Topic 01 (Electrostatic Fund)

8/12/2019 Topic 01 (Electrostatic Fund)

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Electrostatic Fields:

Coulombs Law & the Electric Field Intensity

EE 141 Lecture Notes

Topic 1

Professor K. E. OughstunSchool of Engineering

College of Engineering & Mathematical Sciences

University of Vermont

2009


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Motivation


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Coulombs Law

Let q1 be a stationary point charge with position vector r1 relative toa fixed origin O, and let q2 be a separate, distinct point charge withposition vector r2= r1 relative to the same origin O.

Coulombs Lawthen states that the force F21 exerted on q1 byq2 isgiven by

F21=Kq1q2r2

r (1)

where r |r1 r2| is the separation distance between q1 & q2 andwhere r(r1 r2)/|r1 r2|= (r1 r2)/r is the unit vector in the

direction from q2 to q1. The force is repulsive ifq1 & q2 are of thesame sign and attractive if they are of the opposite sign.


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Coulombs Law

Reciprocityrequires that an equal but oppositely directed force F12 isexerted on q2 byq1; that is

F12=F21. (2)

In the rationalized MKSA (meter, kilogram, second, ampere) system,the unit of force is thenewton (N), the unit of charge is thecoulomb

(C), and the constant appearing in Coulombs law is given by

K= 1

408.988 109N m2/C2.

Permittivity of free space:

0 8.8542 1012F/m,

so thatK 8.988 109m/F

where FaradCoulomb/volt is the unit of capacitance.


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Coulombs Law

Coulombs Lawin MKSA units then becomes

F21 = 1

40

q1q2

r2 r (N) (3)

Coulombs law directly applies to any pair of point charges that aresituated in vacuum and are stationary with respect to each other(Special Theory of Relativity). It also applies in material media ifF21is taken as the direct microscopic force between the two charges q1 &q2, irrespective of the other forces arising from all of the othercharges in the material medium.


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Coulombs Law

Figure: Charles Augustin de Coulomb (17361806)


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Coulombs Law

Figure: Coulombs apparatus (1785)

In 1936, Plimpton & Lawton at Worcester Polytechnic Instituteshowed that the distance dependency in Coulombs law deviated fromthe inverse square law by less than 2 parts in 1 billion; that is, theydetermined that the force varies as r(2+) with||< 2 109.


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Coulombs Law - Principle of Superposition

The Coulombic force satisfies thePrinciple of Superposition:

The electrostatic force exerted on a stationary point charge q1 at r1by a system of stationary point charges qk at rk, k= 1, is given bythevector sumorlinear superpositionof all the Coulombic forcesexerted on q1:

F(r1) =k=1

Fk1 = q1

40

k=1

qk

r21kr1k (4)

where

r1k= r1 rk,

r1k= r1k

r1k,

with r1k=|r1k|.


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Electric Field Intensity E(r)

The staticElectric Field Intensity (or Electrostatic Field Intensity)E(r) = E(x,y, z), at any fixed point r=1xx+1yy+1zz in space isdefined as thelimiting force per unit chargeexerted on a test chargeqat that point as the magnitude of the test charge goes to zero:

E(r) limq0

F(r)

q (N/C =V/m) (5)

The limit q0 is introduced in order that the test charge does not

influence the charge sources that produce the electrostatic field.The electric field is then defined in such a way that it is independentof the presence of the test charge.

( )


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From Eqs. (3) & (5), the electric field intensity at a fixed point r dueto a single point charge q1 situated at r1 is given by

E(r) = 1

40

q1

R2R (6)

where R= r r1 denotes thevector from the source point at r1 tothe field point at r, and where Ris the unit vector along that

direction.

El F ld I E( )


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As a consequence of theprinciple of superposition, the electric fieldintensity at a fixed point r due to a system of fixed, discrete pointcharges qj located at the points rj, j= 1, 2, . . . , n, is given by thevector sum

E(r) = 1

40

nj=1

qj

R2jRj (7)

where Rj= r rjdenotes thevector from the source point at rj to

the field point at r with magnitude Rj, and where Rj is the unitvector along that direction.

Ch D i ( )


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Charge Density(r)

Thecharge density (or net volume charge density) (r) is a scalarfield whose value at any point r in space, given by thesigned net

charge per unit volumeat that point, is defined by the limiting ratio

(r) limV0

q

V (C/m3) (8)

where qis the net charge in the volume element V.

From amicroscopic perspective, the charge density (r) is zeroeverywhere except in those regions occupied by fundamental chargedparticles (electrons & protons).

Ch D i ( )


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Charge Density (r)

From amacroscopic perspective, the abrupt spatial variations in themicroscopic charge density(r), which are on the scale of interparticledistances, are removed through an appropriatespatial averagingprocedureover spatial regions that are small on a macroscopic scalebut whose linear dimensions are large in comparison with the particlespacing. The result is themacroscopic charge density

(r) =(r) (9)

The electric field that is determined from such a macroscopic chargedensity is correspondingly a spatially-averaged field and, as such, isjust what would be obtained through an appropriate laboratorymeasurement.

El i Fi ld I i E( )


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With the introduction of the charge density in Eqs. (8)(9), thevector summation appearing in Eq. (7) may then be replaced (in the

appropriate limit) by a volume integration over the entire region ofspace containing the source charge distribution. Because

q(r) =

(r)V

is the elemental charge contained in the volume element Vat thepoint r, then Eq. (7) may be written as

E(r) = 1

40 (r)

R2RV

1

40

(r)

R2Rd3r asV 0

(10)where R= r r is the vector directed from the source point r tothefield point r with magnitude R, where R is a unit vector alongthat direction, and where d3r =dxdydz denotes the volumeelement at the source point r = (x,y, z).

Notice that the integral in (10) is convergent for r = r.

G L


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Gauss Law

Consider a point charge qat a fixed point in space together with asimple closed surface S. Let rdenote the distance from the point

charge to a point on the surface Swith unit vector r directed alongthe line from the point charge to the surface point, let nbe theoutwardly directedunit normal vectorto the surface Sat that point,and let da be the differential element of surface area at that point.

Theflux ofE passing through the directed element of area da=ndaofSis then given by

E

nda=

1

40q

r n

r2 da (11)

G L


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Gauss Law

Thedifferential element of solid angle d subtended by da at theposition of the point charge is given by

d = r n

r2 da (12)

With this identification, the flux ofEpassing through the directedelement of area da=nda ofS becomes

E nda= 1

40qd (13)

Ga ss La


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Gauss Law

The total flux ofEpassing through the closed surface S in theoutward direction is then given by integrating Eq. (13) over theentire surface S as

S

E nda= 140

qS

d (14)

where

Sd =

4, ifq S0, ifq / S (15)

Gauss Law


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Gauss Law

Gauss Lawfor a single point chargeS

E nda= 1

0

q, ifq S0, ifq / S

(16)

For a system of discrete point charges the principle of superpositionapplies andGauss Lawbecomes

S

E nda= 1

0j

qj (17)

where the summation extends over only those charges that are insidethe region enclosed by the surface S.

Gauss Law


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Gauss Law

If the charge system is described by the charge density (r), one

finally obtains theIntegral Form of Gauss LawS

E nda= 1

0

V

(r)d3r (18)

whereVis the volume enclosed by the surface S.Notice that the derivation of Gauss law depends only upon thefollowing three properties:

1 The inverse square law for the force between point charges, asembodied in Coulombs law.

2 The central nature of the force, also embodied in Coulombs law.

3 The principle of linear superposition.

Gauss Law


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Gauss Law

For asystem of discrete point charges qj located at r= rj, the chargedensity is given by

=j

qj(r rj),

which recaptures the microscopic description, where

(r rj) (x xj)(yyj)(z zj)

is the three-dimensionalDirac delta function.With this substitution in Eq. (18), the integral form ofGauss Lawbecomes

S

E nda= 1

0

j

qj

V

(r rj)d3r=

1

0

j

qj

which is just Gauss law (17) for a system of discrete point charges

with locations rj V.

Gauss Law


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Gauss Law

Figure: Carl Friedrich Gauss (17771855)

Gauss Law - Differential Form


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Gauss Law - Differential Form

With application of theDivergence Theorem

SE nda=

V

( E)d3r (19)

the integral form of Gauss law (18) becomes

V(

E / 0)d

3

r= 0.

Because this expression holds for any regionV, the integrand itselfmust then vanish throughout all of space, so that

E(r) = (r)0

(20)

which is thedifferential form of Gauss law.This single vector differential relation is not sufficient to completely

determine the electric field vector E(r) for a given charge density (r).

Coulombs Law - Differential Form


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Coulomb s Law - Differential Form

Helmholtz Theoemstates that a vector field can be specified almostcompletely (up to the gradient of an arbitrary scalar field) if both its

divergence and curl are specified everywhere.The required curl relation for the electrostatic field follows from theintegral form (10) ofCoulombs law, expressed here as

E(r) =

1

40

(r

)

r r

|r r|3d

3

r

, (21)

where the integration extends over all space. Because(Problem 1)

1

|r r| = r r

|r r|3 (22)

where 1x

x+1y

y +1z

zoperates only on the unprimed

coordinates, then

E(r) = 1

40 (r

)

|r r

|

d3r. (23)

Coulombs Law - Differential Form & The Scalar


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Coulomb s Law Differential Form & The ScalarPotential

Because the curl of the gradient of any well-behaved scalar functionidentically vanishes, then Eq. (23) shows that

E(r) = 0 (24)

Notice that, in general, this expression for the curl ofE holds only for

an electrostatic field.From the form of Eq. (23), define ascalar potentialfor the electricfield as

E(r) =V(r) (25)

where the minus sign is introduced by convention, and

V(r) = 1

40

(r)

|r r|d3r (V) (26)

where the integration extends over all space.

Electrostatic Scalar Potential


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Electrostatic Scalar Potential

The scalar potential at a fixed point Pdue to a point charge qadistanceRaway is given by

V(P) = 1

40

q

R.

By superposition, the scalar potential at a fixed pointPdue to asystem of point charges q1, q2, , qNat distances R1,R2, ,RNfrom P, respectively, is then given by

V(P) = 140

Nj=1

qj

Rj.

Electric Potential & Work


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Consider the work done in transporting a test charge qfrom point Ato point Bthrough an externally produced electrostatic field E(r).

The electric force acting on the test charge qat any point in the fieldis given by Coulombs law as

F(r) =qE(r). (27)



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The work done in moving the test charge qslowly from A to B(negligible accelerations result in negligible energy loss due to

electromagnetic radiation)is given by the path integral

W =

BA

F d= q

BA

E d, (28)

where the minus sign indicates that this is the work done on the testcharge against the action of the field. With Eq. (25) this expressionbecomes

W =q B

A

V d= q B

A

dV =q(VB VA). (29)

This result then shows thatthe quantity V(r) can be interpreted asthe potential energy of the charge q in the electrostatic field.The negative sign in Eq. (25) is then seen to indicate that E pointsin the direction of decreasing potential, and hence, decreasing

potential energy.



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Eqs. (28) & (29) show that the path integral of the electrostatic fieldvector E(r) between any two points is independent of the path and is

the negative of the potential difference between the two points, viz.

VB VA =

BA

E d (30)

If the path is closed, E d= 0 (31)

Application ofStokes theoremto this result then yieldsC

E d=

S

( E) nda= 0 = E= 0

which is just Eq. (24). The electrostatic field E(r) is then seen to beanirrotational vector fieldand is thereforeconservative.

Topic 01 (Electrostatic Fund)

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