Field Theory Electrostatics Ii

Thomas J. Odhiambo Afullo Faculty of Engineering, University of Kwazulu-Natal, Durban 4000, South Africa. E-mail: [email protected]; Tel: +27-31-260 2713; Fax: +27-31-260 2740

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Electrostatic Fields

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ENEL2FT

FIELD THEORY

REFERENCES 1. M.N. Sadiku: Elements of Electromagnetics,

Oxford University Press, 1995, ISBN 0-19-5103688. 2. N.N. Rao: Elements of Engineering Electromagnetics, Prectice-Hall, 1991, ISBN:0-13251604-7. 3. P. Lorrain, D. Corson: Electromagnetic Fields and Waves, W.H. Freeman & Co, 1970, ISBN: 07167-0330-0. 4. David T. Thomas: Engineering Electromagnetics, Pergamon Press, ISBN: 08016778-0.

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ELECTROSTATIC FIELDS  COULOMB’S LAW  The study of electrostatics begins by investigating two fundamental laws: Coulomb’s law and Gauss’s law.  Although Coulomb’s law is applicable in finding the electric field due to any charge configuration, it is easier to use Gauss’s law when charge distribution is symmetrical.  Coulomb’s law is an experimental law formulated in 1785 by the French colonel, Charles Coulomb.  It deals with the force a point charge exerts on another point charge.  By a point charge is meant a charge that is located on a body whose dimensions are much smaller than other relevant dimensions.  For example, the collection of electric charges on a pinhead may be regarded as a point charge.  Charges are generally measured in Coulombs (C).  One Coulomb is approximately equal to 6x1018 electrons; it is a very large unit of charge because the charge of an electron is -1.6019x10-19 C. ENEL2FT Field Theory

Electrostatic Fields

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ELECTROSTATIC FIELDS  COULOMB’S LAW  Coulomb’s law states that the force F between two point

charges Q1 and Q2 is:

 a) Along the line joining the charges  b) Directly proportional to the product Q1Q2 of the charges  c) Inversely proportional to the square of the distance R

between them.

 Mathematically, Coulomb’s law is expressed as: F=

kQ1Q2 R2

 Here, k is the proportionality constant.  In SI units, charges Q1 and Q2 are in coulombs (C), the

distance R is in metres, and the force F is in newtons (N).  A constant ε o is defined as the permittivity of free space (in farads/metre). ENEL2FT Field Theory

Electrostatic Fields

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ELECTROSTATIC FIELDS  COULOMB’S LAW  The constant k is defined as: 1 k=

4πε o

m/ F

10−9 εo = ≈ 8.854 x10−12 F / m 36π

 Then the equation of force becomes: QQ F = 1 22 4πε o R  If point charges Q1 and Q2 are located at points having

position vectors r1 and r2, respectively, then the force Q1Q2 aˆ12 12 = F12 on Q2 due to Q1 is F given 4πεby: R2 o

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ELECTROSTATIC FIELDS  COULOMB’S LAW

F21

Q1

R12

Q2

F12

origin

 Where:

ENEL2FT Field Theory

    R12 = r2 − r1; R = R12  R12 aˆ12 =  R12 Electrostatic Fields

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ELECTROSTATIC FIELDS  COULOMB’S LAW  We may re-write Coulomb’s equation as:    Q1Q2  Q1Q2 ( r2 − r1 ) F12 = R =  3 3 12 4πε o R 4πε o r2 − r1  Also note that:   F21 = − F12

 It noted that like charges (charges of the same sign)

repel each other, while unlike charges attract.  The distance R between the two charged bodies Q1 and Q2 must be large compared with the linear dimensions of the bodies.  Q1 and Q2 must be static (at rest).  The signs of Q1 and Q2 must be taken into account. ENEL2FT Field Theory

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ELECTROSTATIC FIELDS  COULOMB’S LAW  If there are more than two point charges, we can use the

principle of superposition to determine the force on a particular charge.  The principle states that if there are N charges Q1, Q2, ..,QN located respectively at points with position vectors r1,r2,..,r, the resultant force F on a charge Q located at point r is the vector sum of the forces exerted on Q by each of the charges  Q1, Q2, ..,Q  N. Hence:   QQ1 ( r − r1 ) QQ2 ( r − r2 ) QQN ( r − rN ) F= + + .. +  3  3   3 4πε r − r1 4πε r − r2 4πε r − rN    Q N Qk ( r − rk ) ∑ F= 4πε o k =1 r − rk 3

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ELECTROSTATIC FIELDS  COULOMB’S LAW: ELECTRIC FIELD INTENSITY  We define the electric field intensity or electric field

strength as the force per unit charge when placed in the electric field.  1   That is: E= F Q

 Thus the electric field intensity is in the direction of the

force F and is measured in Volts/metre.  The electric field intensity at point r due to a point charge located at r1 is obtained as:  E=

ENEL2FT Field Theory

   Q Q( r − r1 ) R=  3 3 4πε o R 4πε o r − r1

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ELECTROSTATIC FIELDS  COULOMB’S LAW: ELECTRIC FIELD INTENSITY  For N point charges Q1,Q2,..,QN located at positions

r1,r2,..,rN, the electric field intensity at point r is obtained as:      Q1 ( r − r1 ) QN ( r − rN ) Q2 ( r − r2 ) E= + + .. +  3   3   3 4πε r − r1 4πε r − r2 4πε r − rN    1 N Qk ( r − rk ) E= ∑ 4πε o k =1 r − rk 3

 Example:  Point charges of 2mC and 4mC are located at (3,2,1) and

(-1,-2,-3), respectively. Calculate the electric force on a 10 nC charge located at (0,2,4). Also calculate the electric field intensity at that point.

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ELECTROSTATIC ELECTRIC FIELDS DUE FIELDS TO CONTINUOUS CHARGE

DISTRIBUTIONS

 So far, we have only considered forces and electric fields due

to point charges, which are essentially charges occupying very small physical space.  At a macroscopic scale, we can disregard the discrete nature of the charge distribution and treat the net charge contained in an elemental volume ∆v as if it were uniformly distributed within it. ∆q dq ρ v = lim = (C / m3 )  Accordingly, we define∆vthe →0 ∆volume v dv charge density as:

 Where ∆q is the charge contained in ∆v. The variation of ρ v with spatial location is its spatial distribution. The Q called = ∫ ρ v dv Coulombs

total charge contained inv volume v is given by:

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE

DISTRIBUTIONS

 In some cases, particularly when dealing with conductors,

electric charge may be distributed across the surface of a material, in which case the relevant quantity of interest is ∆q dq the surface charge density, ρ = lim ρ s=, defined as: s

∆s →0

∆s

ds

 Where ∆q is the charge present across an elemental

surface area ∆s. Similarly, if the charge is distributed along a line, we characterize the distribution in terms of the line ∆q dq ( C / m) ρ = lim = charge density ρ l, defined as: l ∆l →0 ∆l

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dl

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ELECTROSTATIC FIELDS

 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE

DISTRIBUTIONS

 The electric field intensity due to each of the charge

distributions ρ l,ρ s,and ρ v may be regarded as the summation of the field distributed by the numerous point charges making up the charge distribution.  ρl dl  Thus we replace Q in the E =equations rˆ for E, and integrating, ∫ 2 4πε o R we get:  ρ ds E = ∫ s 2 rˆ 4πε o R  ρ dv E = ∫ v 2 rˆ 4πε o R

 We shall now apply these formulas to specific charge

distributions.

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FIELDS ELECTROSTATIC ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS – AN INFINITE LINE CHARGE  Consider a line charge with a uniform charge density ρ L

extending from -∞ to +∞ along the z-axis, as shown below. dz

z Infinite line charge

ENEL2FT Field Theory

rˆ

 R r

Electrostatic Fields

− zˆ α

rˆ

α

aˆ R

 dE

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO AN INFINITE LINE

CHARGE  The charge element dQ associated with element dz of dQ = ρ L dz the line is:

 The electric field intensity at point P a distance r from  ρ L dz ρ L dz 

= aˆ = R the line, due to thedEelemental 2 R charge 3 ρLdz is given by: 4πε o R 4πε o R   R ⇒ aˆ R = ; R = R R

R = r 2 + z 2 ; z = r tan α ⇒ R = r 2 + r 2 tan 2 α = r secα

 From geometry, we obtain: dz d d  sin α  2 2 =r tan α = r = r sec α ⇒ dz = r sec αdα   dα dα dα  cosα  ENEL2FT Field Theory

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO AN INFINITE LINE

CHARGE  Also, aˆ R = for rˆ costhe α − zˆunit sin α vector we have:

 ρ L  ( rˆ cosα − zˆ sin α ) r sec 2 αdα  ρL [ ( rˆ cosα − zˆ sin α ) dα ] ∴ dE = =   2 2 4πε o  4 πε r r sec α o 

 If we now integrate over the entire line, then α varies

from / 2 +π/2 as z variesρfrom -∞ to +∞; thus:  –π/2 ρ πto E=

L

∫ ( rˆ cosα − zˆ sin α ) dα =

4πε o r −π / 2

{ [ rˆ sin α ] 4πε r L

o

π /2 −π / 2

+ [ zˆ cosα ]π−π/ 2/ 2

}

 ρL ∴E = rˆ 2πε o r

 In normal cylindrical coordinates, the expression  ρL becomes: E= ρˆ 2πε o ρ ENEL2FT Field Theory

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO AN INFINITE LINE

CHARGE  Alternatively, one can see from the expression for dE  that: ρ L  ( rˆ cosα − zˆ sin α ) r sec2 αdα  ρL

dE =

 4πε o 

r sec α 2

2

[ ( rˆ cosα − zˆ sin α ) dα ] = rˆdEr + zˆdEz =  4πε o r

 One observes that at the observation point P, the

contribution to Ez due to the element dz at point +z on  by the contribution due to the line charge is cancelled E z = 0; ⇒ E = rˆEr + zˆE z = rˆEr the element dz at position –z along the line charge. Therefore, we could just conclude that:

Fieldsargument for surface charge. We shall useElectrostatic a similar

 Field Theory ENEL2FT

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO CIRCULAR RING OF

CHARGE

zˆ

 z dE  αR

h Circular ring of charge

ϕ

α

aˆ R

− ρˆ y

ρ

dl

x

 Consider a circular ring of charge of radius ρ, having

uniform charge density ρ l C/m. The ring is placed on the x-y plane.

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO CIRCULAR RING OF

CHARGE  We are required to determine the total electric field at the point P along the z-axis, located a height h above the x-y plane.  Consider an elemental length dl on the ring. The electric  this dQ elemental charge is given by: field arising from dE = aˆ 2 4πε o R

R

dQ = ρ L dl = ρ L ( ρdϕ )  R = ρ 2 + h 2 ; aˆ R = − ρˆ sin α + zˆ cosα  ∴ dE =

ρ L ( ρdϕ ) ( − ρˆ sin α + zˆ cosα ) 2 2 4πε o ρ + h

(

)

 ⇒ dE = ρˆdE ρ + zˆdE z

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO CIRCULAR RING OF

CHARGE  Thus dE has both a z-component and a ρ−component.  However, from symmetry considerations, for every element dl in the direction ρ giving rise to an elemental field strength dEρ, there is a corresponding opposite element –dl giving rise to an opposite elemental electric field strength –dEρ. Therefore the ρ components of dE  dE hasρ Lonly ( ρdϕ )a z-component. cancel; thisρˆdE implies that ( zˆ cosα ) = 0 , ⇒ d E = z dE = ˆ ρ z Thus: 4πε o ( ρ 2 + h 2 )  2π ρ L ( zˆ cosα )( ρdϕ ) ρ L ( zˆρ cosα ) E= ∫ = 2 2 4πε o ρ + h 2ε o ρ 2 + h 2 0

(

 ρ L ( zˆρ cosα ) zˆρhρ L  Simplifying, E= = we obtain: 2ε o ρ 2 + h 2 2ε o ρ 2 + h 2

(

ENEL2FT Field Theory

)

(

Electrostatic Fields

)

)

3/ 2

(

=

)

zˆ ( 2πρρ L ) h

(

4πε o ρ + h 2

)

2 3/ 2

=

(

zˆQh

4πε o ρ + h 2

)

2 3/ 2

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ELECTROSTATIC FIELDS

 ELECTRIC FIELDS DUE TO AN INFINITE SURFACE

CHARGE  Let us consider an infinite plane sheet of charge in the xyplane with uniform surface charge density ρ s C/m2. We are  required to find the electric field intensity due to it z dE sheet. everywhere above the α

h

ENEL2FT Field Theory

Electrostatic Fields

zˆ

aˆ R α

ϕ x

 R

y ρ dϕ

− ρˆ

dρ

dA 21

FIELDS ELECTROSTATIC ELECTRIC FIELDS DUE TO AN INFINITE SURFACE CHARGE

 Consider the point P(0,0,h) on the z-axis. The sheet of

surface charge is thus placed a distance h below P. The charge contribution due to an elemental area dA is given by: dQ = ρ s dA; dA = ( ρdϕ ) dρ ⇒ dQ = ρ s ( ρdϕ ) dρ  R = R = h 2 + ρ 2 ; ρ = h tan α ; ⇒ R = h 1 + tan 2 α = h secα

 We also derive the following relationships from the dρ d sin α  2 2 = h sec α ; ⇒ d ρ = h sec αdα sketch: = h  dα dα  cosα  aˆ R = zˆ cosα − ρˆ sin α

 dQaˆ R ρ s ( ρdϕdρ ) ρ s  (h tan αdϕ )(h sec 2 αdα )[ zˆ cosα − ρˆ sin α ]  dE = = aˆ R =   2 2 2 2  Then the electric intensity arising from this 4πε o R 4πε o R field4πε h sec α o  

elemental charge is:

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ELECTROSTATIC FIELDS

 ELECTRIC FIELDS DUE TO AN INFINITE SURFACE CHARGEdE = ρ s { [ zˆ tan α cosα − ρˆ tan α sin α ] dαdϕ } 4πε o   ρ s  sin 2 α  ˆ ∴ dE = z sin α − ρ d α d ϕ ˆ   = zˆdE z + ρˆdE ρ  4πε o  cosα  

 The total electric field is obtained from the integration of

dE over the entire surface. Here, ϕ varies from (0,2π), while α varied from (0,π/2).  Note that dE has two components: one, dEz in the zdirection, and the other is dEρ in the ρ direction.

 For the ρ component of dE, for each value of dEρ , there

is a canceling value, -dEρ , from the opposite element. Thus theρρ components cancel each other out, and we 2π π / 2 π /2 ρ ρ     s = only α ) dα  dϕ = s  ∫ ( zˆ sin α ) dα  = s zˆ = zˆE z ∫  the ∫ ( zˆ sinz-component: haveEleft 4πε 2ε 2ε o 0

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

0

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

o



0



o

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ELECTROSTATIC FIELDS  ELECTRIC FIELDS DUE TO AN INFINITE SURFACE

CHARGE

 For a point located below the charge sheet, the electric field intensity is:  ρ E = − zˆ s 2ε o  If we consider two infinite parallel, oppositely-

charged charge sheets, one with charge density ρ s, 2 and the other with charge density –ρ C/m , the s  opposite ρs  ( − ρs )  zˆ + − zˆthe two total electric fieldE =between plates is given by: 2ε  2ε  o



o



 ρ ∴ E = zˆ s εo

 This would therefore be the total electric field between

two plates of a parallel-plate capacitor with (approximately) infinite dimensions.

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ELECTROSTATIC FIELDS  ELECTRIC FLUX DENSITY  Let us define a vector field, D, as:   D = εE

 Where ε is the electrical permittivity of the medium.

Thus D is independent of the medium. Define the electric flux, Ψ, as:   Ψ = ∫ D.dS

 The electric flux is measured in Coulombs, and therefore

the vector D is called the electric flux density, measured in C/m2.  Thus all formulas derived for E from Coulomb’s law can be used in calculating D, except we have to multiply those results by εo. Thus for a volume charge  ρ dv distribution, D = v aˆ ∫

ENEL2FT Field Theory

Electrostatic Fields

4πR 2

R

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ELECTROSTATIC FIELDS  GAUSS’S LAW  Gauss’s law states that the total electric flux, Ψ, flowing

out of a closed surface S equals to the total charge enclosed by the surface.  That is:   Ψ = Qenc ⇒ Ψ = ∫S D.dS = Qenc

 Where Qenc=total charge enclosed.   Q = ∫ ρ v dv ⇒ ∫ D.dS = ∫ ρ v dv v

s

v

 Gauss’s law is thus an alternative statement of

Coulomb’s law.  Gauss’s law provides an easy means of finding E or D for symmetrical charge distributions such as a point Charge, an infinite line charge, an infinite surface charge, and a spherical charge distribution.

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ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A POINT

CHARGE  Suppose that a point charge Q is located at the origin.  To determine the flux density D at a point P, it is seen that choosing a spherical surface containing P will satisfy symmetry conditions.  Thus a spherical surfacez centered at the origin is the Gaussian surface in this case, as shown below.

P

Q

x ENEL2FT Field Theory

Electrostatic Fields

r

 D

y Gaussian Surface

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ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A POINT

CHARGE  Applying Gauss’s law, with a spherical surface as the Gaussian surface, we  have:  Q = ∫ D.dS = ∫ Dr rˆ.dS V V  dS = rˆ( rdθ )( r sin θdϕ ) π 2π

  ∴ Q = Dr ∫  ∫ r 2 dϕ  sin θdθ = Dr 4πr 2  0 0  Q ∴ D = rˆ 4πr 2

 From this, we can determine be:    E to Q D = ε o E ⇒ E = rˆ 4ε oπr 2

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ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A LINE CHARGE  Suppose the infinite line of uniform charge ρL C/m lies

along the z-axis.  To determine D at a point P a distance ρ from the line, we choose a cylindrical surface containing P to satisfy symmetry conditions as shown in the figure below.

z

Line charge ρ L C/m

ρ

L

P

Gaussian surface

 D y

x ENEL2FT Field Theory

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ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A LINE CHARGE  D is constant on and normal to the cylindrical Gaussian

surface. Thus,

 D = ρˆDρ

 If we apply Gauss’s law to an arbitrary length L of the

line, we have:   Qenc = ρ L L = ∫ D.dS  dS = ρˆ ( ρdϕdz ) ∴ Qenc = ρ L L = Dρ 2 ρπL   ρL ρL ∴ D = ρˆ ⇒ Eρˆ 2 ρπ 2ε o ρπ

 Note that the evaluation of D.dS on the top and bottom

surfaces of the cylinder is zero since D has no zENEL2FT Field Theory Electrostatic Fields component.

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ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A UNIFORMLY

CHARGED SPHERE

Gaussian surface

r

a

r≤a

r r≥a

a

 Consider a sphere of radius a with a uniform charge ρ v

C/m3.

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ELECTROSTATIC FIELDS

 APPLICATION OF GAUSS’S LAW TO A UNIFORMLY

CHARGED SPHERE  To determine D everywhere, we construct Gaussian surfaces for cases r≤a, and r≥a, separately.  Since the charge has spherical symmetry, it is obvious that a spherical surface is an appropriate Gaussian surface.  For r≤a, the total charge enclosed by the spherical surface of radius r is: 2π

Qenc = ∫ ρ v dv = ∫

π

∫

r

2 ∫ r sin θdrdθdϕ

ϕ = 0θ = 0 r = 0

Qenc

4πr 3 = ρv 3

(

 The total flux is given   by:2π π 2 Ψ = ∫ D.dS = Dr ∫ ∫ r sin θdθdϕ = Dr 4πr 2 ϕ =0 θ = 0

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) 32

ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A UNIFORMLY

CHARGED SPHERE  Thus we have: 4πr 3 Ψ = Qenc ⇒ 4πr Dr = ρv 3  r ∴ D = rˆ , 0 ≤ r ≤ a 3 2

 For r≥a, the charge enclosed by the Gaussian surface is

the entire charge in this case, that is: Qenc

4πa 3 = ∫ ρ v dv = ρ v ∫ ∫ ∫ r sin θdrdθdϕ = ρv 3 ϕ =0 θ =0 r =0 2π

π

a

2

 Similarly, the flux is given by:

(

)

  Ψ = ∫ D.dS = 4πr 2 Dr

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ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A UNIFORMLY

CHARGED SPHERE  Hence we obtain,

(

)

4πa 3 4πr Dr = ρv 3  a3 ρv ⇒ D = rˆ 2 , r ≥ a 3r 2

 Thus from the foregoing, D is everywhere given by:  rρ v rˆ , r≤a   3 D= 3 rˆ a ρ v , r ≥ a  3r 2

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ELECTROSTATIC FIELDS  APPLICATION OF GAUSS’S LAW TO A AN INFINITE

SHEET OF CHARGE  Consider the infinite sheet of uniform charge with charge density ρ s C/m2 lying on the z-0 plane (xy-plane).

z Infinite sheet of charge, ρ s C/m2

 D P

y Area A

x ENEL2FT Field Theory

 D Electrostatic Fields

Gaussian surface 35

FIELDS ELECTROSTATIC APPLICATION OF GAUSS’S LAW TO A AN INFINITE SHEET OF CHARGE  To determine D at point P, we choose a rectangular box that is cut symmetrically by the sheet of charge and has two of its sides parallel to the sheet as shown in the figure.   As D is normal toDthe = zˆDzsheet, we have, when applying Gauss’s law:     ∫ D.dS = Q = Dz  ∫ dS + ∫ dS  bottom  top

 Note that D has no x- and y- components, hence Dx=0, Q = ρ s A = Ψ = Dz ( A + A) = 2 ADz

Dy=0.

   ρs D ρ ∴ D = zˆ of ⇒the E = pillbox = zˆ s each has area A,  If the top and bottom 2 εo 2ε o

then we get:

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ELECTROSTATIC FIELDS  ELECTRIC POTENTIAL  In electric circuits, we work with voltages and

currents.  The voltage V between two points in the circuit represents the amount of work, or potential energy, required to move a unit charge between the two points.  In fact, the term “voltage” is a shortened version of the term “voltage potential” and is the same as electric potential.  Even though when we solve a circuit problem we usually do not consider the electric fields present in the circuit, in fact it is the existence of an electric field between two points that gives rise to the voltage difference between them, such as across a resistor or capacitor.  The relationship between the electric field, E, and the electric potential, V, is the subject of this section. ENEL2FT Field Theory

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ELECTROSTATIC FIELDS  ELECTRIC POTENTIAL

y dy

 E

 E

q

x

 Consider the case of a positive charge q in a uniform

electric field

 E = − yˆ E

 Which is parallel to –y direction, as shown in the figure.  The presence of the field E exerts a force F on the

charge, given by: ENEL2FT Field Theory

Electrostatic Fields

38

ELECTROSTATIC FIELDS ELECTRIC POTENTIAL   Fe = qE = − yˆ qE

 The force exerted is in the negative y-direction.  If we attempt to move the charge along the positive y-

direction, against the force Fe, we will need to provide an external force Fext to counteract Fe, which requires an expenditure of energy.  To move q without any acceleration (at a constant speed), it is necessary that the net force acting on the charge be zero. This means that:    Fext = − Fe = − qE

 The work done, or energy expended, in moving any

object a vector differential distance dl under the     influence of force Fext is: dW = F .dl = − qE.dl ext

ENEL2FT Field Theory

Electrostatic Fields

39

ELECTROSTATIC FIELDS ELECTRIC POTENTIAL  If the charge is moved a distance dy along y, then: dW = −q( − yˆ E ) . yˆdy = qEdy

 The differential electric potential energy dW per unit

charge is called the differential electric potential, or differential voltage, dV.  That is,   dW dV =

q

= − E.dl ( J / C or V )

 The unit of V is the volt (V), and therefore the electric

field is expressed in volts per metre (V/m).

ENEL2FT Field Theory

Electrostatic Fields

40

ELECTROSTATIC ELECTRIC POTENTIAL

FIELDS

 Thus the potential difference between any two points P2

and P1 is obtained by integrating dV along the path between P1 and P2V. =That ∫ dV is:   V21 = V2 − V1 = ∫ dV = ∫ E.dl P2

P2

P1

P1

 Where V1 and V2 are the electric potentials at points P1

and P2, respectively.  The result of the line integral above should be

independent of the specific path of integration between points P1 and P2. P P

    V22 = V2 seen − V2 = ∫ dV = ∫ E .dl = integral ∫ E.dl = 0  It is also readily that the of the P2 P2 C electrostatic field closed contour is   E around     any  But ∫ E.dl = ∫ ∇xE.dS ⇒∇xE = 0 zero: 2

C

ENEL2FT Field Theory

2

S

Electrostatic Fields

41

ELECTROSTATIC FIELDS ELECTRIC POTENTIAL  We now define what is meant by the electric potential V

at a point in space.  Whenever we talk of a voltage V in a circuit, we do so in reference to a voltage of some conveniently chosen point to which we have assigned a reference voltage of zero, which we call ground.  The same principle applies to electric potential V. Usually, the reference potential point is chosen to be at infinity. That is, if we assume V1=0 when P1 is at infinity, the electric potential at any point P is given by: P   V = − ∫ E.dl ∞

ENEL2FT Field Theory

Electrostatic Fields

42

 ELECTROSTATIC FIELDS ELECTRIC POTENTIAL DUE TO POINT CHARGES  For a charge q located at the origin of a spherical coordinate system, the electric field at a distance R is given by:  E = aˆ R

q (V / m) 4πε o R 2

 As indicated before, the choice of the integration path

between two points in determining the potential V is quite arbitrary. Hence we conveniently choose the path to be along the radial direction R, in which case we have: R   R q  q .( aˆ R dR ) = V = − ∫ E.dl = − ∫  aˆ R (V ) 2 4 πε R 4πε o R  ∞ ∞ o

 If the charge q is at a location other than the origin,

specified by a source position vector R1, then the potential V at observation position vector R becomes: q V=

ENEL2FT Field Theory

Electrostatic Fields

  4πε o R − R1

(V )

43

ELECTROSTATIC FIELDS ELECTRIC POTENTIAL DUE TO POINT CHARGES The principle of superposition that has been applied

previously to the electric field E also applies to the electric potential V. For N discrete point charges q1, q2, ..,qN having position vectors R1, R2, .., R , the electric potential 1 N qNi V= ∑   (V ) is: 4πε i =1 R − R o

ENEL2FT Field Theory

Electrostatic Fields

i

44

ELECTROSTATIC FIELDS ELECTRIC POTENTIAL DUE TO CONTINUOUS

CHARGE DISTRIBUTIONS  For a continuous charge distribution specified over a

given volume V, across a surface S, or along a line l, we replace the qi with: ρ dv; ρ ds; ρ dl v

s

l

 Then, converting the summation into integration, we 1 ρv V ( R) = obtain: ∫ dv (volume distribution) 4πε o V R 1 ρs V ( R) = ∫ dS ( surface distribution) 4πε o S R 1 ρl V ( R) = ∫ dl (line distribution) 4πε o L R

ENEL2FT Field Theory

Electrostatic Fields

45

ELECTROSTATIC FIELDS  ELECTRIC FIELD AS A FUNCTION OF ELECTRIC

POTENTIAL  We have seen that:   dV = − E.dl

 If we resolve E and dl into rectangular coordinates, we 



have: E = xˆE x + yˆ E y + zˆE z ; dl = xˆdx + yˆ dy + zˆdz

  ∴ E.dl = ( xˆE x + yˆ E y + zˆE z ).( xˆdx + yˆ dy + zˆdz ) = E x dx + E y dy + E z dz ∂V ∂V ∂V dx + dy + dz ∂x ∂y ∂z ∂V ∂V ∂V ∴ Ex = − ; Ey = − ; Ez = − ; ∂x ∂y ∂z dV =

  E = −∇V

 Thus ENEL2FT Field Theory

Electrostatic Fields

46

ELECTROSTATIC FIELDS

EXAMPLE:

Given the potential function:

V=

10 r

2

sin θ cos φ

Determine: A) The electric field strength and the electric

flux density at (2,π/2, 0) The work done in moving a 10-µC charge from point A (1,30o, 120o) to B(4,90o,60o)

ENEL2FT Field Theory

Electrostatic Fields

47

ELECTROSTATIC FIELDS

 SOLUTION:

   ∂V 1 ∂V ˆ 1 ∂V ˆ E = −∇V = −  rˆ + θ+ φ r ∂θ r sin θ ∂φ   ∂r 20 10 10 = 3 sin θ cos φrˆ − 3 cosθ cos φθˆ + 3 sin φφˆ r r r   20  20 E =  rˆ − 0θˆ + 0φˆ  = rˆ V / m = 2.5rˆ V / m ( 2,π / 2, 0 )  8  8   10 − 9  20  −11 D = εoE = C / m2  rˆ  = 2.21x10 36π  8  B  W = QV AB = −Q ∫ E.dl = Q(VB − V A ) A

 10  10 = Q  2 sin θ cos φ − 2 sin θ cosφ   r ( 4,90 o ,60 o ) r (1,30 o ,120 o )  10 10  = 10 x10 − 6  sin 90 o cos 60 o − sin 30 o cos120 o  = 10 − 5 1 16 

(

)

(

)1032 − −410 

∴W = 2.8125x10 -5 J

ENEL2FT Field Theory

Electrostatic Fields

48

ELECTROSTATIC FIELDS  THE ELECTRIC DIPOLE  An electric dipole is formed when two point charges of

equal but opposite sign are separated by a small distance, as shown below. z P

r1 θ r2 +Q

r

d

y -Q

x

ENEL2FT Field Theory

An Eectric Dipole

Electrostatic Fields

49

ELECTROSTATIC FIELDS  THE ELECTRIC DIPOLE  The potential at point P(r,θ,φ) is given by: V=

Q 4πεo

1 1  Q  r2 − r1  − =      r1 r2  4πεo  r1r2 

 Where r1 and r2 are the distances between P and +Q and –Q,

respectively. 2  If r>>d,2 then: 2 d  r =r + − 2r (d / 2) cosθ ≈ r 2 − 2r (d / 2) cosθ   2

1

∴ r1 = r 2 − 2r (d / 2) cosθ = r 1 − (d / r ) cosθ ≈ r − (d / 2) cosθ r22

2

d  = r +   + 2r (d / 2) cosθ ≈ r 2 + 2r (d / 2) cosθ 2 2

∴ r2 = r 2 + 2r (d / 2) cosθ = r 1 + (d / r ) cosθ ≈ r + (d / 2) cosθ ∴ r2 − r1 ≈ d cosθ

(

)(

)

r1r2 = r 1 − (d / r ) cosθ r 1 + (d / r ) cosθ = r 2

((

(1 − (d / r ) cosθ ) (

1 + (d / r ) cosθ

) ))

= r 2  1 − [ (d / r ) cosθ ] 2  ≈ r 2   ∴V =

ENEL2FT Field Theory

Q 4πε o

 r2 − r1  Qd cosθ  rr = 2  1 2  4πε o r Electrostatic Fields

50

ELECTROSTATIC FIELDS

 THE ELECTRIC DIPOLE  Define the dipole moment p as:   p = Qd  Qd cosθ p.rˆ V= = 2 4πε o r 4πε o r 2

 The electric field due to the dipole with centre at the

origin, is:

  1 ∂V ˆ  ∂V E = −∇V = −  rˆ + θ ∂ r r ∂ θ   Qd cosθ Qd sin θ ˆ ˆ+ = r θ 3 3 2πε o r 4πε o r  p ˆ + sin θθˆ ∴E = 2 cos θ r 4πε o r 3

[

ENEL2FT Field Theory

Electrostatic Fields

] 51

ELECTROSTATIC FIELDS

THE ELECTRIC DIPOLE Notice that a point charge is a monopole, and its electric filed varies inversely as r2, while its potential varies inversely as r. For the dipole, we notice that the electric field varies inversely as r3, while its potential varies inversely as r2. The electric fields due to the presence of a quadrupole (consisting of two dipoles) vary inversely as r4, while the corresponding potential varies inversely as r3. EXAMPLE:  Two dipoles have dipole moments p1 and p2 p1 = −5 x10 − 9 zˆ Cm; p2 = 9 x10 − 9 zˆ Cm

are located at points (0,0,2) and (0,0,3), respectively. Find the potential at the origin if:

ENEL2FT Field Theory

Electrostatic Fields

52

ELECTROSTATIC FIELDS

THE ELECTRIC DIPOLE  SOLUTION:  The potential is given by: 2

V= ∑

  pk .rk

3 k =1 4πε o rk

 p1 = −5 x10 − 9 zˆ  p2 = 9 x10 − 9 zˆ; ∴V =

1 10 − 9 4π 36π

ENEL2FT Field Theory

    1  p1.r1 p2 .r2  =  3 + 3  4πε o  r1 r2    ; r1 = (0,0,0) − (0,0,−2) = 2 zˆ; r1 = r1 = 2   ˆ r2 = (0,0,0) − (0,0,3) = −3 z; r2 = r2 = 3  − 10 x10 − 9 27 x10 − 9   10  − = 9 − − 1 = −20.25V    8 27   8  

Electrostatic Fields

53

ELECTROSTATIC FIELDS  EXAMPLE:  An electric dipole of dipole moment p is located at the

origin, where:

 p = 100 x10 −12 Cm  Find the electric filed intensity E and potential V at the

following points:  A) (0,0,10).  B) (1,π/3,π/2)  ANS: 

A) E = 1.8 x10 −3 rˆV / m; V = 9 x10 −3V  B) E = 0.9rˆ + 0.78θˆ x10 −3V / m; V = 0.45V

(

ENEL2FT Field Theory

)

Electrostatic Fields

54

ELECTROSTATIC FIELDS

 ENERGY DENSITY IN ELECTROSTATIC FIELDS  To determine the energy present in an assembly of

charges, we must first determine the amount of work necessary to assemble them.  Suppose we wish to position three point charges Q1, Q2, and Q3 in an initial empty space shown below.

P1

Q1

P2 P3

∞

Q2 Q3

ENEL2FT Field Theory

Electrostatic Fields

55

ELECTROSTATIC FIELDS  ENERGY DENSITY IN ELECTROSTATIC FIELDS  No work is required to transfer Q1 from infinity to P1

because the space is initially charge free and there is no electric field.  The 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.  Similarly, the work done in positioning Q3 at P3 is equal to Q3 (V32+V31), where V32 and V31 are the potentials at P3 due to Q2 and Q1, respectively. WE work = W1 + W  Hence the total 2 + W3in positioning the three done charges is: = 0 + Q2V21 + Q3 (V31 + V32 )

WE = W3 + W2 + W1

 If the charges=were ) reverse order, then: 0 + Q2Vpositioned 23 + Q1 (V12 + V13in ENEL2FT Field Theory

Electrostatic Fields

56

ELECTROSTATIC FIELDS ENERGY DENSITY IN ELECTROSTATIC

FIELDS

 Here, V23 is the potential at P2 due to Q3, V12 and V13 are

respectively the potentials at P1 due to Q2 and Q3. Thus the two give: 2Wequations E = Q1 ( V12 + V13 ) + Q2 ( V21 + V23 ) + Q3 ( V31 + V32 )

= Q1V1 + Q2V2 + Q3V3 1 ∴WE = ( Q1V1 + Q2V2 + Q3V3 ) 2

 Where V1, V2, and V3 are the potentials at P1, P2, and P3,

respectively. 1 n  In general, if there WE =are n point QkVk charges, the above equation becomes: 2 k =1

∑

ENEL2FT Field Theory

Electrostatic Fields

57

ELECTROSTATIC FIELDS  ENERGY DENSITY IN ELECTROSTATIC FIELDS  If, instead of point charges, the region has a continuous

charge distribution, the above summation becomes as integration: 1 WE = ∫ ρ LVdl ( line ch arg e ) 2 1 WE = ∫ ρ SVdS ( surface ch arg e ) 2 1 WE = ∫ ρV Vdv ( volume ch arg e ) 2

 We can further refine the expression using volume

 the  vector identities: charge density by using

ρ v = ∇.D      ∇.VA = A.∇V + V ∇. A      ∴V ∇. A = ∇.VA − A.∇V

( )

ENEL2FT Field Theory

Electrostatic Fields

( )

58

ELECTROSTATIC FIELDS  ENERGY DENSITY IN ELECTROSTATIC FIELDS Therefore we obtain:

(

)

1 1   ρ Vdv = ∇.D Vdv V 2∫ 2∫ 1   1   1  = ∫ ∇.D Vdv = ∫ ∇.VD dv − ∫ D.∇V dv 2 2 2

WE =

(

)

(

)

(

)

By applying the divergence theorem to the first

term on the right-hand side of the equation, we have:   1  1 WE = ∫ (VD ).dS − ∫ ( D.∇V ) dv 2 2 S

V

For point charges, V varies as 1/r, and D varies

as 1/r2; for dipoles, V varies as 1/r2 and D varies ENEL2FT Field Theory Fields as 1/r3; andElectrostatic so on.

59

ELECTROSTATIC FIELDS  Hence VD in the first term on the rhs must vary at least as 1/r

3

while dS varies as r2.  Consequently the first integral must tend to zero as the surface dS becomes large.  Therefore WE reduces to:

(

)

(

)

1  1   WE = − ∫ D.∇V dv = ∫ D.E dv 2V 2V ∴WE =

ENEL2FT Field Theory

1 2 ε E dv o ∫ 2V

Electrostatic Fields

60

ELECTROSTATIC FIELDS

 EXAMPLE:

 Three point charges, -1nC, 4nC, and 3nC, are located at

(0,0,0), (0,0,1), and (1,0,0), respectively. Find the energy in the system.  SOLUTION: WE =

1 n 1 ∑ QkVk = [ Q1V1 + Q2V2 + Q3V3 ] 2 k =1 2

V1 = V12

Q3 Q2 + V13 = + = 4πε o (1) 4πε o (1)

4 x10 − 9 10 − 9 (1) 4π 36π

+

3 x10 − 9 10 − 9 (1) 4π 36π

= 63V

Q3 Q1 − 1x10 − 9 3 x10 − 9 V2 = V21 + V23 = + = + = 10.09V 4πε o (1) 4πε o 2 10 − 9 10 − 9 4π (1) 4π 2 36π 36π

( )

Q1 Q2 − 1x10 − 9 4 x10 − 9 + = + = 16.46V 4πε o (1) 4πε o 2 10 − 9 10 − 9 (1) 4π 4π 2 36π 36π 1 1 ∴WE = [ Q1V1 + Q2V2 + Q3V3 ] = − 1x10 − 9 63 + 4 x10 − 9 10.09 + 3 x10 − 9 16.46 2 2 V3 = V31 + V32 =

( )

[(

)

(

)

(

)

]

WE = 13.36x10 - 9 J ENEL2FT Field Theory

Electrostatic Fields

61

ELECTROSTATIC FIELDS

 DIELECTRICS IN AN ELECTRIC FIELD  An ideal dielectric or insulator is a material with no free

electrons in its lattice structure.  That is, all the electrons associated with an ideal dielectric are strongly bound to its constituent molecules.  These electrons experience very strong internal restraining forces that oppose their random movements.  Therefore when an electric field is maintained within a dielectric by an external source of energy, there is no current.  However, under the influence of an electric field, the molecules of a dielectric material experience distortion in the sense that the centre of a positive charge of a molecule no longer coincides with the centre of a negative charge.  We then say the dielectric material is polarized; in such ENEL2FT Field Theory Electrostatic Fields a polarized state, the dielectric material contains a

62

ELECTROSTATIC FIELDS  DIELECTRICS IN AN ELECTRIC FIELD

Fig.A: A dielectric in its normal state where the centre of a positive charge coincides with that of a negative charge

 A schematic diagram of a dielectric slab in its normal

state is shown above. In this case, there is no influence of an external field.

ENEL2FT Field Theory

Electrostatic Fields

63

ELECTROSTATIC FIELDS

 DIELECTRICS IN AN ELECTRIC FIELD  The figure below shows the same dielectric material

under the influence of an external electric field.

 E

Fig.B: A polarized dielectric showing the separation between charge pairs

ENEL2FT Field Theory

Electrostatic Fields

64

ELECTROSTATIC FIELDS  DIELECTRICS IN AN ELECTRIC FIELD  The potential at a point P outside a polarized dielectric

material is shown in the figure below.  r

   R = r − r'

P

O

 r'

S’ dv’

ENEL2FT Field Theory

Electrostatic Fields

65

ELECTROSTATIC FIELDS

 DIELECTRICS IN AN ELECTRIC FIELD  Define the polarization vector, as the number of dipole

moments per unit volume:

  ∆p P = lim ∆v ∆v → 0

 Therefore for the volume dv’ in the figure, we can

represent the dipole moment as:

  dp = Pdv'

 The potential at point P due to the dipole moment is

given by:

 P.Rˆ dV = dv' 4πε R 2

o

ENEL2FT Field Theory

Electrostatic Fields

66

ELECTROSTATIC FIELDS

 DIELECTRICS IN AN ELECTRIC FIELD  But we have:

1 1 ∇'   = Rˆ R R   ˆ P.R P.∇' (1 / R ) ∴ dV = dv' = dv' 4πε R 4πε 2

2

o

o

 Using the vector identity:

     P.∇' (1 / R ) = ∇'.( P / R ) − ( ∇'.P ) / R     P.∇' (1 / R ) 1    P  ( ∇'.P )  ∴ dV = dv' = ∇'.  − dv'   4πε 4πε   R  R  o

ENEL2FT Field Theory

Electrostatic Fields

o

67

ELECTROSTATIC FIELDS 

DIELECTRICS IN AN ELECTRIC FIELD  Now integrating over the volume v’ of the polarized dielectric, we obtain the potential at point P as:

   ( 1   P ∇'.P )  V = ∫ dV = dv' ∫ ∇'. dv'− ∫  4πε   R  R  v'

v'

o

v'

 Applying the divergence theorem to the first term on the

right-hand side, we obtain:

   ( 1  ( P.aˆ ) ∇'.P )  V = ∫ dV = ds '− ∫ dv' ∫  πεpoint R  Thus the potential4at  PRdue to the polarized  n

v'

o

s'

v'

dielectric is the algebraic sum of two terms: a surface term and a volume term.

ENEL2FT Field Theory

Electrostatic Fields

68

ELECTROSTATIC FIELDS  DIELECTRICS IN AN ELECTRIC FIELD  If we define the bound surface charge density, ρsb,

 ρ = P.aˆ   ρ = −∇'.P

and the bound volume charge density, ρvb, as: sb

n

vb

ρ  ρ  ∴V =  ∫ ds '+ ∫ dv' R  R  sb

s'

vb

v'

 Thus the polarization of a dielectric material results

in bound charge distributions.  These bound charge distributions are not like free charges: they are created by separating the charge ENEL2FT Field Theory Electrostatic Fields pairs mentioned earlier.

69

ELECTROSTATIC FIELDS

DIELECTRICS IN AN ELECTRIC FIELD If a dielectric region contains the free charge

density in addition to the bound charge density, the contribution due to the free charge density must also be considered to determine   the  dielectric electric field in the ρ v + ρ vb region. ρ v − ∇.PThat is:

∇.E =

ε

=

   ∴ ∇.( ε E + P ) = ρ o

ε

o

o

v

The right-hand side is simply the free charge

density. But the divergence of D is also the free charge density, and it would be true as P goes    to 0. Therefore in D general = ε E we + Phave:

ENEL2FT Field Theory

Electrostatic Fields

o

70

ELECTROSTATIC FIELDS

 DIELECTRICS IN AN ELECTRIC FIELD  We can also express the polarization vector in terms of

the susceptibility of the medium, χ:

  P = ε χE     ∴ D = ε ( 1 + χ ) E = ε ε E = εE o

o

o

r

 Here, (1+ χ) is called the relative permeability or the

dielectric constant, ε, of the medium. Therefore, in any medium, the electrostatic fields satisfy the following equations:  

∇XE = 0   ∇.D = ρ   D = εE v

ENEL2FT Field Theory

Electrostatic Fields

71

ELECTROSTATIC FIELDS  PROPERTIES OF MATERIALS

 Just as electric fields can exist in free space, they can exist

in material media.  Materials are therefore broadly classified in terms of their electrical properties as conductors and non-conductors.  Non-conducting materials are usually referred to as insulators or dielectrics.  A material with high conductivity has very high electrical conductivity σ>>1. This is the case for most metals.  Insulators or dielectrics, on the other hand, are materials with low conductivity (σ<<1).  A material whose conductivity lies between those of metals and insulators is called a semi-conductor.  The conductivity of metals generally increases with decrease in temperature. At temperatures near absolute zero (T=0oK), some conductors exhibit infinite conductivity, and are called superconductors.

ENEL2FT Field Theory

Electrostatic Fields

72

ELECTROSTATIC FIELDS CONVECTION CURRENT  Electric current is generally caused by the motion of

electric charges.  Convection current, as distinct from conduction current, does not involve conductors and consequently does not satisfy Ohm’s law.  Convection current occurs when current flows through an insulating medium such as liquid, rarefied gas, or a vacuum.  A beam of electrons in a vacuum tube, for example, is a convection current.  The current through a given area is defined as the electric charge passing through the area per unit time.  That is:

dQ I= dt

ENEL2FT Field Theory

Electrostatic Fields

73

ELECTROSTATIC FIELDS

CONVECTION CURRENT

 Thus, in a current of one ampere, charge is being

transferred at a rate of one coulomb per second.  Consider the current filament shown below. ∆S

ρv

u

∆l

ENEL2FT Field Theory

Electrostatic Fields

74

ELECTROSTATIC FIELDS

CONVECTION CURRENT

 If there is a flow of charge, of density ρv, at velocity:  u = u yˆ y

∴ ∆I =

∆Q ∆l = ρ ∆S = ρ ∆Su ∆t ∆t v

v

y

 If we define the current density at a given point as the

current through a unit normal area at that point, then the y-directed current density Jy is given by:

∆I J = =ρu ∆S   ⇒J =ρu y

v

y

v

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75

ELECTROSTATIC FIELDS

CONVECTION CURRENT

 The current I is the convection current, and J is the

convection current density in Amperes per square meter.

 The total current through a prescribed surface S is

  I = ∫ J .dS

therefore given by:

S

 Compared with the general definition of flux, the

above equation shows that the current I through S is merely the flux of the current density J.

ENEL2FT Field Theory

Electrostatic Fields

76

ELECTROSTATIC FIELDS  CONDUCTION CURRENT

 When an electric filed is applied to a conductor,

conduction current flows due to the drift motion of electrons.  As the electrons move, they encounter some damping forces called resistance.  The average drift velocity of the electrons is directly proportional to the applied field.  Thus, for a conductor, we have:

  J = σE

 Here, σ is the conductivity of the material in siemens

per meter (S/m), and J is known as the conduction current density. The above equation is referred to as Ohm’s law. ENEL2FT Field Theory Electrostatic Fields

77

ELECTROSTATIC FIELDS

CONTINUITY EQUATION AND

RELAXATION TIME

Due to the principle of conservation of

charge, the time rate of decrease of charge within a given volume must be equal to the net outward flow through the closed surface of the volume.   dQ Thus the current I out = ∫Iout J .dcoming S = − outin of the closed surface is: dt

Here, Qin

    is theJ total by the dv ∫ .dS =charge ∫ ∇.Jenclosed

closed surface. V Invoking the divergence theorem, we have: ENEL2FT Field Theory

Electrostatic Fields

78

ELECTROSTATIC FIELDS CONTINUITY EQUATION AND RELAXATION TIME  But we also have:

dQ d ∂ρ − = − ∫ ρ dv = − ∫ dv dt dt ∂t in

v

V

v

v

 Thus we have:

    ∂ρ ∂ρ −∫ dv = ∫ ∇.Jdv ⇒ ∇.J = − ∂t ∂t v

V

v

V

 This is called the continuity equation, which is derived from  

J =0 the principle of conservation of∇.charge.  For steady currents, d/dt=0, and thus , showing that the total charge leaving a volume is the same as the total charge entering it.  Kirchhoff’s current law follows from this. ENEL2FT Field Theory Electrostatic Fields

79

ELECTROSTATIC FIELDS CONTINUITY EQUATION AND RELAXATION TIME  To determine the relaxation time, we use Ohm’s law

 and Gauss’s laws:

 J = σE     ρ ∇.D = ρ ⇒ ∇.E = ε

v

v

 Substituting these into the continuity equation, we

have:

  σρ ∂ρ ∇.(σE ) = =− ε ∂t ∂ρ σ ∴ + ρ =0 ∂t ε v

v

v

v

ENEL2FT Field Theory

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80

ELECTROSTATIC FIELDS

CONTINUITY EQUATION AND RELAXATION

TIME

 The solution to the above first-order differential

equation is:

ρ =ρ e ε T= σ v

− t /T

vo

 In the equation, ρvo is the initial (or introduced) charge

density at t=0.  The equation shows that as a result of introducing volume charge ρ vo at some point of the material, there is a decay of volume charge density, ρ v.  Associated with this decay is charge movement from the interior point at which it was introduced to the surface of the material.  The decay time constant T is known as the relaxation ENEL2FT Field Theory Electrostatic Fields

81

ELECTROSTATIC FIELDS 

CONTINUITY EQUATION AND RELAXATION TIME

 The relaxation time, T, is the time it takes a (volume)

charge placed in the interior of a material to drop to 1/e =0.368 of its initial value.  T is small for good conductors (metals) and very large for insulators (dielectrics).  For example, for copper, σ=5.8x107, and εr=1, while for quartz, with σ=10-17, and εr=5.0, we have: −12

T

Copper

ε ε ε (1)( 8.84 x10 ) = = = 1.53 x10 sec σ σ 5.8 x10 ε ε ε (5.0)( 8.84 x10 ) = = = = 4.42 x10 sec = 51.2 days σ σ 10 =

r

−19

o

7

−12

T

Quartz

ENEL2FT Field Theory

r

6

o

−17

Electrostatic Fields

82

ELECTROSTATIC FIELDS CONTINUITY EQUATION AND

RELAXATION TIME Thus we have a rapid decay of charge within copper, which shows that for good conductors, the relaxation time is so short that that most of the charge will vanish from the interior point and appear on the surface as surface charge. On the other hand, for good insulators like quartz, one may consider the introduced volume charge to remain ENEL2FT Field Theory Electrostatic Fields

83

ELECTROSTATIC FIELDS BOUNDARY CONDITIONS  So far we have considered the existence of the electric field

in a homogeneous medium.  If the field exists in a region consisting of two different media, the conditions that the field must satisfy at the interface separating the media are called boundary conditions.  The boundary conditions are helpful in determining the field on one side of the boundary if the field on the other side of the boundary is known.  To determine the boundary   conditions, we need to use .dlelectrostatic =0 ∫ Efor Maxwell’s equations fields:

  ∫ D.dS = Q  E = tˆE + nˆ E enc

t

n

 Here, the electric field intensity at the interface has been ENEL2FTdecomposed Field Theory Electrostatic into twoFields components:

the tangential component,

84

ELECTROSTATIC FIELDS

DIELECTRIC-DIELECTRIC BOUNDARY

CONDITIONS  Consider the electric field E existing in a region

containing two different dielectrics characterized by electric permitivities ε1=εoεr1, and ε2=εoεr2, as shown below. Medium 1, ε 1

E1

E1n E1t

a

b ∆h

E2t d E2n

∆w

c

E2 Medium 2, ε 2

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Electrostatic Fields

85

ELECTROSTATIC FIELDS  DIELECTRIC-DIELECTRIC BOUNDARY CONDITIONS  Consider the electric field intensity existing in the region

consisting of two different dielectrics, shown in the figure.  The fields in medium 1 and medium 2 can be decomposed  as: E1 = tˆE1t + nˆ E1n

 E = tˆE + nˆ E 2

2t

2n



  the first of the two∆hMaxwell’s  WE apply ∆h equations∆to h the ∆h l =0⇒ 0 = of E1t ∆ w −figure, E1n −assuming E2 n − E2 tthe ∆w +path E2 n is+very E1n ∫ E.dpath closed abcd the 2 2 2 2   small. We obtain: lim ∫ E.dl = 0 ⇒ E ∆w − E ∆w = 0 1t

∆h →0

∴E = E 1t

ENEL2FT Field Theory

2t

2t

Electrostatic Fields

86

ELECTROSTATIC FIELDS

DIELECTRIC-DIELECTRIC BOUNDARY

CONDITIONS  Consider now the pillbox (Gaussian surface) in the figure

below.

Medium 1, ε 1 D1

D1n

∆S

D1t

∆h

D2t D2n

D2

Medium 2, ε 2

ENEL2FT Field Theory

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87

ELECTROSTATIC FIELDS 

DIELECTRIC-DIELECTRIC BOUNDARY CONDITIONS

If we ∆h →0, we have:   allow

∫ D.dS = Q = 2πr∆h( D + D ) + ∆S ( D − D   lim ∫ D.dS = Q = ∆S ( D − D ) = ρ ∆S enc

1t

enc

∆h →0

∴( D − D 1n

2n

)=ρ

1n

2t

2n

1n

2n

)

s

s

Here ρs is the surface charge density placed at

the boundary. If no surface charges exist at the interface, then the normal component of D is continuous D that = D at the interface; is: 1n

ENEL2FT Field Theory

Electrostatic Fields

2n

88

ELECTROSTATIC FIELDS

 DIELECTRIC-DIELECTRIC BOUNDARY CONDITIONS  We now use the boundary conditions to determine the

“refraction” of the electric field across the interface.  Consider, in the figure below, D1 or E1 and D2 or E2 making angles θ1 and θ2 with the normal as illustrated D below. Medium 1, ε 1

1

E1 D1n θ1

E1n D1t

θ2

E1t

E2

E2n

E2t D2n

D2

Medium 2, ε 2

D2t

ENEL2FT Field Theory

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89

ELECTROSTATIC FIELDS  DIELECTRIC-DIELECTRIC BOUNDARY CONDITIONS  Using the equation for the tangential components of E, we have:

E = E sin θ = E = E sin θ 1t

1

1

2t

2

∴ E sin θ = E sin θ 1

1

2

2

2

 Similarly, applying the equation relating the normal

components of D, we have (assuming there is no surface charge at the interface):

D = ε E = ε E cosθ = D = ε E = ε E cosθ 1n

1

1n

1

1

1

2n

∴ ε E cosθ = ε E cosθ 1

1

1

2

2

2

2n

2

2

2

2

 Dividing these two equations, we have the law of refraction ε E cosθ ε E cosθ tan θ ε ε 1 1 1 2 2 at a boundary free of =charge:

E sin θ 1

ENEL2FT Field Theory

1

Electrostatic Fields

E sin θ 2

2

2

⇒

1

tan θ

2

=

1

ε

2

=

r1

ε

r2

90

ELECTROSTATIC FIELDS

 CONDUCTOR-DIELECTRIC BOUNDARY CONDITIONS  If we let medium 2 to be a perfect conductor (σ →∞), from

Ohm’s law,

  J = σE

 Good conductors like copper, silver, aluminium, etc, can be

regarded as perfect conductors, since σ is very large (σ∼107).  To maintain a finite current density, J, for infinite conductivity, it is required that as σ →∞, E →0.  If some charges are introduced within the conductor, the charges will move to the conductor surface and redistribute themselves such that the field inside the conductor vanishes.  the divergence of E is also 0. According to Gauss’s law, if E=0, Qenc =conductor D.dno S =volume ∫v ρ v dv = ∫shas ∫v ∇.Ddvcharge distribution. Thus a good

    ∴ ∇.D = ε∇.E = ρ = 0 v

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91

ELECTROSTATIC FIELDS

 CONDUCTOR-DIELECTRIC BOUNDARY CONDITIONS  To determine the boundary conditions, we incorporate the fact

that E →0, inside a good conductor.  Therefore as ∆h →0, we have:

E =E =0 1t

2t

 Thus the tangential component of E vanishes at the interface.  Similarly, considering the normal component of D, we again

incorporate the fact that E →0, inside a good conductor. We have

(D

1n

−D

2n

)=ρ

s

⇒ ε E −ε E = ρ 1

∴ D = −ρ ⇒ D = ρ n2

s

n

1n

2

2n

s

s

 An important application of the fact that E=0 inside a good

conductor is in electrostatic screening or electrostatic shielding.

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92

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS INTRODUCTION  The procedure for determining the electric field E when

using Coulomb’s law or Gauss’s law assumes a known charge distribution or potential V in the region.  In most practical situations, however, neither the charge distribution nor the potential distribution is known.  In the case of practical electrostatic problems, only electrostatic conditions (charge and potential) at some boundaries are known. It is then desired to find E and V throughout the region.  Such problems are usually tackled using Poisson’s or Laplace’s equation.  They are referred to as boundary-value problems. ENEL2FT Field Theory

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93

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS

POISSON’S AND LAPLACE’S

EQUATIONS  Poisson’s and Laplace’s equations are easily

 law and the gradient of the derived from  Gauss’s ∇.D = ∇.εE = ρ v potential field:   E = −∇V

( ) v gives, for a homogeneous medium, Poisson’s 

 Substituting the second equation into the first ∇. − ε∇V = ρ

ρv ∴∇ V = − ε

equation, namely: 2

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94

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS

POISSON’S AND LAPLACE’S EQUATIONS  A special case of Poisson’s equation occurs when the

volume charge distribution ρv=0. Then we have Laplace’s equation, namely: 2

∇ V =0

 The Laplacian operator, ∇ 2, is defined in Cartesian, cylindrical, coordinates as: ∂ 2Vand ∂ 2Vspherical ∂ 2V 2 ∇ V = 2 + 2 + 2 =0 ∂x ∂y ∂z 1 ∂  ∂V  1 ∂ 2V ∂ 2V ∇ V= + 2 =0 ρ + 2 2 ρ ∂ρ  ∂ρ  ρ ∂φ ∂z 2

1 ∂  ∂V ∇ V = 2 r2 r ∂r  ∂r 2

ENEL2FT Field Theory

1 ∂  ∂V  + sin θ   ∂θ  r 2 sin θ ∂θ 

Electrostatic Fields

1 ∂ 2V  =0 + 2 2 2  r sin θ ∂φ 95

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS GENERAL PROCEDURE FOR SOLVING

POISSON’S OR LAPLACE’S EQUATIONS

 The following procedure may be used in solving a given

boundary-value problem involving Poisson’s or Laplace’s equations (see Sadiku):  Solve Laplace’s or Poisson’s equation either using direct

integration (when V is a function of a single variable like x or y, or r), or using the separation of variables (if V is a function of more than one variable). The solution at this stage is not unique.  Apply the boundary conditions to determine a unique solution for V. Imposing the given boundary conditions makes the solution unique.  Having obtained V, find E from E=-∇V, and D=εE.  If desired, find the charge Q induced in a conductor using Q=∫ρsdS where ρs=Dn, and Dn is the component of D normal to the conductor.



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96

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS GENERAL PROCEDURE FOR SOLVING

POISSON’S OR LAPLACE’S EQUATIONS  EXAMPLE:

x  In a one-dimensional ρ vdevice, = ρ o the charge density is a given by: V=

(

)

ρo 3 a − x3 ; 6εa

2  ρ o xat  If E=0 at x=0 and V=0 x=a, find V and E. E= xˆ

2εa

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97

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS

GENERAL PROCEDURE FOR SOLVING

POISSON’S OR LAPLACE’S EQUATIONS  Solution:

ρv ρ o x ∂ 2V ρ x ∇ V =− =− ⇒ 2 =− o ε εa εa ∂x 2

ρo x 2 ∂V ∴ =− + K1 = − E ⇒ E x = 0 = K1 = 0 ∂x 2εa  ρo x 2 ∴E = xˆ 2εa ρo x3 ρo x3 ρo a 2 V ( x) = − + K1 x + K 2 = − + K 2 ⇒ V ( x) x = a = 0 ⇒ K 2 = 6εa 6εa 6ε ρ ∴V ( x) = o a 3 − x 3 6εa

(

ENEL2FT Field Theory

)

Electrostatic Fields

98

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS

 EXAMPLE:

 Two semi-infinite conducting planes φ=0 and φ=π/6 are

separated by an infinitesimal insulating gap as shown below. If V(φ=0)=0 and V(φ=π/6)=100V, calculate E and V in the region between the planes. z

gap

φo 0V y

x Vo

ENEL2FT Field Theory

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99

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS  SOLUTION  As V depends only on φ, Laplace’s equations in

cylindrical coordinates becomes:

1 ∂  ∂V  1 ∂ V ∂ V 1 ∂ V ∇V = + = =0 ρ + ρ ∂ρ  ∂ρ  ρ ∂φ ∂z ρ ∂φ 2

2

2

2

2

2

2

2

2

 Multiplying both sides by ρ2 and integrating twice, we

obtain:

dV = 0 ⇒ V = Aφ + B dφ 2

2

 To evaluate the integration constants A and B, we

apply the boundary conditions.

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100

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS  SOLUTION  WE obtain:

V φ =B=0 =0

π 600 V = A = 100 ⇒ A = 6 π  600  ∴V = Aφ =  φ  π    1 dV ˆ  600  ˆ E = −∇V = − φ = φ ρ dφ  πρ  φ =π / 6

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101

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS

 EXAMPLE

 Two conducting plates of size 1X5 m are inclined at φo=45o to

each other with a gap of 4 mm separating them as shown below. Determine an approximate value of the charge per plate if the plates are maintained at a potential difference of Vo=50V. Assume that the medium between them has εr=1.5. z

gap

φo 0V y

x Vo

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102

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS SOLUTION  We can determine V and E, as before. But at the interface,

Dn=ρs; that is, the surface charge density equals the normal component of D. We then have: V φ =B=0 =0

π 200 = 50 ⇒ A = 4 π  200  ∴V = Aφ =  φ  π    1 dV ˆ  200  ˆ E = −∇ V = − φ = φ ρ dφ πρ      200ε   200ε  ˆ D = εE =  φ ; D = D =    πρ πρ     Vφπ =A = /4

n

ENEL2FT Field Theory

Electrostatic Fields

φ

103

ELECTROSTATIC FIELDS – BOUNDARY VALUE PROBLEMS  SOLUTION  To obtain the charge per plate, we need to get the 1 5 the total charge enclosed: total flux,  which  equals

Ψ = ∫ D.dS = ∫ ∫ Ddzdρ ρ = 0.004 z = 0

S

 200ε   200ε  = ∫ ∫ dzdρ =  5ln ρ ρ  π   πρ  = 1.555xε 10 C = 2.3325x10 C 1

5

= 0.004 z = 0

-8

1 0.004

-8

r

= 23.3 nC ENEL2FT Field Theory

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104