,……….QN

1, 2,...... Nr r r Er

1 1 2 23 3 3

0 1 0 2 0

( ) ( ) ( )4 4 4

N N

N

Q r r Q r r Q r rEr r r r r rπε πε πε− − −

= + + ⋅ ⋅ ⋅ ⋅ ⋅ +− − −

310

1 ( )4

Nk k

k k

Q r rEr rπε =

−=

−∑


Example 2Point charges 1mC and -2mC are located at (3,2,-1) and (-1,-1,4) respectively. Calculate the electric force on a 10 nC charge located at (0,3,1) and the electric field intensity at that point.

310

1 ( )4

Nk k

k k

QQ r rFr rπε =

−=

−∑

:Solution

1 1 2 23 3

0 1 2

( ) ( )4

Q Q r r Q r rFr r r rπε

⎛ ⎞− −= +⎜ ⎟⎜ ⎟− −⎝ ⎠

[ ] [ ]93 3

3 30

(0,3,1) (3,2, 1) (0,3,1) ( 1, 1,4)10 10 10 2 104 (0,3,1) (3,2, 1) (0,3,1) ( 1, 1,4)

Fπε

−− −

⎧ ⎫− − − − −× ⎪ ⎪= − ×⎨ ⎬− − − − −⎪ ⎪⎩ ⎭

( ) ( )

3 9

9 3/2 3/210 10 10 ( 3,1,2) 2(1,4, 3)

10 9 1 4 1 16 9436

ππ

− −

−

⎧ ⎫× × − −⎪ ⎪= −⎨ ⎬+ + + +⎪ ⎪⎩ ⎭


Example 2

2 ( 3,1,2) ( 2, 8,6)9 1014 14 26 26

F − − − −⎧ ⎫= × +⎨ ⎬⎩ ⎭

( )ˆ ˆ ˆ6 .5 0 7 3 .8 1 7 7 .5 0 6x y zF a a a m N= − − +

FAt this point E =Q

( )ˆ ˆ ˆ6 5 0 .7 3 8 1 .7 7 5 0 .6 /x y zE a a a K V m= − − +

( ) 3

9

ˆ ˆ ˆ6 .507 3.817 7.506 1010 10

x y za a a −

−

− − + ×=

×


Electric fields due to continuous charge distributions

+ + + ++ + + ++ + + +

+++++++

+

+ ++ + + +++ + ++++ + +++ +

POINTCHARGE

LINECHARGE

SURFACECHARGE

VOLUMECHARGE

Lρ

Sρ Vρ

Q


Electric fields due to continuous charge distributions

A charge may be located on a point, along a line, on a surface or in a volume. Accordingly we have four types of charge distributions:

Point chargesLine chargesSurface chargesVolume charges.

Point Charge: A charge that is located on a body whose dimensions are much smaller than other relevant dimensions is called point charge. A collection of charges on a pinhead may be considered as a point charge.

ii

Total Charge Q Q=∑


Charge distributionsLine Charge: A charge that is distributed along a fine line, as in the case of a sharp electron beam in a cathode ray tube, is considered as a line charge distribution. It is convenient to associate a line charge density with a line charge distribution.

Surface Charge: A charge that is distributed over a surface is considered as a surface charge distribution. It is convenient to associate a surface charge density with a surface charge distribution.

LCharge eleme t dQ dln ρ=

Lρ

LLTotal charge Q dlρ= ∫

Sρ

SCharge eleme t dQ dSn ρ= SSTotal charge Q dSρ= ∫


Charge distributionsVolume Charge: A charge that is distributed throughout a specified volume is considered as a volume charge distribution. It is convenient to associate a volume charge density with avolume charge distribution.

Lρ

VCharge eleme t dQ dVn ρ= VVTotal charge Q dVρ= ∫


Electric field intensity of Charge distributions

Electric field intensity of a point charge is given by

By replacing the charge Q by charge elements and integrating we get the electric field intensity of various charge distributions.

204

ˆ RQ aE

Rπ ε=

20

ˆ 4

LRF or line charge distributions, d lE

Raρ

πε= ∫

20

4

ˆ RSF or surface charge distrib dSution E as,

Rρπε

= ∫

20

4

ˆ RVF or volum e charge distribu dVtions E a,

Rρπε

= ∫


Electric field intensity of a finite line charge+++++++++++++++++

1dE

2dE

dE

LdQ dzρ= dz

dzLdQ dzρ=

/L C mρ

2Lz = +

2Lz = −

Z

R

R

zρ

aρ

αα

αα

P

1ZdE

2ZdE


Electric field intensity of a finite line charge

dzTake two elemental lengths of the line charge that are symmetr ical about the origin.

Ldzz dQd ρ=The charge associated with each elemental length is and can be treated as a point charge.

1 2dE dE The vertical components of and gets cancelled leaving only the aρradial component along

1 2dE dEThe radial components of and gets add ed .

φThere is no variation of the field along direc tion .

aρSo we need to calculate only the radial component along

Align the line charge along the Z axis symmetrically with respect to theorigin.



1 2dE dE dE= + Radial components of ( )

20

2 c o s4

ˆd QR

a ραπ ε

= ×

c o s P u tRρα =

30

24

ˆd Qd ER

a ρρ

π ε= ×

2 2 R =P u t zρ + ( )3 22 204

ˆ2 L d zd Ez

a ρρ ρ

π ε ρ= ×

+

The electric field intensity at P due to the entire line charge is

( )/ 2

3 22 200

124

ˆL

L

z

E d zz

a ρρ ρπ ε ρ=

=+

∫


Electric field intensity of a finite line chargeUsing the standard in tegral

( )3 / 2 2 2 22 2

1 xd xa a xa x

=++

∫/ 2

2 2 20 0

2ˆ

L

L zEz

a ρρ ρπ ε ρ ρ

⎡ ⎤= × ⎢ ⎥

+⎢ ⎥⎣ ⎦

2 2 20

ˆ2 4

L L aL

ρρ ρπ ε ρ ρ

⎡ ⎤= × ⎢ ⎥

+⎢ ⎥⎣ ⎦

2 202 4

ˆL L aEL

ρρπ ε ρ ρ

⎡ ⎤= ⎢ ⎥

+⎢ ⎥⎣ ⎦



2 202 4

ˆ L LEFor a finite line char aL

ge ρρπε ρ ρ

⎡ ⎤= ⎢ ⎥

+⎢ ⎥⎣ ⎦

→∞For an infinite line charge, L and

2 20

1lim2 4 1

ˆLL

aEL

ρρπ ε ρ ρ→ ∞

⎡ ⎤= ⎢ ⎥

+⎢ ⎥⎣ ⎦

0

ˆ2

LE a ρρπ ε ρ

=

0

ˆ2

LFor an infinite line cha Erge a ρρπε ρ

=


Electric field intensity of an infinite surface charge

••

X

Y

Z

αα

αα

R

R

P

1dE

2dE

dE

2( / )S C mρL Sdzρ ρ=

'L Sdzρ ρ=

d

•


Electric field intensity of a infinite surface charge

Assume that the infinite sheet charge is located in the x - z plane.

Assume that the infinite sheet charge is composed of line cha rge /L Sdz C mρ ρ=distributions with density

The electric field intensity of an infinite line charge is radially direct ed away from the line charge and its magnitude is

02Ld E

Rρπ ε

=02

S d zR

ρπ ε

=

' LρConsider one more line charge symmetrically located with respect to the origin

1 2dE dE The vertical components of and gets cancelled leaving only the ˆ yaradial component along



1 2dE dE dE= + y components of ( )0

ˆ2 c o s2

Sy

d z aR

ρ απ ε

= ⋅

c o s dP u tR

α =

20

ˆSy

dd E d z aR

ρπ ε

=

The electric field intensity at P due to the entire sheet charge is

200

ˆSyz

dE d z aR

ρπ ε

∞

== ∫

2 2 R =P u t d z+



( )2 200

ˆSyz

dE d z ad zρ

π ε∞

==

+∫

( )2 200

1 ˆSyz

d d z ad z

ρπ ε

∞

==

+∫

Using the standard in tegral ( )1

2 2

1 1 ta n zd xa aa x

− ⎛ ⎞= ⎜ ⎟+ ⎝ ⎠∫

1

00

1 ˆta nSy

d zE ad d

ρπ ε

∞−⎡ ⎤= ⎢ ⎥⎣ ⎦ 0

ˆ2

Syaρ π

π ε= ⋅

0

ˆ2

Syaρ

ε=

0

ˆ 02

SyE yaρ

ε= >

0

0ˆ 2

SyE yaρ

ε= − <



The electric field of a sheet charge is normal to the plane of the sheet and is independent of the distance between the sheet and the point of observation.In a parallel plate capacitor, the electric field existing between two plates having equal and opposite charges is given by

0

ˆ2

SyE aρ

ε=

( )0 0

ˆ ˆ 2 2

S Sy yE a aρ ρ

ε ε−

= + −0

ˆ Syaρ

ε=

0

ˆ SyE aρ

ε=


Electric fluxIf a positive test charge is brought near another positive charge it will be repelled in a direction along the line joining the two charges.If the second charge is moved around the first one, it will be repelled in radially outward direction at all points.The lines drawn to trace the force on a test charge is called lines of force.Such lines are called electric flux in electrostatics. A charge of Q Coulombs produces an electric flux of ψ Coulombs.

-Q+Q

Qψ = Qψ =


Electric flux densityThe electric field around a point charge is given by

If this equation is multiplied by ε0, we get

The RHS of the above equation is independent of permittivity ε0, and so ε0E is a quantity that is independent of the medium.

is the area of an imaginary sphere around the charge Q.

is then charge per area or surface charge density.

204

ˆ RQ aE

Rπ ε=

0 24ˆ R

QER

aεπ

=

24 Rπ

24Q

Rπ


Electric flux densitySince Q Coulombs of charge produces ψ Coulombs of flux ,

This quantity is represented by and its unit is Coulombs/m2

In the case of a point charge

All the equations derived for from Coulomb’s law can be used for calculating by multiplying with

0 2 2 4 4

QER R

ψεπ π

= = is th e f lu x d e n s ity

D

0D Eε=

2 ˆ4 R

QD aRπ

=

ED 0ε


Gauss’s LawThe total electric flux passing through any closed surface is equal to the total charge enclosed by that surface.Or, the surface integral of the normal component of the electric flux density over any closed surface is equal to the charge enclosed.Consider a cloud of point charges surrounded by a closed surface of any arbitrary shape as shown below.

dS

SD dSθ


Gauss’s LawIf the total charge inside the surface is Q then Q Coulombs of electric flux must pass through the surface.At every point on the surface the electric flux density vector will have a valueConsider a small element of the surface having area dS.dS is fully specified only if its magnitude and orientation in space is specified.The only unique direction that may be associated with dS is the direction of the outward normal to the plane which is tangential to the surface dS. Let make an angle θ with .The flux crossing normally is then

DSD

SD dSdS

cosSd D dSψ θ= × SD dS= ⋅


Gauss’s LawThe total flux passing through the closed surface is

The enclosed charge might be point charges, line charge, surfacecharge or a volume charge. Accordingly,

, Charge enclosed.SSD dS Qψ = ⋅ =∫

The surface integral of the normal component of the electric flux density over any closed surface is equal to the charge enclosed.

SSQ D dS= ⋅∫

Point chargesnQ Q= ⇒∑Line chargelL

Q dlρ= ⇒∫Surface chargeSS

Q dSρ= ⇒∫Volume chargeVV

Q dVρ= ⇒∫


Gauss’s LawIn general,

Applying divergence theorem to to the LHS

Comparing (1) and (2)

This is the first of the four Maxwell’s Equations

(1)S VS VD dS dVρ⋅ = − − − −∫ ∫

(2)S sS VD dS D dV⋅ = ∇ ⋅ − − − − −∫ ∫

VD ρ∇⋅ =

Divergence of the electric flux density is the same as the volume charge density: First Maxwell’s Equation.

VD ρ∇⋅ =


Gaussian SurfacesGauss’s Law may be applied to evaluate electric field intensity.For that we have to assume a Special Gaussian Surfacesurrounding the charge distribution. The closed surface should be selected such that the electric field has a normal component or tangential component on these surface.To evaluate the electric field intensity of a point charge or spherical charge cloud we may select a spherical Gaussian surface surrounding it.The electric field is every where in radial direction and is normal to the spherical surface, or it is along dS


Gaussian SurfacesIn order to evaluate the electric field intensity of a line charge, we may select a cylindrical Gaussian surface.The electric field is in radial direction on the curved surface and is in the direction of The electric field is in tangential direction on the top and bottom surfaces and is normal to the

dSE

dS


Applications of Gauss’s Law: Example 1Electric field of a point charge.

Let a point charge be placed at the centre of a spherical coordinate system.Select a sphere of radius r as the Gaussian surface.

rS

Q


Applications of Gauss’s Law: Example 1Electric field of a point charge.

Since has only component,On a spherical surfaceApplying Gauss’s Law

E rE r rE E a=

rdS dSa=

SSD dS Q⋅ =∫

0 ˆr r rSE a dSa Qε ⋅ =∫

0 rSE dS Qε =∫

0 r SE dS Qε =∫

20 4rE r Qε π =

204r

QErπε

=

r rE E a=

20

ˆ4 r

QE arπε

=

20

ˆ4 r

QE arπε

=


ρ

l

dS

dS

dS

SD

SD

SD

Applications of Gauss’s Law: Example 2Electric field of a long line charge.

Z



Let a line charge of uniform density ρL C/m2 be placed along the z axis. Place the line charge along z axis of a cylindrical coordinate systemSelect a cylinder of radius ρ and length l as the Gaussian surface.Using Gauss’s Law

S LS lD dS dlρ⋅ =∫ ∫

S S S Lside top bottom lD dS D dS D dS dlρ⋅ + ⋅ + ⋅ =∫ ∫ ∫ ∫

( )ˆ ˆ ˆ ˆ. 0S z ztop top topD dS D a dSa D dS a aρ ρ ρ ρ⋅ = ⋅ = ⋅ =∫ ∫ ∫

( )ˆ ˆ ˆ ˆ( ) . ( ) 0S z zbottom bottom bottomD dS D a dS a D dS a aρ ρ ρ ρ⋅ = ⋅ − = ⋅ − =∫ ∫ ∫



( )ˆ ˆ ˆ ˆ.Sside side side sideD dS D a dSa D dS a a D dSρ ρ ρ ρ ρ ρ ρ⋅ = ⋅ = ⋅ =∫ ∫ ∫ ∫

Lside lD dS dlρ ρ=∫ ∫

0 2 L lE l dlρε πρ ρ= ∫

0 2 LE l lρε πρ ρ=

02LEρ

ρπε ρ

=

0

ˆ2

LE aρρπε ρ

=

ˆE E aρ ρ=

0

ˆ2

LE aρρπε ρ

=


Applications of Gauss’s Law: Example 3Electric field of a uniformly charged sphere.

++

++

++

+++ +

++

+++

+

++ +

+++

++

+

+

+++

+

+++

+

++

Spherical

Charged

Cloud

a

r

Gaussian

Surface

: Case I r a≥



Consider a spherical charged cloud having volume charge densityρv C/m2 radius a A concentric sphere of radius r may be selected as the Gaussian surface.The electric field has got only Er component.Applying Gauss’s Law

ˆ ˆr r r VS VD a dSa dVρ⋅ =∫ ∫

S VS VD dS dVρ⋅ =∫ ∫

( )ˆ ˆr r r VS VD dS a a dVρ⋅ =∫ ∫

r VS VD dS dVρ=∫ ∫

2 3443r VD r aπ ρ π=3

23V

raDr

ρ=



3

23V

raDr

ρ=

3

0 23V

raEr

ρε =

3

203

Vr

aEr

ρε

=

ˆr rE E a=

3

20

ˆ3

Vr

aE ar

ρε

=

3

20

ˆ3

Vr

aE ar

ρε

=



+

+++

++

+++ +

++

+++

+

+

+

++++

++

+

+

++

+

+

+++

+

++

Spherical

Charged

Cloud

a

r

Gaussian

Surface

: Case II r a≤

+ +

+

+

+++

++

+++ +

++

+++

+

++

+++

++

++

+++

+

++

+ +

+

+

++

++

+++ +

++

+++

+

++

+++

++

++

+++

+

++

+ ++

+++

++

+++ +

++

+++

+

++

+++

++

++

+++ +

+ +

+

+++

++

+++ +

++

++

+

++

+++

++

++

++ ++

+ +

+

+

+++

+

+++ +

++

+++

+

+++

++

++

+++

+

++

+

++ ++++

+++

++ ++

++

+ + +

+++

+

+

+++

+ ++

+++ +

+++

++ +

++ ++++

+++ +

+++

++

+++

++ +++

++

++ ++

++

+ + +

+++

+

++

+ ++

++ +

+++

++ +

++

+ +++

+

+++



Consider a spherical charged cloud of having volume charge density ρv C/m2 radius a A concentric sphere of radius r may be selected as the Gaussian surface.The electric field has got only Er component.Applying Gauss’s Law

ˆ ˆr r r VS VD a dSa dVρ⋅ =∫ ∫


( )ˆ ˆr r r VS VD dS a a dVρ⋅ =∫ ∫

r VS VD dS dVρ=∫ ∫

2 3443r VD r rπ ρ π=3

23V

rrD

rρ

=



3V

rrD ρ

=

0 3V

rrE ρε =

03V

rrE ρε

=

ˆr rE E a=

0

ˆ3

Vr

rE aρε

=

0

ˆ3

Vr

rE aρε

=


Electric field between two conducting spherical shells

sρ−sρ+

a

b

r


ˆ ˆr r r SS SD a dSa dSρ⋅ =∫ ∫

( )ˆ ˆr r r SS SD dS a a dSρ⋅ =∫ ∫

r SS SD dS dSρ=∫ ∫

r SS SD dS dSρ=∫ ∫

2 24 4r SD r aπ ρ π=


Electric field between two conducting spherical shells

2

2S

raD

rρ

=2

0 2S

raE

rρε =

2

20

Sr

aEr

ρε

=2

20

ˆSr

aE ar

ρε

=

2 , 4S

QOn the surface of the inner spherea

ρπ

=

2

2 20

ˆ4 r

Q aE aa rπ ε

= ⋅

20

ˆ4 r

QE arπε

= 20

ˆ4 r

QE arπε

=


Electric field between two conducting cylindrical shells

ab

ρ

Lρ+ Lρ−


ˆ ˆ LS LD a dSa dlρ ρ ρ ρ⋅ =∫ ∫

( )ˆ ˆ. LS LD dS a a dlρ ρ ρ ρ⋅ =∫ ∫

LS LD dS dlρ ρ=∫ ∫

2 L LD l dlρ πρ ρ= ∫

l

2 LD l lρ πρ ρ=


Electric field between two conducting cylindrical shells

2 LD l lρ πρ ρ=

2LDρ

ρπρ

=

0 2LEρ

ρεπρ

=

02LEρ

ρπε ρ

=

0

ˆ ˆ2

LE E a aρ ρ ρρπε ρ

= =

0

ˆ2

LE aρρπε ρ

=

LQ lρ=L

Ql

ρ =

0

ˆ2

QE al ρπε ρ

=

0

ˆ2

QE al ρπε ρ

=

0

ˆ2

LE aρρπε ρ

=


Work done and Electric PotentialAn electric charge produces an electric field and if a test charge is brought to this region it experiences a force. If the test charge is moved against this field equal and opposite forces will have to be exerted by the field and this needs work to be done or energy to be spent.If the charge is moved against the field, work done is positive and if it is moved in the direction of the field work done is negative.Consider a uniform field in space as shown belowE

initial

final

dLθcosdL θ

E


Work done and Electric PotentialLet a positive test charge is moved through a small distance dLthrough electric field in a direction opposite to the field.The field exerts a force on the charge and some work must be done to move the charge.This work is equal to the product of the force and the distance through which the charge has to be moved in the direction of theforce.

The incremental work dW done in moving the positive charge through the distance dL is

⇒Force experienced on charge Q by the field E F = Q E

⇒ aApplied Force F = -Q E = -QE

=F cosadW dL θ


Work done and Electric Potential

The total work required to move the charge from an initial point to a final point is

=F cosadW dL θ = QE dL− ⋅=Fa dL⋅

W=final

initialQE dL− ⋅∫

=final

initialQ E dL− ⋅∫

W=final



Electric Potential-Potential differenceThe work done in moving a unit positive charge from an initial point to a final point is called the potential difference between these points.Work done in moving a charge of Q Coulombs is

Work done in moving a unit positive charge is

If B is the initial point and A is the final point

W=final


W V= =final

initialE dL

Q− ⋅∫Potential difference

V =A

AB BE dL− ⋅∫Potential difference V =

A

AB BE dL− ⋅∫


Electric Potential-Potential difference

If A and B are the same points, the limits of integration are the same and

Work done in carrying in carrying a unit positive charge from aninitial point to the same final point is zero, whatever be the path taken for this travel.The potential difference around any closed path is zero, irrespective of the geometry of the path.

V =A

AB BE dL− ⋅∫

V = 0A

AB AE dL− ⋅ =∫ 0or E dL⋅ =∫

0E dL⋅ =∫ A E


Electric Potential- Absolute potentialThe absolute potential of a point or simply potential of a point A is defined as the work done in moving a unit positive charge from infinity or from zero potential to that point.

V =VA

A A E dL∞ ∞= − ⋅∫ V =

A

A E dL∞

− ⋅∫


Potential Difference between two points in the field of a point charge

Let a point charge Q be placed at the centre of a spherical coordinate system.Let two points in the field of this charge are at radial distances rA and rB from the point charge.

Ar Br• •Q A B


Potential Difference between two points in the field of a point charge

The electric field intensity of a point charge is given by

20

ˆ4 r

QE arπε

= V =A

AB BE dL− ⋅∫

20

ˆ ˆ=4

A

B

r

r rr

Q a drarπε

− ⋅∫ ˆ = rSince dL dra

20

1=4

A

B

r

r

Q drrπε

− ∫0

1=4

A

B

r

r

Qrπε

⎡ ⎤− −⎢ ⎥⎣ ⎦

0

1 1=4 A B

Qr rπε⎡ ⎤

−⎢ ⎥⎣ ⎦ 0

1 1V =4A B

A B

Qr rπε

⎡ ⎤−⎢ ⎥

⎣ ⎦


Potential Difference between two points in the field of a line charge

• •Ar BrA B

c/mLρ

Let a line charge distribution having density ρL is placed at the centre of a cylindrical coordinate system.Let two points A and B in the field of this charge is at radial distances rA and rB from the point charge.


Potential Difference between two points in the field of a line charge

The potential difference between A and B is

0

ˆ ˆ=2

A

B

rL

ra d aρ ρ

ρ ρπε ρ

− ⋅∫

V = A

B

r

AB rE dL− ⋅∫

0

ˆ=2

A

B

rL

ra dLρ

ρπε ρ

− ⋅∫

0

1=2

A

B

rL

rdρ ρ

πε ρ− ∫

[ ]0

= ln ln2

LA Br rρ

πε− −

[ ]0

= ln ln2

LB Ar rρ

πε−

0

= ln2

L B

A

rr

ρπε

⎡ ⎤⎢ ⎥⎣ ⎦

A B0

V = ln2

L B

A

rr

ρπε

⎡ ⎤⎢ ⎥⎣ ⎦

[ ]0

= ln2

A

B

rLr

ρ ρπε

−


Potential difference between two conducting spherical shells

a

bQ+Two spherical shells having radius a and b carrying charges +Q and Q on the innerand outer shells

−

V =a

AB bE dL− ⋅∫

20

ˆ ˆ=4

a

r rb

Q a drarπε

− ⋅∫

20

1=4

a

b

Q drrπε

− ∫0

1 1=4

Qa bπε⎡ ⎤−⎢ ⎥⎣ ⎦

AB0

1 1V =4

Qa bπε⎡ ⎤−⎢ ⎥⎣ ⎦

0

1=4

a

b

Qrπε

⎡ ⎤− −⎢ ⎥⎣ ⎦

Q−


Potential difference between two conducting cylindrical shellsTwo cylindrical shells carrying charge density on the inner shell and having radii a and b

Lρ

V =a

AB bE dL− ⋅∫

0

ˆ ˆ=2

aL

ba d aρ ρ

ρ ρπε ρ

− ⋅∫

A B0

V = ln2

L ba

ρπε

⎛ ⎞⎜ ⎟⎝ ⎠

a

b

/L c mρ

0

1=2

aL

bdρ ρ

πε ρ− ∫

[ ]0

= ln2

aLb

ρ ρπε

−

[ ]0

= ln ln2

L a bρπε

− −

0

= ln2

L ba

ρπε

⎡ ⎤⎢ ⎥⎣ ⎦


Absolute potential of a point in the field of a point charge

If the position vector of the point A is and the charge is located at point whose position vector is the potential at point A is

For N point charges Q1, Q2, Q3,……. QN, located at points whose position vectors the potential of the point A whose position vector is is given by

r'r

'04AQVr rπε

=−

1 2 3, , ,.........., Nr r r rr

1 2

0 1 0 2 04 4 4N

AN

Q Q QVr r r r r rπε πε πε

= + + ⋅⋅ ⋅ ⋅ ⋅ +− − −

10

14

Nk

Ak k

QVr r

Orπε =

=−∑


Absolute potential of a point in the field of charge distributions

For continuous charge distributions, the above equation can be modified by incorporating appropriate charge distributions and changing the summation to integration.

'0

14

LA L

For line chdlV ar

rr

geρπε

= ⇒−∫

'0

1 4

SA S

For surface cdSVr r

hargeρπε

= ⇒−∫

'0

1 4

VA V

For vdV olume ch rger

aVr

ρπε

= ⇒−∫


Example 1Two point charges -4μC and 5μC are located at (2,-1,3) and (0,4,-2) respectively. Find the potential at (1,0,1) assuming zero potential at infinity.

1 2

0 1 0 24 4AQ QV

r r r rπε πε= +

− −1 4Q Cμ= −

2 5Q Cμ=

1 (1,0,1) (2, 1,3) 1,1, 2 6r r− = − − = − − =

2 (1,0,1) (0,4, 2) 1, 4,3 26r r− = − − = − =

6 6

0 0

4 10 5 104 6 4 26AVπε πε

− −− × ×= +

5.872AV kV= −


sE dSε= ⋅∫

QCV

=

lV E dl= − ⋅∫

s

l

E dsC

E dl

ε ⋅=− ⋅∫∫

Capacitance

CONDUCTOR 1

CONDUCTOR 2

+ + + +++++++

− − −− − −

−−−−−−−

−

E

Q+

Q−

sQ D dS= ⋅∫

s

l

E ds

E dl

ε ⋅=− ⋅∫∫

s

l

E dsC

E dl

ε ⋅=− ⋅∫∫


Capacitance of a parallel plate capacitor

+ + + + + + + + + + + + + + + + + + + + +

− − − − − − − − − − − − − − − − −E

X

1

2

Dielectric

Plate area AQ+

Q−

d



SΔ Ed

A

A

Plate1

Plate2

sρ+

sρ−

Q+

Q−

/s Q Aρ =


Capacitance of a parallel plate capacitorEach plate of the parallel plate capacitor has an area A and areseparated by a distance d.The plates 1 and 2 carries charges +Q and -Q uniformly distributed on them. The distance between plates is assumed to be very small when compared to the plate dimensions such that the fringing between plates is small enough to be ignored.Applying Gauss’s law on the surface of a small pill box shaped volume as shown in figure,

Only the bottom face of the cylindrical gaussian surface contributes towards the electric flux.Let the electric flux density on the bottom face of the gaussiancylinder is

enclQψ =s Sρ= Δ

D D Sψ = Δ



Electric field intensity on the bottom face of the gaussian cylinder is

sD S SρΔ = Δ

( )ˆS xD aρ= −

DEε

= ˆSxaρ

ε= − ˆS

xE aρε

= −

SQA

ρ =

ˆxQE aAε

= −

sD ρ= ˆS xD aρ= −


Capacitance1

2 0ˆ ˆ

d

x xQV E dL a dxaAε

= − ⋅ = − − ⋅∫ ∫

0

dQ dxAε

= ∫

QdAε

=

C Q ACapacitanceV d

ε= =

C Adε

=


Coaxial capacitor

Dielectric

a

b

L

1 212


Coaxial capacitorConsider a length of two coaxial conductors of inner radius a and outer radius b. The space between the conductors is filled with a homogeneous dielectric with permittivity ε.Assume that the conductors 1 and 2 carry +Q and –Q charges uniformly distributed on them.By applying Gauss’s law we obtain the electric field between the cylinders.

ˆ2

QE aL ρπερ

=

1

2ˆ ˆ

2a

b

QV E dl a d aL ρ ρρ

περ= − ⋅ = − ⋅∫ ∫

12

a

b

Q dL

ρπε ρ

= − ∫


Coaxial capacitor

[ ]ln2

a

b

QVL

ρπε

= −

ln2

Q bL aπε⎛ ⎞= ⎜ ⎟⎝ ⎠

C QCapacitanceV

=2

ln

Lba

πε=⎛ ⎞⎜ ⎟⎝ ⎠

2

l n

LCba

π ε=

⎛ ⎞⎜ ⎟⎝ ⎠


Spherical capacitor

+

+

++++

++

−−

−

−−−

−

−

a

b1

2Dielectric


Spherical capacitorConsider the capacitance formed by two spherical shells of radii a and b as shown in figure.The spheres are separated by a dielectric medium with permittivity ε.Charges +Q and –Q are distributed on the inner and outer spheres respectively.By applying Gauss’s law we can find out the electric field of such a system.

The potential difference between the conductors is

2 ˆ4 r

QE arπε

=

2 ˆ ˆ4

a

r rb

Q a drarπε

= − ⋅∫

1

2V E dl= − ⋅∫


Spherical capacitor

21

4a

b

QV drrπε

= − ∫1 1

4Q

a bπε⎛ ⎞= −⎜ ⎟⎝ ⎠

C QCapacitanceV

=

41 1

C

a b

πε=⎛ ⎞−⎜ ⎟⎝ ⎠

41 1

C

a b

π ε=⎛ ⎞−⎜ ⎟⎝ ⎠


Relationship between E and VThe potential around any closed path is zero

Applying Stokes’ Theorem

Equations (1) and (2) are referred to as Maxwell’s second equation for static electric fields.

A vector field that satisfies equation (1) or (2) is called a conservative field or irrotational field

0 -----(1)E dL⋅ =∫

( ) 0S

E dL E dS⋅ = ∇× ⋅ =∫ ∫0 ------(2)E∇× =

0E∇× =

0 E dL⋅ = ⇒∫ Maxwell's second equation in integral form

0 E∇× = ⇒Maxwell's second equation in differential form


Relationship between E and VThe general expression for potential is

Comparing the two expressions

V= E dL− ⋅∫dV= E dL− ⋅

dV= x y zE dx E dy E dz− − −

V V VBut dV= dx dy dzx y z

∂ ∂ ∂+ +

∂ ∂ ∂

V=xEx

∂−∂

V=yEy

∂−∂

V=zEz

∂−∂


Relationship between E and V

Electric field intensity is the gradient of potential VThe negative sign shows that the direction of is opposite to the direction in which V increases , it is directed from higher to lower levels of V .Also it satisfies the equation

since the curl of the gradient of a scalar is always zero

ˆ ˆ ˆ= x y zV V VE a a ax y z

∂ ∂ ∂− − −∂ ∂ ∂

=E V−∇=E V−∇

E

0 E∇× =

0 V∇×−∇ =


Example 1Given the potential (a) Find the electric flux density at (2,π/2,0)(b) Calculate the work done in moving a 10μC charge from the point

B(1,300,1200) to A(4,900,600)

210 sin cosVr

θ φ=

:Solution

E V= −∇

1 1ˆ ˆ ˆsinr

V V VE a a ar r rθ φθ θ φ

⎡ ⎤∂ ∂ ∂= − + +⎢ ⎥∂ ∂ ∂⎣ ⎦

210 sin cosVr

θ φ=

3 3 320 10 10ˆ ˆ ˆsin cos cos cos sinrE a a ar r rθ φθ φ θ φ φ= − +


Example 1(2, / 2,0)At π

20 ˆ ˆ ˆ0 08 rE a a aθ φ= − +

20 ˆ8 ra=

0 020 ˆ8 rD E aε ε= =

2ˆ22.1 /rD a pC m=

AB

A

B

Work done W Q E dl QV= − ⋅ =∫( )A BW Q V V= −

6 10 1010 10 sin90 cos60 sin30 cos12016 1

o o o o− ⎛ ⎞= × −⎜ ⎟⎝ ⎠

28.125 JW μ=

(4,90 ,60 )o oA (1,30 ,120 )o oB

210 sin cosVr

θ φ=


Electric dipoles: Potential and Electric field

Q+

Q−

dθ

cosd θ

P

1r

2rr

Z

Y

X

0 1 0 24 4Q QV

r rπε πε= −

22 1 1 2cos , r r d rr rθ−

+Q -Qd


Electric dipoles: Potential and Electric fieldAn electric dipole is formed when two point charges of equal magnitude but opposite sign are separated by a small distance.Referring to figure (1) the potential at point P is given by

If r>>d we can make the following assumptions2

2 1 1 2 and cos r r d rr rθ−

20

cos 4

Q dThen Vr

θπε

=

0 1 2

1 14

QVr rπε⎡ ⎤

= −⎢ ⎥⎣ ⎦

ˆ ˆ ˆcos where r z rd d a d da a is a unit vector in r directionθ = ⋅ =

0 1 0 24 4Q QV

r rπε πε= − 2 1

0 1 24Q r r

rrπε⎡ ⎤−

= ⎢ ⎥⎣ ⎦



20

ˆ

4rQd aThen V

rπε⋅

=

20

ˆ

4rp aVrπε

⋅=

, By defining Qd p dipole moment=

The magnitude of dipole moment is equal to the product of chargeand distance and its direction is from Q to +Q−

The electric field due to the dipole with centre at the origin is given by

E V= −∇1ˆ ˆr

V Va ar r θθ

∂ ∂⎡ ⎤= − +⎢ ⎥∂ ∂⎣ ⎦There is no field variation along directionφ



20

cos 4

Q dWhere Vr

θπε

=1ˆ ˆr

V VE a ar r θθ

∂ ∂⎡ ⎤= − +⎢ ⎥∂ ∂⎣ ⎦

3 30 0

cos sinˆ ˆ2 4r

Qd QdE a ar r θθ θ

πε πε= + ( )3

0

ˆ ˆ2cos sin4 r

p a ar θθ θ

πε= +

( )30

ˆ ˆ2cos sin4 r

pE a ar θθ θ

πε= +

20

ˆ

4rp aVrπε

⋅=When the dipole centre is at the origin,

'When the dipole centre is not at the origin, but at r( )'

3'0

4

p r rV

r rπε

⋅ −=

−



2 A point charge is a monopole and its field varies inversely as r♣and its potential varies inversely as r

3 The electric field due to a dipole varies inversely as r and its♣2potential varies inversely as r

The electric field due to successive higher order multipoles♣4 5 6, , ,...... varies inversely as r r r while their potential varies

3 4 5, , ,...... inversely as r r r


Flux lines and equipotential surfacesAn electric flux line is an imaginary line or path drawn in such a way that its direction at any point is the direction of the electric field at that point.A surface on which the potential is the same throughout is called equipotential surface.The intersection of an equipotential surface and a plane results in a path or line called equipotential line.No work is done in moving a charge from one point to another along an equipotential line or surface.

This implies flux lines (direction of E) are always normal to equipotential surfaces.

0E dl⋅ =∫


Electric dipoles and flux lines

+

Flux lines

Equipotential surface

Equipotential surfaces of a point charge


Electric dipoles and flux linesZ

Flux lines

Equipotential surface

0V >

0V <

Equipotential surfaces of a dipole


Energy density in electrostatic fieldsConsider a region free of electric fields. Let there be three point charges Q1,Q2,Q3 at infinityTo determine the energy present in the assembly of charges, we have to determine the amount of work necessary to assemble them.No work is required to transfer Q1 from infinity to P1 because the space is initially charge free.

•

•

•∞

1P

2P

3P

1Q

2Q

3Q


Energy density in electrostatic fieldsThe work done in transferring Q2 from infinity to P2 is equal to the product of Q2 and the potential V21 at P2 due to Q1.The work done in positioning Q3 at P3 is equal to Q3(V32+V31). The total work in positioning the three charges is

If the charges were positioned in the reverse order

Adding (1) and (2)

1 2 3EW W W W= + +( )2 21 3 31 320 ------(1)Q V Q V V= + + +

3 2 1EW W W W= + +

( )2 23 1 12 130 ------(2)Q V Q V V= + + +

( ) ( ) ( )1 12 13 2 21 23 3 31 322 EW Q V V Q V V Q V V= + + + + +


Energy density in electrostatic fields

If there are n point charges

If the region has a continuous charge distribution, the summation becomes integration

( )1 1 2 2 3 312EW QV Q V Q V= + +

1 2 3 1 2 3V ,V ,V Total potentials at P , P and P⇒

1

12

n

E k kk

W Q V=

= ∑

1 2E LL

W Vdl Line Chargeρ= ⇒∫1 2E SS

W VdS Surface Chargeρ= ⇒∫1 2E VV

W VdV Volume Chargeρ= ⇒∫

1 1 2 2 3 32 EW QV Q V Q V= + +



,VApplying D ρ∇⋅ =

( )12E V

W D VdV= ∇⋅∫

( )Using the identity A V VA A V∇⋅ = ∇⋅ − ⋅∇

( )12E V

W VD D V dV= ∇⋅ − ⋅∇∫

( ) ( )1 12 2V V

VD dV D V dV= ∇⋅ − ⋅∇∫ ∫Applying divergence theorem to the first term on the RHS

( )1 12 2E S V

W VD dS D V dV= ⋅ − ⋅∇∫ ∫

12E VV

W VdVρ= ∫



2 3 1 / 1 / 1 /V varies as r and D as r so that VD varies as r2 The surface area increases as r

.the surface area goes to infinity

1 The quantity VD dS decreases effectively asr

⋅

, If we allow the volume v to increase to include all space

S The surface integral goes to zero as r and→∞ → ∞

( )212

1SE VVD dSW D V dV= − ⋅⋅ ∇∫ ∫ 0=

( )1 12 2E S V

W VD dS D V dV= ⋅ − ⋅∇∫ ∫



20

12

Eε=

( )12EdW D E dV= ⋅

( )12

EdW D EdV

= ⋅

is defined as the energyEE

dW wdV

=

( )2

20

0

1 12 2 2E

Dw D E Eεε

= ⋅ = =

E EW w dV= ∫

( )012 V

E E dVε= ⋅∫2

012 V

E dVε= ∫

( )12E V

W D V dV= − ⋅∇∫

20

12

E dVε=

Energy density is a quantity which when integrated on overall space yields the

total energy

' ' All space is the volume . containing the entire field

density in electrostatic field

Putting E V= −∇

( )12E V

W D E dV= ⋅∫

E EW w dV= ∫


Example 1Three point charges -1nC, 4nC,and 3nC are located at (0,0,0), (0,0,1), and (1,0,0). Find the energy in the system.

3

1

12 k k

k

Q V=

= ∑1

12

n

E k kk

W Q V=

= ∑ [ ]1 1 2 2 3 312

QV Q V Q V= + +

2 3 1 3 1 21 2 3

0 0 0 00 0

1 1 12 4 (1) 4 (1) 2 4 (1) 2 4 (1)4 ( 2) 4 ( 2)

Q Q Q Q Q QQ Q Qπε πε πε πεπε πε

⎡ ⎤ ⎡ ⎤⎡ ⎤= + + + + +⎢ ⎥ ⎢ ⎥⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

•

•

•

•

•

•1nC−

4nC

3nC

4nC

1nC−3nC

1

1

2

x

z

y x

z

1Q

2Q

3Q


Example 1

2 31 2 1 3

0

14 2

Q QW Q Q Q Qπε

⎛ ⎞= + +⎜ ⎟⎝ ⎠

0

1 124 34 2

nJπε

⎛ ⎞= − − +⎜ ⎟⎝ ⎠

13.37nJ=


Example 2 A charge distribution with spherical symmetry is given by

determine the energy stored in the region r < R

0 00 V

r Rr R

ρρ

≤ ≤⎧= ⎨ >⎩ RVρ

The electric field of spherical charge distribution is

0

0

ˆ 3

rrEor r aF R ρε

≤ =

( )12E V

W D E dV= ⋅∫ 20

12 V

E dVε= ∫


Example 2

0

0

ˆ3 r

rE aρε

= 0

03rE ρε

=2 2

2 02

09rE ρ

ε=

20

12E V

W E dVε= ∫2 2

020

0912 V

dr Vρεε

= ∫0

22

0 0218 V

r dVε ρε

= ∫

0

22

0 2 si8

n1 V

r r drd dθε

θρ φ= ∫2 4

20

0 0 00

s18

inR

rr drd d

π π

φ θθ θ φρ

ε = = == ∫ ∫ ∫

2 4

0 0 0

20

0

si18

nR

rd d r dr

π π

φ θφρ

εθ θ

= = == ∫ ∫ ∫

2 50

0 0

418 5

Rrρ π

ε⎡ ⎤

= ⎢ ⎥⎣ ⎦

2 50

0

245

Rπρε

=2 5

0

0

245E

RW πρε

=


Conduction and convection currentsThe current through a given area is the electric charge passing through the area per unit time. Its unit is Amperes.

One ampere current is produced if charge is transferred at the rate of one coulomb per second.The concept of current density is useful in defining the events occurring at a point. If current ΔI flows through a surface ΔS, then current density is

dQIdt

=

nIJS

Δ=Δ

nAnd Assuming the current density is perpendicΔ uI = J ΔS lar to the surface.


Conduction and convection currentsIf the current density is not normal to the surface

The total current flowing through the surface is

ΔI = Jcos ΔSθ = J ΔS⋅

SI = J d S ⋅∫ SΔ

Jθ

SΔ


Convection currentsConvection current is produced when current flows through an insulating medium such as liquid, rarefied gas, or a vacuum.It does not involve conductors and hence does not satisfy Ohm’s law. A beam of electrons in a vacuum tube is an example of convectioncurrent.

lΔ

SΔ Vρ

u

y


Convection currentsConsider a beam of electrons with volume charge density ρV flowing in the y direction with a velocity uy.

The current through the filament is

The y directed current density Jy is given by

In general

ˆy yu u a=

QIt

ΔΔ =

Δ( )V S l

tρ Δ Δ

=Δ

VlSt

ρ Δ⎛ ⎞= Δ ⎜ ⎟Δ⎝ ⎠ V ySuρ= Δ

y V yIJ uS

ρΔ= =Δ

VJ uρ=

VConvection current density J uρ=


Conduction currentsConduction current requires a conductor as a medium.A conductor contains a large number of free electrons that provide conduction current due to an impressed electric field.When an electric field is applied the force on an electron with charge –e is

Since the electron is not in free space it will not be accelerated under the influence of the electric field.It suffers constant collisions with the atomic lattice and drifts from one atom to another.If the electron with mass m is moving in an electric field with an average drift velocity the average change in momentum of thefree electron must match the applied force.

F eE= −

E

Eu


Conduction currentsAccording to Newton’s law,

This indicates that the drift velocity is directly proportional to the applied field.If there are n electrons per unit volume the electronic charge density is given by Convection current density is

mu eEτ

= −eu Emτ

= − Average time intervalτ ⇒

V neρ = −VJ uρ=

2neJ Emτ

= J Eσ=

J Eσ=

between collisions

2

is the conductn i ityem

vτσ =

POINT FORM OF OHM’S LAW


ConductorsA conductor contains free electrons which accounts for its conductivity.In an isolated conductor, when an external electric field is applied, the positive charges moves in the same direction as theapplied field.The negative charges moves in the opposite direction.These free charges accumulate on the surface of the conductor and form an induced surface charge.The induced charges set up an internal induced field which cancels the externally applied field

eE

iEeE

Thus a perfect conductor cannot contain an electrostatic field within it.


Conductors++++++++

________

eE

eE

eE

iE

iE

++++++++

________

0E =

0Vρ =

A perfect conductor cannot contain an electrostatic field within it.

eE

eE

eE


ConductorsInside a conductorThus a conductor is an equipotential body. Also, in a conductor, and as per the equation, the electric field intensity According to Gauss’s law, If the charge density

0 0 V=0E V= −∇ =and which implies

σ →∞ J Eσ=0E →

0Vρ =VS V

E dS dVε ρ⋅ =∫ ∫0E =

Inside a perfect conducter V ab,E = 0, ρ = 0, V = 0


ConductorsWhen the two ends of the conductor are maintained at a potentialdifference V, the electric field is not zero inside the conductor.In this case there is no static equilibrium, since the applied voltage prevents the establishment of such equilibrium.An electric field must exist inside the conductor to sustain the flow of current.The opposition to the flow is called resistance.

lE

V

I- - - - -


ConductorsThe magnitude of the electric field is given by Since the conductor has a uniform cross section,By Ohm’s law

VEl

=IJA

=J Eσ=

I VEA l

σσ= =

V lI Aσ=

l lRA A

ρσ

= =l Resistivityρσ

= ⇒

lRAρ

=


ConductorsResistance of a conductor having non-uniform cross section is

E dlVRI E dSσ

− ⋅= =

⋅∫∫


Example 1If calculate the current passingthrough

A hemispherical shell of radius 20 cmA spherical shell of radius 10 cm

33

1 ˆ ˆ(2cos sin ) A/mrJ a ar φθ θ= +

SI J dS= ⋅∫ 2 si ˆn radS r d dθ θ φ=

33

1 ˆ ˆ(2cos sin ) A/mrJ a ar φθ θ= +

( ) ( )23

1 ˆ ˆ ˆ2cos sin sinr rSI a a r d d a

r φθ θ θ θ φ= + ⋅∫1 2cos sin

SI d d

rθ θ θ φ= ∫


Example 1

31.4 A=

, 0 0.1In the second case and rθ π≤ ≤ =

1 2cos sinS

I d dr

θ θ θ φ= ∫2 /2

0 0

1 2sin cos d dr

π π

φ θθ θ θ φ

= == ∫ ∫

/2

0

2 2sin cos dr

π

θ

π θ θ θ=

= ∫/2

031.4 2sin cos d

π

θθ θ θ

== ∫

0

2 2sin cosI dr

π

θ

π θ θ θ=

= ∫ 062.8 2sin cos d

π

θθ θ θ

== ∫ 0 =


Continuity equation and relaxation time

dSJ

dS

Bounding Surface S

inQ


Continuity equation and relaxation timeElectric charges can be neither created nor destroyed according to law of conservation of electric charges.Consider an arbitrary volume V bounded by surfaces S as shown in figure.A net charge Qin exists within this region.If a net current I flows across this surface the charge in the volume must decrease at a rate that equals the current.If a net current flows across the surface in to the volume, the charge in the volume must increase at the rate equal to the current.The current leaving the volume is the total outward flux of the current density vector through the surface S.

----(1)inOUT S

dQI J dSdt

= ⋅ = −∫



Substituting (2) and (3) in (1)

This equation is called continuity of current equation or continuity equation.

---(2)S V

Using divergence theorem, J dS = Jdv⋅ ∇ ⋅∫ ∫----(3)in v

vV V

dQ d= dv =dt dt t

ρρ ∂− − −

∂∫ ∫

= vV V

Jdvtρ∂

∇ ⋅ −∂∫ ∫

= vJtρ∂

∇ ⋅ −∂

= vJtρ∂

∇ ⋅ −∂


Continuity equation and relaxation timeIt basically states that there can be no accumulation of charge at any point. For steady currents and hence The total current leaving a volume is the same as the total current entering it.Kirchhoff’s current law follows from it.

=0v

tρ∂∂ =0J∇ ⋅


Continuity equation and relaxation timeIf we introduce charge at some interior point of a given material with constants σ and ε its effect can be obtained using the continuity equation.From Ohm’s law From Gauss’s lawSubstituting in the continuity equation,

=J Eσ

vD ρ∇ ⋅ = vand hence E ρε

∇ ⋅ =

= vJtρ∂

∇ ⋅ −∂

= vEtρσ ∂

∇ ⋅ −∂

= v v

tσρ ρε

∂−

∂

=0 vvt

ρ σ ρε

∂+

∂

=0 vvt

ρ σ ρε

∂+

∂



This equation is a homogeneous linear ordinary differential equation.Separating the variables,

Integrating both sides

=0 vvt

ρ σ ρε

∂+

∂

/0 = rt T

v v eρ ρ −

v

v

tρ σρ ε∂

= − ∂

0ln ln v vtσρ ρε

= − +

0ln vρWhere is a constant of integration

/0 = rt T

v v eρ ρ − rWhere T ε

σ=


Continuity equation and relaxation timeWhen we introduce a volume charge density at an interior point in a material, it decays resulting in a charge movement from the interior point at which it was introduced to the surface of the material.The time constant Tr of this decay is called relaxation time or rearrangement time.Relaxation time is the time it takes for a charge placed in the interior of a material to drop to e-1 or 36.8 percent of its initial value. It is very short for good conductors and very long for good dielectrics.For a good conductor the relaxation time is so short that most of the charge will vanish from the interior point and appear at the surface within a short time.For a good dielectric the relaxation time is very long that the introduced charge remains at the same point.

191.53 10Cu s−⇒ × 51.2Quartz days⇒


Boundary conditionsWhen the field exists in a medium consisting of two different media, the conditions the field must satisfy are called boundary conditions.For the electrostatic field the following boundary conditions are important.

Dielectric – dielectric interface. Conductor – dielectric.Conductor – free space.


Dielectric-dielectric Boundary conditionsConsider the boundary between two dielectrics with permittivities

1 21 0 2 0 r rε ε ε ε ε ε= =and

ab

cd

WΔ

hΔ

1ε

2ε

1tE

1nE1E

2tE

2nE2E

1

2


Dielectric-dielectric Boundary conditionsThe fields in the two media can be expressed as

Apply the equation to the path abcda in the figure.

Tangential components of are equal at the boundary.undergoes no change on the boundary and it is continuous

across the boundary.

1 1 1t nE E E= + 2 2 2t nE E E= +

0lE dl⋅ =∫

1 1 2 2 2 1 02 2 2 2t n n t n nabcda

h h h hE dl E w E E E w E EΔ Δ Δ Δ⋅ = Δ − − − Δ + + =∫

1 2 0t tE w E wΔ − Δ =

1 2t tE E= 1 2t tE E=E

tE


Dielectric-dielectric Boundary conditions

The tangential component of under goes some change across the boundary.So is said to be discontinuous across the boundary. The boundary conditions for the normal components are obtained by applying Gauss’s law on a small pill box shaped volume as in the figure.

1 2t tE E= 1 2

1 2

t tD Dε ε

= 2 1 1 2t tD Dε ε=

D

D

SD dS Q⋅ =∫

1 2n n SD S D S Q SρΔ − Δ = Δ = Δ

1 2n n SD D ρ− =

0hΔ →Assuming



hΔ

1ε

2ε

1tD

1nD1D

2tD

2nD2D

1

2

SΔ



If no free charge exists at the boundary,

Normal components of are equal at the boundary.undergoes no change on the boundary and it is continuous

across the boundary.

The normal component of under goes some change across theboundary.So is said to be discontinuous across the boundary.

1 2n n SD D ρ− =0Sρ =

1 2n nD D=

1 2n n SD D ρ− =

1 2n nD D=D

nD

1 2n nD D= 1 1 2 2n nE Eε ε=1 2

2 1

n

n

EE

εε

=

D

nD



1θ1tD

1nD

1D

2tD

2nD2D

1tE1nE

1E

2tE

2nE2E

2θ

1ε

2ε

1

2



1 2t tE E=

1 1 2 2sin sin - - (1)E Eθ θ=

1 2n nD D=

1 1 2 2cos cosD Dθ θ=

1 1 1 2 2 2cos cos - - - (2)E Eε θ ε θ=

1 1 2 2

1 1 1 2 2 2

(2)co

sin sis cos

n) /(1 EE

EE

θ θε θ ε θ

⇒ =

2 1 1 2tan tanε θ ε θ=

1 1 0 1 1

2 2 0 2 2

tantan

r r

r r

θ ε ε ε εθ ε ε ε ε

= = = 1 1

2 2

tantan

r

r

θ εθ ε

=

This is called law of refraction


Conductor-Dielectric boundary conditions

ab

cd

WΔ

hΔ

0 rDielectric ε ε ε=

tEnE

E

0Conductor E =


Conductor-Dielectric boundary conditions

0 rDielectric ε ε ε=D

nD tD

0Conductor E =

hΔ

SΔ


Conductor-Dielectric boundary conditionsThe interface between a perfect conductor and a dielectric shown in figure.Apply the equation to the path abcda in the figure.

The boundary conditions for the normal components are obtained by applying Gauss’s law on a small pill box shaped volume as in the figure.

0lE dl⋅ =∫

0 0 0 02 2 2 2t n nabcda

h h h hE dl E w E w EΔ Δ Δ Δ⋅ = Δ − − ⋅ − ⋅Δ + ⋅ + =∫

0tE wΔ =

0tE =

SD dS Q⋅ =∫ 0nD S S QΔ − ⋅Δ = Δ 0hΔ →Assuming

n SQDS

ρΔ= =Δ 0n r n SD Eε ε ρ= =

0tD = 0 0t r tD Eε ε= = 0 0t r tD Eε ε= =


Conductor-Dielectric boundary conditionsNo electric field exists inside a perfect conductor.External electric field, if any, is normal to the conductor surface as given by n SD ρ=


Conductor-Free space boundary conditionsThis is a special case of conductor-dielectric conditions.It can be obtained by replacing in the equation

As in the earlier case

0n r n SD Eε ε ρ= =

1rε =

0n n SD Eε ρ= =

0 0t r tD Eε ε= =

0 0t tD Eε= =

0n n SD Eε ρ= =


Conductor-Free space boundary conditions

0 rFree space ε ε ε=D

nDtD

0Conductor E =

nE tEE


Poisson’s and Laplace’s equationGauss’s law in point form is given by

Substituting in the above equation,

We know that

This equation is known as Poisson’s equation.

VD ρ∇⋅ =D Eε=

VEε ρ∇⋅ =E V= −∇

( ) VVε ρ∇⋅ − ∇ =

VV ρε

∇⋅∇ = − 2. ., Vi e V ρε

∇ = −

2 VV ρε

∇ = −


Poisson’s and Laplace’s equation

ˆ ˆ ˆ ˆ ˆ ˆx y z x y zV V VV a a a a a a

x y z x y z⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂ ∂ ∂

∇⋅∇ = + + ⋅ + +⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦V V V

x x y y z z⎡ ⎤⎛ ⎞∂ ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞= + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦

2 2 2

2 2 2V V Vx y z

∂ ∂ ∂= + +∂ ∂ ∂

2 2 22

2 2 2VV V VV

x y zρε

∂ ∂ ∂∇ = + + = −

∂ ∂ ∂

2 2 2

2 2 2VV V V

x y zρε

∂ ∂ ∂+ + = −

∂ ∂ ∂


Poisson’s and Laplace’s equationIn the case of a charge free region, Poisson’s equation reduces to Laplace’s equation as given below.

In cylindrical and spherical coordinate systems

2 0V∇ =2 2 2

2 2 2 0V V Vx y z

∂ ∂ ∂+ + =

∂ ∂ ∂

2 22

2 2 21 1 0V V VV

zρ

ρ ρ ρ ρ φ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂

∇ = + + =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

22 2

2 2 2 21 1 1sin 0

sin sinV V VV r

r r r r rθ

θ θ θ θ φ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞∇ = + + =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠


Poisson’s and Laplace’s equationLaplace’s equation is very useful in finding out the potential V of a set of conductors maintained at different potentials as in the case of capacitor plates.Electric field can be obtained once we obtain potential V fromE

E V= −∇


General Procedure for solving Poisson’s and Laplace’s equations to find capacitance

Solve Laplace’s ( if ρv = 0 ) or Poisson’s equation ) if ( ρv ≠ 0 ) using Direct integration if V is a function of one variableSeparation of variables if v is a function of more than one variable.The solution at this point is not unique but expressed in terms of unknown integration constants.

Apply boundary conditions to determine a unique solution for V.

After obtaining V, find using and using

Find the charge induced on a conductor using where ρs=Dn and Dn is the component of normal to the conductor.

If required find the capacitance between the conductors using C=Q/V

E E V= −∇ D D Eε=

DsQ dSρ= ∫


Uniqueness Theorem

Proof: ( by contradiction )Proof: ( by contradiction )

Assume that there are two solutions V1 and V2 of Laplace’s equations both of which satisfies the given boundary conditions.

V1 and V2 must reduce to the same potential along the boundary

If a solution to Laplace’s Equation can be found that satisfies the boundary conditions, then the solution is unique.

21 0V∇ = 2

2 0V∇ =

( )22 1 0That is V V∇ − =

( )2 1 dLetting V V V− = 2 0dV∇ =

( )2 1 0dV V V− = =


Uniqueness TheoremFrom divergence theorem The above equation is true for any vector, so let be the vector

V SAdV A dS∇⋅ = ⋅∫ ∫

d dA V V= ∇

A

( ) ( ) ------(1)d d dV S dV V V VdV dS∇ ⋅ =∇ ∇ ⋅∫ ∫Using the vector identity ( )A A Aψ ψ ψ∇⋅ = ∇⋅ + ⋅∇

( ) ( ) ( )d d d d d dV V V V V V∇⋅ ∇ = ∇⋅ ∇ + ∇ ⋅∇

2 0dPutting V∇ =

( )22d d dV V V= ∇ + ∇

( ) ( )2 ----(2)d d dV V V∇ ⋅ ∇ = ∇


Uniqueness TheoremPutting eq (2) in eq (1)

( )2

d d dV SdV dSV V V=∇ ⋅∇∫ ∫

0dPutting V = ( )20dV

V dV∇ =∫2

0dVV dV∇ =∫

0dV∇ = ( )2 1 0V V∇ − =

2 1 .Which implies V V is a constant every where−

2 1 0At the boundary we have seen that V V− =

2 1 1 2 0 .So V V or V V everywhere− = =

1 2 .So V and V cannot be different solutions of the same problem .Uniqueness theorem is proved


Example 1Using Laplace’s theorem obtain the potential distribution between two spherical conductors separated by a single dielectric. The inner spherical conductor of radius a is at a potential V0 and the outer conductor of radius b is at potential zero. Also evaluate the electric field.

0V

ab

24sbQb

ρπ−

=

24saQ

bρ

π=


Example 1

22 2

2 2 2 21 1 1sin 0

sin sinV V VV r

r r r r rθ

θ θ θ θ φ∂ ∂ ∂ ∂ ∂⎛ ⎞ ⎛ ⎞∇ = + + =⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

Laplaces equation in spherical coordinates is

Since V is a function of r only2 2

21 0VV rr r r

∂ ∂⎛ ⎞∇ = =⎜ ⎟∂ ∂⎝ ⎠

21 0 Since we getr

≠ 2 0Vrr r∂ ∂⎛ ⎞ =⎜ ⎟∂ ∂⎝ ⎠

Integrating with respect to r 21

Vr Kr

∂=

∂2

1, VrThat Kisr

∂=

∂


Example 1 Again integrating with respect to r 1

2KV Kr

= − +

1 2 Constants K and K are found out by applying boundary . conditions

0( ) ,

( ) , 0

i r a V V

ii r b V

= =

= =

10 2

KV Ka

= − +

120 K K

b= − +

⎫⎪⎬⎪⎭

1eqations

1Solving eqations

1 0 ( )abK V

b a= −

−0

2 ( )V aK

b a−

=−

0 0

( (

) )V ab V aVb b

Tr

ha a

en = −− −


Example 1

0 0

( ) ( )V ab V aVb a r b a

= −− −

E V= −∇

1 1ˆ ˆ ˆsinr

V V VE a a ar r rθ φθ θ φ

⎡ ⎤∂ ∂ ∂= − + +⎢ ⎥∂ ∂ ∂⎣ ⎦

Since V is a function of r only ˆrVE ar

∂= −

∂0 0

( ) ( )ˆr

V ab V ab a rr b

E aa

−⎡ ⎤∂

= − ⎢ − ⎥∂ ⎣ ⎦−

02 ˆ

( ) rV abE a

b a r=

−

02 ˆ

(V/m

) rV abE a

b a r=

−


Example 2Find the potential at any point between the plates of a parallel plate capacitor and electric field.

++++++++

+ −−−−−−−−−−−

0y = y d=

z

y

0V


Example 2Laplace’s equation in rectangular coordinates is

Since V is a function of y only2

2 0Vy

∂=

∂

2 2 2

2 2 2 0V V Vx y z

∂ ∂ ∂+ + =

∂ ∂ ∂

Integrating with respect to y 1V Ky

∂=

∂ Again integrating with respect to y 1 2V K y K= +


( ) , 0ii y d V= =0( ) y 0, i V V= =


Example 21 2 Substituting in V K y K= + 2 0K V= 0

1VKd

= −

00

VV y Vd

= − +

E V= −∇ ˆ yV ay

∂= −

∂0

0 ˆ yV y V a

y d∂ ⎛ ⎞= − − +⎜ ⎟∂ ⎝ ⎠

0 ˆ yV ad

=

0 ˆ yVE ad

=0 ˆ y

VE ad

=


Example 3Find the potential distribution between the conductors, and the capacitance per unit length.

ab

0V

1m


Example 3Laplace’s equation in cylindrical coordinates is

2 22

2 2 21 1 0V V VV

zρ

ρ ρ ρ ρ φ⎛ ⎞⎛ ⎞∂ ∂ ∂ ∂

∇ = + + =⎜ ⎟⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠ ⎝ ⎠

Since V is a function of onlyρ1 0Vρρ ρ ρ

⎛ ⎞∂ ∂=⎜ ⎟∂ ∂⎝ ⎠

1 0 Since we getρ≠ 0Vρ

ρ ρ⎛ ⎞∂ ∂

=⎜ ⎟∂ ∂⎝ ⎠

Integrating with respect to ρ 1V Kρρ

∂=

∂

Again integrating with respect to ρ 1 2lnV K Kρ= +


Example 3


0( ) ,

( ) , 0

i a V V

ii b V

ρ

ρ

= =

= =

⎫⎪⎬⎪⎭

(1)eqations

( ) 1Solving eqations

0 1 2lnV K a K= +

1 20 lnK b K= +

01 ln( / )

VKb a

= − 02 ln

ln( / )VK bb a

=


Example 3

0 0ln lnln( / ) ln( / )

V V V bb a b a

ρ +−=0

ln( / )ln( / )

bVa bρ

=

0ln( / ) Voltsln( / )

bV Va bρ

=

E V= −∇ ˆV aρρ∂

= −∂ 0

ln( / ) ˆln( / )

bV aa b ρρ

ρ⎛ ⎞∂

= − ⎜ ⎟∂ ⎝ ⎠

0 ˆln( / )

V ab a ρρ

=

0 ˆ V/mln( / )

VE ab a ρρ

=

' ,Applying gauss s law for the inner conductor

SD dS Q⋅ =∫


Example 3

0

2C= ln( / )

QV b a

πε=

2C= F/mln /

( )b aπε

0 ˆ ˆ ln( / )S

V a d dza Qb a ρ ρε ρ φ

ρ=∫

0

ln( / )S

V d dz Qb aε φ =∫

0

ln( / ) S

V d dz Qb aε φ =∫ 1 2

00 0ln( / )

V dz d Qb a

πε φ =∫ ∫02

ln( / )V Q

b aπε

=

EMT Electromagnetic Theory MODULE II

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