Field Theory Electrostatics

ELECTROSTATIC FIELDS Dr. Thomas Afullo, UKZN, Durban

ENEL2FT Field Theory

Electrostatic Fields

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ENEL2FT • • • • •

FIELD THEORY

REFERENCES 1. M.N. Sadiku: Elements of Electromagnetics, Oxford University Press, 1995, ISBN 0-19-510368-8. 2. N.N. Rao: Elements of Engineering Electromagnetics, Prectice-Hall, 1991, ISBN:0-13-251604-7. 3. P. Lorrain, D. Corson: Electromagnetic Fields and Waves, W.H. Freeman & Co, 1970, ISBN: 0-7167-0330-0. 4. David T. Thomas: Engineering Electromagnetics, Pergamon Press, ISBN: 08-016778-0.

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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 charge of an electron is ENEL2FT Fieldlarge Theory unit of charge because the 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 kQ Q is expressed as: F=

1 2 2

R

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 εo is defined as the permittivity of free space (in farads/metre).

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ELECTROSTATIC FIELDS • •

COULOMB’S LAW The constant k is defined as: k=

1 m/ F 4πε o

10−9 εo = ≈ 8.854 x10−12 F / m 36π

•

Then the equation of force becomes: F=

•

Q1Q2 4πε o R 2

If point charges Q1 and Q2 are located at points having position vectors r1 and r2, respectively, then the force F12 on Q2 due to Q1 is given by:  QQ F12 =

ENEL2FT Field Theory

1 2

4πε o R 2

aˆ12

Electrostatic Fields

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ELECTROSTATIC FIELDS •

COULOMB’S LAW

F21

Q1

R12

Q2

F12

origin •

Where:     R12 = r2 − r1; R = R12  R12 aˆ12 =  R12

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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:

•

•

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.

•

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  F21 = − F12

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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. That is:  1  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 FIELDS •

ELECTRIC FIELDS DUE 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. Accordingly, we define the volume charge density as:

•

•

∆q dq = (C / m3 ) ∆v →0 ∆v dv

ρ v = lim

•

Where ∆q is the charge contained in ∆v. The variation of ρv with spatial location is called its spatial distribution. The total charge contained in volume v is given by: Q = ∫ ρ v dv Coulombs v

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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 the surface charge density, ρs, defined as: ∆q dq = ∆s →0 ∆s ds

ρ s = lim

•

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 charge density ρl, defined as: ∆q dq ρl = lim

∆l →0

ENEL2FT Field Theory

∆l

=

dl

( C / m)

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ELECTROSTATIC FIELDS •

ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS

•

The electric field intensity due to each of the charge distributions ρl,ρs,and ρv may be regarded as the summation of the field distributed by the numerous point charges making up the charge distribution. Thus we replace Q in the equations for E, and integrating, we get:

•

 ρ dl E = ∫ l 2 rˆ 4πε o R  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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ELECTROSTATIC FIELDS •

ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS – AN INFINITE LINE CHARGE

•

Consider a line charge with a uniform charge density ρL extending from -∞ to +∞ along the z-axis, as shown below. dz

z Infinite line charge

ENEL2FT Field Theory

rˆ

 R r

− 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 the line is: dQ = ρ L dz

•

The electric field intensity at point P a distance r from the line, due to the elemental charge ρ Ldz is given by:  dE =

ρ L dz

aˆ = 2 R

ρ L dz

4πε o R 4πε o R   R ⇒ aˆ R = ; R = R R

•

3

R

R = r 2 + z 2 ; we z = obtain: r tan α ⇒ R = r 2 + r 2 tan 2 α = r secα From geometry, dz d d  sin α  2 2 =r tan α = r = r sec α ⇒ dz = r sec αdα   dα dα dα  cosα 

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ELECTROSTATIC FIELDS • •

ELECTRIC FIELDS DUE TO AN INFINITE LINE CHARGE Also, for the unit vector we have: aˆ R = rˆ cosα − zˆ sin α  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 – π/2 to +π/2 as z varies from -∞ to +∞; thus:

{

•

 ρL π / 2 (∫ rˆ cosα − zˆ sin α ) dα = ρ L [ rˆ sin α ]π−π/ 2/ 2 + [ zˆ cosα ]π−π/ 2/ 2 E= 4πε o r −π / 2 4πε o r  ρL ∴E = rˆ 2πε o r

}

In normal cylindrical coordinates, the expression becomes:  E=

ENEL2FT Field Theory

ρL ρˆ 2πε o ρ Electrostatic Fields

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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 sec 2 αdα  ρL [ ( rˆ cosα − zˆ sin α ) dα ] = rˆdEr + zˆdEz dE = =   2 2 4πε o  r sec α  4πε o r

•

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

•

We shall use a similar argument for surface charge.

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ELECTROSTATIC FIELDS •

ELECTRIC FIELDS DUE TO CIRCULAR RING OF CHARGE

h Circular ring of charge •

zˆ

 z dE  αR ϕ

α

aˆ R

− ρˆ y

ρ

dl

Consider a circular ring of charge of radius ρ, having x uniform charge density ρl 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 field arising from this elemental charge is given by:  dE =

dQ

 2 aˆ R 4πε o 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 cancel; this implies that dE has only a z-component. Thus:  ˆ ρdE ρ = 0, ⇒ dE = zˆdE z =

ρ L ( ρdϕ ) ( zˆ cosα ) 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

(

•

)

(

)

)

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

(

ENEL2FT Field Theory

)

(

)

3/ 2

=

zˆ ( 2πρρ L ) h

(

4πε o ρ + h 2

)

2 3/ 2

=

Electrostatic Fields

(

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 ρs C/m2. We are required to find the electric field intensity due to it everywhere above thesheet. dE z α

h

ENEL2FT Field Theory

zˆ

aˆ R α

ϕ x

 R

y ρ dϕ

− ρˆ

dρ

dA Electrostatic Fields

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ELECTROSTATIC FIELDS •

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

•

We also derive the following relationships from the sketch:  R = R = h 2 + ρ 2 ; ρ = h tan α ; ⇒ R = h 1 + tan 2 α = h secα

dρ d  sin α  2 2 =h = h sec α ; ⇒ d ρ = h sec αdα dα dα  cosα  aˆ R = zˆ cosα − ρˆ sin α

•

Then the electric field intensity arising from this elemental charge is:

 dQaˆ R ρ s ( ρdϕdρ ) ρ s  (h tan αdϕ )(h sec 2 αdα )[ zˆ cosα − ρˆ sin α ]  dE = = aˆ R =   2 2 4πε o  4πε o R 4πε o R h 2 sec 2 α  ENEL2FT Field Theory

Electrostatic Fields

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ELECTROSTATIC FIELDS •

ELECTRIC FIELDS DUE TO AN INFINITE SURFACE CHARGE  ρ dE = 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 the ρ components cancel each other out, and we have left only  the z-component: ρ s 2π π / 2 ρ s π / 2   ρs E=

4πε o

ENEL2FT Field Theory

zˆ = zˆE z ∫  ∫ ( zˆ sin α ) dα  dϕ = ∫ ( zˆ sin α ) dα  =  2 ε 2 ε   0  0 o  0 o Electrostatic Fields

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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 ρs, and the other with opposite charge density –ρs C/m2, the total electric field between the two plates is given by:   ( − ρs )  ρ E = zˆ s + − zˆ 2ε o  2ε 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 ε o. Thus for a volume charge distribution,  ρ dv D = ∫ v 2 aˆ R 4πR

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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:  

•

Where Qenc=total charge enclosed.  

Ψ = Qenc ⇒ Ψ = ∫S D.dS = Qenc

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 surface centered at the origin is the Gaussian surface in thiszcase, as shown below.

P Q

x ENEL2FT Field Theory

r

 D y Gaussian Surface

Electrostatic Fields

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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 E to be:    D = ε o E ⇒ E = rˆ

ENEL2FT Field Theory

Q 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 ρ L 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

Electrostatic Fields

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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, Qenc = ρ L L = ∫ D.dS we have:  dS = ρˆ ( ρdϕdz ) ∴ Qenc = ρ L L = Dρ 2 ρπL   ρ ρL ∴ D = ρˆ L ⇒ 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 ρv 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   Ψ = ∫ D.dS = Dr ∫ ∫ r 2 sin θdθdϕ = Dr 4πr 2

(

ϕ =0 θ = 0

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Electrostatic Fields

) 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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33

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 ρs 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

ELECTROSTATIC FIELDS • •

•

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 to the sheet, we have, when applying  Gauss’s law: D = zˆDz     ∫ D.dS = Q = Dz  ∫ dS + ∫ dS  bottom  top

•

Note that D has no x- and y- components, hence Dx=0, Dy=0.

•

If the top and bottom the pillbox each has area A, then Q = ρ s Aof =Ψ =D ( A + A ) = 2 AD z z  we get:   ∴ D = zˆ

ENEL2FT Field Theory

ρs D ρ ⇒ E = = zˆ s 2 εo 2ε o

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36

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.

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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:

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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 ydirection, 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

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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 FIELDS • •

ELECTRIC POTENTIAL Thus the potential difference between any two points P2 and P1 is obtained by integrating dV along the path between P1 and P2. That is: V = ∫ dV   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.

•

It is also readily seen that Pthe integral P2   of the  electrostatic field E 2 V22 = Vcontour dV zero: = ∫ E.dl = ∫ E.dl = 0 2 − V2 = ∫ is around any closed P2

P2

C

       But ∫ E.dl = ∫ ∇xE.dS ⇒∇xE = 0 C

ENEL2FT Field Theory

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   q  q  ( ˆ ) V = − ∫ E.dl = − ∫  aˆ R . a dR = (V ) R 2 4 πε R 4πε o R  ∞ ∞ o R

•

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: V=

ENEL2FT Field Theory

q   4πε o R − R1

(V )

Electrostatic Fields

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, ..,N RN, the electric potential is: V=

ENEL2FT Field Theory

1 q ∑  i  (V ) 4πε o i =1 R − Ri

Electrostatic Fields

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

ENEL2FT Field Theory

V ( R) =

1 ρs ∫ dS ( surface distribution) 4πε o S R

V ( R) =

1 ρl ∫ dl (line distribution) 4πε o L R

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 =

•

Thus

ENEL2FT Field Theory

  E = −∇V 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 cos φ  r  r ( 4,90 o ,60 o ) (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 

(

)

(

− 10  )10 − 32 4 

∴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 4πε o

1 1  Q  r2 − r1   r − r  = 4πε  r r   1 2 o  1 2 

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 =

Q 4πε o

ENEL2FT Field Theory

 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 are located at points (0,0,2) and Find the potential at  (0,0,3),−respectively.  the origin if: p1 = −5 x10 9 zˆ Cm; p 2 = 9 x10 − 9 zˆ Cm

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 =

ENEL2FT Field Theory

1 10 − 9 4π 36π

    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 p is located at the  dipole moment −12 origin, where: p = 100 x10 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. WEdone = W1 +inWpositioning 2 + W3 Hence the total work the three charges is: = 0 + Q2V21 + Q3 (V31 + V32 )

+ W2 + W1 If the chargesWwere in reverse order, then: E = W3positioned

ENEL2FT Field Theory

= 0 + Q2V23 + Q1 (V12 + V13 )

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 equations 2WEgive: = 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 are WEn point = charges, QkVk 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 1 integration: 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 charge   identities: density by using the vector

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

(

•

)

(

)

(

)

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 2S 2V

•

For point charges, V varies as 1/r, and D varies as 1/r2; for dipoles, V varies as 1/r2 and D varies as 1/r3; and so on.

ENEL2FT Field Theory

Electrostatic Fields

59

ELECTROSTATIC FIELDS • • •

Hence VD in the first term on the rhs must vary at least as 1/r3 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 ∑ Qk Vk = [ Q1V1 + Q2V2 + Q3V3 ] 2 k =1 2

V1 = V12

Q3 Q2 + V13 = + = 4πε o (1) 4πε o (1)

4 x10 − 9 10 − 9 4π (1) 36π

+

3 x10 − 9 10 − 9 4π (1) 36π

= 63V

Q3 Q1 − 1x10 − 9 3 x10 − 9 V2 = V21 + V23 = + = + = 10.09V 4πε o (1) 4πε o 2 10 − 9 10 − 9 (1) 4π 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 + 3x10 − 9 16.46 2 2 V3 = V31 + V32 =

( )

[(

)

(

)

(

)

]

WE = 13.36x10 - 9 J

ENEL2FT Field Theory

Electrostatic Fields

61