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Elsevier Editorial System(tm) for Journal of Electrostatics

Manuscript Draft

Manuscript Number: ELSTAT-D-11-00096R3

Title: A Generalization Of Gauss's Theorem In Electrostatics

Article Type: Full Length Article

Keywords: Forces and Fields; Gauss's Theorem; Surface Charge; Method of Imag

Dielectrics in Constant Electric Field

Corresponding Author: Mr. Ishnath Pathak,

Corresponding Author's Institution: IIT Guwahati

First Author: Ishnath Pathak

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A Generalization Of Gausss Theorem In Electrostatics

Ishnath Pathak

Department of Civil Engineering, Indian Institute of Technology, Guwahati 781039, India

ABSTRACT

Gausss Theorem of electrostatics states that the flux of the electrostatic field over a close

enclosed divided by the permittivity of free space. In the derivation it is assumed that n

question. Consider the problem of evaluation of the flux of the electrostatic field due to

surface of radius R, carrying a net charge Q, through the surface itself. Due to the reason

apply Gausss Theorem In this paper we prove a generalization of Gausss Theorem wh

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for point, line and surface charges. We, in this paper, will frequently distinguish between

charges - the meanings of which are obviously understood.

2. Flux of the Electrostatic Field of a Uniformly Charged Spherical Surface

There isnt any trouble with Gausss Theorem - the only issue is that it assumes tha

over which the flux is to be evaluated. If charges lie on the surface, then - as a rule

applied. We consider as an example, the case of a net charge Q smeared uniformly over

using Coulombs law and invoking the principle of superposition, we can evaluate

E(rz

will definitely provide us the values 0 and Q

40r2z, respectively. Here, z denotes the unit

of increase of the cartesian coordinate z and throughout this paper we adopt a conventi

pointing in the direction of steepest increase of any scalar function u of coordinates by the

the variable. Also we can take note here that since the situation exhibits spherical symmet

of z is arbitrary and our final results about the magnitude of the field strength stand as e

position at an equal distance Now we expect the process of evaluation to boil the integra

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Now, Q(r) = (4/3)(r3 R3) = Q(r3 R3)

(3R2 + 3R+2). So,

=

3R2Q

0 R+

R (r3

R3)dr

2(3R2 + 3R+2)2 =

3R2Q

40 6R2 + 4R+

(3R2 + 3R+

Letting 0 in this expression, we get = Q/20. As we shall see in the coming seTheorem, which is to be enunciated, states that the flux of the electrostatic field, throug

charges might as well lie, due to a source charge configuration consisting of only volum

equals

E da= Qenc/0+Qcon/20, where Qenc is the net charge lying inside the openthe boundary, and Qcon is the net charge lying on the closed surface itself. In case of our s

Qcon = Q, so the flux is = Q/20. In conclusion, our plan in this paper is to generaliz

can tell the flux of the electrostatic field through a surface carrying charges as easily as Ga

tell the same for surfaces free of charges; and thus avoid operating with actual integrals an

It can be mentioned here in passing that the flux 4 R2E0,E0being the constant (having

field) described in the beginning of this section, also equals Q/20. A general area element

d it i h Q i dd/4 Thi l t f l d t th i t R

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We evaluated the flux of a net charge Q smeared uniformly over the spherical surfac

E0 = Q

80R2. Let us now model the surface charge as a net surface charge Q0 smeare

surfacer = R plus a net surface charge Q0smeared uniformly over the spherical surfacer =

Q1 spilled uniformly in the enclosed spherical shell, i.e. the open regionR < r < R + , bou

kind of describing a surface charge in terms of surface charge may be considered recursive

ofQ0 andQ0, as we shall do now, we shall use the Generalized Gausss Theorem derived

as the limit of a volume charge. That other descriptions of a surface charge cannot lead

the flux is corroborated in the next three paragraphs. Let us set Q0+ Q1+ Q0 to be Q an

write

E(r) =E0(r)r+E1(r)r+E0(r)r, where E0(r)r, E1(r)r and E0(r)r are the electroNow, the flux of the net charge Q through the spherical surface r = R, when modeled b

three charge distributions, takes the value 0+ 1+ 0 = 4R

2E0+ 4R2 < E1 > +4(R

mean value of the scalar field E1(r) over the spherical surface. Now as 0, 1 4R2

(4)). AlsoR + R as 0 and we get 0 Q0/20 and 0 Q0/20. Thus, the fluis =Q/20.

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1

2

R+R

r2(r)dr 3R2

(R+)3 R3R+R

rR

u2(u)du

dr as

3. Flux of the Electrostatic Field of a Point Charge

We prove in this section that the flux of the electrostatic field

Ei of a charge qi ove

regionRenc is given by

S

Ei da=

0, ifqi lies outside S;

qii40

, ifqi lies on S at a pointr i and

S(da )/2, being

qi0

, ifqi is enclosed by S.

This statement is axiomatic but still we will write a satisfactory proof soon. Yet, be

expatiate a little. An analogy is often drawn[3] between the actual situation and an ima

h di ti ti l t t t t i h i ll t i Si th

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S

Ei da=

S1

Ei da+

S2

Ei da

because when we add the flux

= qi40

area ofS2

of

Ei through S to the second integral on the right of equality and subtract the s

identity. Now according to the result of case A, the second integral must vanish. The flu

easily obtained

S

Ei da= lim

0

S

Ei da= lim

0

S1

Ei da= lim

0 =

qi

4

Finally we consider the case C whenr i lies enclosed by S. We consider any plane passS0 the part of the plane that is enclosed by S. We denote again by R1 and R2 the regions i

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(as shown in Figure-5), of base area A and height 2d extending an equal distance d abovE = E(d)n, where n is a unit vector normal to the plane and directed away from the pl

Theorem 2

E(d)

A = A/0. So, in this manner, the electrostatic field is evaluated

Theorem. If the Generalized Gausss Theorem is applied to a Gaussian surface of the s

Figure-6), which has a base of area A lying on the plane and extends to a distance d abo

invoked to argue that the normal component of the field is E(d) throughout the face on th

face at the bottom, we get E(d) A= A/20, i.e.E = 20

n, which is the same result.

If the surface charge density on the infinite flat plane is not uniform, then we cannot a

electrostatic field near the plane is normal to the plane. But, still, at any point on the pl

near the plane directed along the normal to the plane pointing away from the plane, can

that of last example, to be

20, where is the local surface charge density. Inherent in th

that just near the plane, the upward normal component of the field is the same in magnit

but opposite in sign. Symmetry is invoked to deduce this assumption, riding on which

using Gausss Theorem. If the derivation is to be done using the Generalized Gausss T

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5. The Simplest Form of the Generalized Gausss Theorem

We in this section consider those cases in which the Generalized Gausss Theorem ta

shall call as the simplest form of the theorem. We begin by confining ourselves to anconfiguration of charges not containing any point or line charges. In this special caseQc

interpreted as the net charge contained by S, for the net charge lying on the discontinuiti

S now contains no point or line charges - which were the only varieties which could acc

assembling only at discontinuities. Also d vanishes in case of an electrostatic field caused

needs some explanation. Let us imagine a different source charge configuration - the one in

by a corresponding positive one of an equal magnitude. The fact that the net charge ly

is zero whenever the charge configuration does not contain any point or line charges impl

configuration well have discon

| qi|= 0.Now, since | qi| qi| qi|, we have

| qi| i qii | qi| i

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is the algebraic sum (with due consideration to the signs) of all point and line charges res

recent most restriction holds if S has no discontinuity at all, (23) holds whenever S is th

call equation (23) as the simplest form of the Generalized Gausss Theorem and see that

6. Results

In the abstract it was mentioned that for the majority of cases the statement of our Ge

be assumed to be this: the flux of electrostatic field over a closed surface equals 1/0 ti

enclosed by S and half the value of net charge contained by S. Let us now enlist the cases wh

1. The surface of integration does not contain point or line charges at any of the corne

2. The surface of integration does not have any edge or corner, i.e. it is throughout co

3. The source charge configuration consists of only volume charges and surface charges

7. Discussion

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and the remaining part extending inside the conductor to an infinitesimal distance from

around r is supposed so small that the surface charge density in that area is essentially uapplication of the Generalized Gausss Theorem, which applies here in its simplest form, g

S

E da= Qenc

0+

1

2

Qcon0

=(r)A

20

Now the flux

E da is simplyE(r) Az as the field inside the conductor is zerocomponent of electrostatic field is zero on the surface[8], and it follows that

E(r) = ((r)

Coulombs law (see Figure-8) that for any point r on the conducting plane, we have Eq(rHence,

q

40

(r hz)(r2 +h2)3/2

+ES(

r) = (r)z

20

Asr is orthogonal to z, we can take a dot product of both sides of the above equatio

() h

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For any minute sphere of radius , drawn centered atr, carries a net dipole mo(2/3)3

P2 . If the interface is between a dielectric and vacuum, the polarization at a

P =

Pin/2, where

Pin equals the polarization just inside the dielectric at the point

r oSo, for our dielectric placed at R in vacuum[11],

S

P da= 1

2

S

Pin da= Qbcon

2

From the above two equations we see

SD da= Qfenc+ Qbenc+

1

2 Qfcon+ Qbcon

where

D =0E+

P denotes the electric displacement. As the net bound chargeQben

S

D da= Qfenc+

1

2Qfcon

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To arrive at this result using Gausss Theorem for electric displacement, we have to

charge enclosed by the pillbox, the z component of the electric displacement just abov

well as sign to that just below r . If we denote these components by Dz1 and Dz2 respDz2 = 0(1 + e)(Ez (r)/20). Only this time, well have no contribution to Dz1 fDz1/0 will simply be the z component of the electrostatic field just above

r. Just as in thEz+ (

r)/20, so that once again we haveDz1 equal to 0Ez+ (r)/2, and get the same

Acknowledgements

Its a pleasure for the author to acknowledge that the inception of the simplest form of thin the mind of the author occurred in 2006 in a conversation with his younger brother Sha

integrating the flux of the electrostatic field of a point charge placed on an ellipsoid at t

foci. Therefore, from the viewpoint of the author - as opposed to his brothers - this work

takes pleasure in thanking the reviewers for many fruitful suggestions, which in each criti

to c eate e a g e ts to i cl de i the disco ses a s e s to c io s a big ities left e

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References

[1] E.M.Purcell, Electricity and Magnetism, first ed., McGraw-Hill, New York, 1965, p. 2

[2] M.Zahn, Electromagnetic Field Theory: a problem solving approach, John Wiley and

[3] See, for example, the last paragraph on p.24, reference [1].

[4] J.D.Jackson, Classical Electrodynamics, third ed., John Wiley and Sons, 1999, pp. 27

[5] D.J.Griffiths, Introduction to Electrodynamics, third ed., Prentice-Hall, Upper Saddl

2.10.

[6] R.P.Feynman, R.B.Leighton, and M.Sands, The Feynman Lectures on Physics, Ad

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FIGURES

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Figure-3

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Figure-5

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Figure-7

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Figure Captions

Figure-1: The field of a uniformly charged spherical surface

x2

+y2

+z2

=R is to be evhere in the figure is the case where ris greater than R.

Figure-2: A cross section (of the closed surface S) showing the point chargeqi and the i

solid curve is the outer surface of S and the thick dotted curves are the inner surfaces of

which represent the cavities. The part of the outer surface drawn with a thin dotted cu

surfaceS2, while the remaining part on the left representsS1. The thin solid arc represen

surface of radius centered at qi.Figure-3: A point charge Q lies at a corner of a cube. The flux through the shaded face is

Figure-4: The same charge shown at the centre of a cube of sides of length twice as tha

bigger cube, the original surface on which the flux was obtained, lies as a quadrant of one

Figure-5: A Gaussian pillbox of base area A straddling the infinite flat plane. The cuboid e

sides of the la e *Detailed Response to Reviewers

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Dear Reviewer,

I have read the reviewers review and found many questions, wh

raised, helpful in improving the presentation of ideas. The issues raised in t

and I have tried my best to resolve them in this revised manuscript. I am

reviewer finds them appropriate. The point-by-point response to the r

follows below:

In response to earlier critical reviews by this reviewer, the autpaper but this has not improved the content or clarity of the work.

of definition but the last sentence of the abstract is the esscontribution and it is incorrect.

Yes, the assertions of the paper may turn out to be incorre

charge contained by a set of points in a manner different from intudo not think it meaningful to examine the situations, which may ari

conventions that will go on affecting many major equations of scien

change in my definitions of the net charge enclosed by a surfac

p

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conQ , we have encQ equal to zero and QQcon = . In the non-stand

somehow occurred to the reviewer and which statesenc

Q as equal to

the statement00

+

=

conenc QQshall turn out to be correct. I could

such confusion could occur to anyones mind and after having point

to me, the reviewer must have realized by now why the Generalize

needed.

The paradoxical result most likely comes from not being definitcharge is distributed on the surface of zero thickness. One can as

charge density rho(r) is uniformly distributed from R torho(r)=3Q/[4*pi{(R+delta)^3-R^3}] and the total charge in the lay

allow any surface charge at r=R and r=(R+delta) the electric field

are:

E(r

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solutions above with his in eqs. 1-5.

These remarks have been analyzed next in 2 from the

paragraph of 2 to equation (11). That the assumptions regardindensities at Rr= and Rr += do not alter the statement of t

discussed prior to the last paragraph of 2. In the last paragraph, i

non-uniform spherically symmetric volume density functions, when

thickness with the same net charge, also give the same Generalizwhich uniform spherically symmetric volume density gives. Physic

as convincing as a mathematical proof to me that if the result ismodeling a surface charge by the limit of a uniform volume ch

essentially the same - if surface charge can in principle be modevolume charge - when the surface charge Q is modeled by the lim

symmetric non-uniform volume charge distribution of a net charshell. I think it is convincing enough to state this line based o

physicist like me it would feel exhilarating and astonishing to

operator of this tool called integral calculus has turned him down, bby means of a proof by contradiction or by an example, when he wa

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it is free from errors to state that the suggestions of the reviewer co

of words only enhance the presentation of ideas.

Throughout the paper the permittivity of the outside medium iepsilon0. However, the results are equally valid if the outside m

permittivity of epsilon

Yes, I have considered microscopic electrostatics until make the discussions prior to that section count only free charges

charges. When I want to talk of Gausss Theorem for the electrosta

both free charges and bound charges as simply charges withou

between different kinds of media and without considering all the ch

not just as charges in all space, I have to consider permittivity of foutside medium is filled with a dielectric, Gausss Theorem or the

Theorem (whichever is applicable) will contain only permittivity of

encQ or

conQ we are referring to the net charge, which is not just th

the net free charge plus the net bound charge. This is just what

whenever microscopic electrostatics is discussed. The results using medium are valid I know that but since I have seen many great

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as suggested. Whenever I say, when a volume charge distribution

do not mean the assertion when a volume charge distribution exisa surface. A surface cannot have a finite thickness and by defi

value, which the thickness of a surface can take is zero. Since I m

distribution to exist on a surface, its volume density is an ordinary m

taking finite values and there is no need to describe its densityvolume density function. By definition, a volume charge distr

volume charge density cannot reside only on a surface and nowhere

of the surface. Volume charge must have some finite thickness andnice example will be the problem of evaluation of the flux

distribution

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HIGHLIGHTS

Gausss theorem applies only for a surface free of any charges Here the idea is that it can be generalized The generalized theorem only contains some new terms in the equation It is shown that for most of the cases only one missing term appears The equation 0enc

S

/Q= daE is changed to0

con

0

enc

S

Q

2

1

Q+= daE

This equation is shown to have wide applicability