Phys 230 Winter 2010 - Chapter 22

Gauss’s LawCharge and Electric Flux

Electric flux computations

Gauss‟s Law

Using Gauss‟s Law to Calculate the E-field

in high-symmetry situations

spherical symmetry

cylindrical symmetry

planar symmetry

Charges on conductors

2

0

1ˆ , ...

4P

qE r

r

0

ˆ2

PE rr

02E const

0enc

S

QAdE

AdEE

1

Recall our chapter 21/22 roadmap… Two basic problems in electrostatics:

1) Given a charge distribution determine the electric field at point P

Coulomb‟s

Law

Electric

Fields

Gauss‟s

Law

We used Coulomb‟s law for a (point charge) & superposition

principle to determine this question in Chapter 21.

2) Given the electric field in a region, determine the charge distribution

that produced it.

In this chapter we discuss a law that answers this question in principle,

but also allows us to easily answer the first question (i.e. the

determination of E itself) in situations with sufficient SYMMETRY.

Problem-solving tip: We use Coulomb‟s Law/Superposition in situations of low symmetry,

and Gauss‟s Law in situations of high symmetry. As we will show, the two laws are equivalent

to each other. The situation is analogous to calculating integrals via Riemann sums or using

antiderivatives and the fundamental theorem of calculus. 2

Gauss‟s Law - OverviewGauss‟s law, like Coulomb‟s law, is a relationship between electric charge and the electric field charge sets up.

Central to Gauss‟s Law is a hypothetical, closed1 surface in space called a

Gaussian surface. The closed surface is arbitrary – YOU CHOOSE IT.

4 Gaussian surfaces:

1The fact that it is closed means there is a well-defined outside and inside.

A encloses the positive charge.

B encloses the negative charge.

C encloses both and D encloses neither.

Gauss’s Law tells us how the fields at and over the Gaussian surface are

related to the charges contained within that surface.

To make this idea quantitative requires us to introduce the notion of an electric flux through the surface.

3

Electric Flux – Fluid Flow Analogy

Suppose you dip a rectangular loop enclosing an area A in a uniformly flowing stream

with velocity ) so the loop is perpendicular to the current stream. The rate at which water

flows through the loop (in say m3/s) is simply:v

dVAv

dt This is the called the flux.

If the loop‟s cross-sectional area is not perpendicular to the velocity, then less water

flows through the loop. If we DEFINE the vector ˆA An

which is perpendicular to the plane defined by the cross-sectional area, then the

volume flow rate, or the flux, through the loop can be written as

ˆ

A

n

area of loop

unit normal vector

cosA v Av

For an open surface, like the loop, there is an ambiguity

in the direction of n. With closed surfaces, we will always

define n to be the OUTWARD unit normal vector.

is the angle between

ˆ (or ) and .n A v

4

Electric Flux - In the fluid analogy the collection of velocity vectors of each fluid particle collectively form the

„velocity field‟, and streamlines are the unique particle trajectories formed by „connecting‟ velocity vectors from

point to point in such a way that the velocity vectors at each point are tangent to the streamline.

The notion of flux through a surface depends

only on the existence of a vector field. Thus, we

similarly define electric flux of a uniform E-field

through an open rectangular surface as

cosE E A EA

even though nothing is actually „flowing‟ in this case, and again remind ourselves that E-field lines are not

trajectories but lines of force. Here the electric field plays the role of the velocity field in the fluid case.

5

YF22.2 A flat sheet is in the shape of a rectangle with sides of lengths

0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of

magnitude 75.0 N/C that is directed at 20° from the plane of the sheet.

Find the magnitude of the electric flux through the sheet.

Note carefully how the angle was measured.

As we mentioned, Gauss‟s law involves closed surfaces which intuitively divide

three-dimensional space into an „inside‟ and „outside‟ (and the surface itself). The normal

is taken to be the outward unit normal (no Klein bottles!). We are then interested in the

flux through the entire surface. As in integral calculus the idea will be to break the surface

into an infinite number of infinitesimal rectangles and define the electric flux provisionally as:

A Klein bottle is a non-orientable surface

that has no inside/outside, but is not

realizable in 3D without self-intersections.

~ cosE i i i i iE A E A

6

Electric Flux – General DefinitionIn the simultaneous limit of a larger and larger number of smaller and smaller rectangles

that cover the surface S we define the electric flux as the surface integral:

This allows us to deal with both non-uniform fields, and curved surfaces.

In general this multiple integral is very difficult to compute, but in every example we will consider, the integral will

reduce to at worst a (set of) single ordinary integral(s), usually because of the symmetry of the problem.

θ Direction of E Sign of E·dA

>90° Into surface Negative

=90° Parallel to surface Zero

<90° Out of surface Positive

„closed‟

ˆdA ndA

SSSE dAEdAEAdE cos

7

YF22.4(a) (Simple closed surface, non-uniform field)

A cube has sides of length L = 0.300 m. It is placed with one corner at the origin as

shown. The electric field is not uniform but is given by

ˆ ˆ( 5.00 / ) (3.00 / )E N C m x N C m z i k

Calculate the electric flux through each of the six cube faces, and find the total electric flux.

E E E

YF22.5 A hemispherical surface with radius r in a region of uniform electric field

has its axis aligned parallel to the direction of the field. Calculate the flux through

the open hemispherical surface.E

Hint: Given the field is uniform what is the total flux through the closed surface consisting of the hemisphere and its endcap?

(A case where we actually have to integrate to get a flux, b/c the field is nonuniform over a surface.)

8

Enclosed Charge and Electric Flux

Gauss‟s Law relates the electric flux through a Gaussian surface to the charge enclosed by this fictitious surface.

More enclosed positive charge,

more outward flux of field lines...More enclosed negative charge,

more inward flux of field lines...

Positive charge = source of field lines. Negative charge = sink of field lines.

9

Enclosed Charge and Electric Flux

Doubling the charge doubles the flux, but doubling the

dimensions of the Gaussian surface (here the box), does

NOT change the flux as long as we enclose no additional

charge: the same number of field lines cross the surface.

Intuitively, although the box is bigger the field lines are „less dense‟ in an exactly

compensating manner. We will make this quantitatively precise shortly.

It is crucial to understand that while the presence of charge OUTSIDE the Gaussian surface affects the

electric field on the surface of the Gaussian surface, it does not affect the NET flux through the box.

Positive charges act as sources of field lines while negative charges act as sinks of field lines. If the net charge

enclosed by a Gaussian surface is zero, then there is no net flow of field lines into or out of the surface.

10

Gauss‟s Law for a Point Charge (from Coulomb‟s law)

Consider a positive point charge and a Gaussian sphere of

radius r centered on the point charge. Last chapter we saw

the E-field is given by:2

0

1ˆ( )

4

qE r r

r

This expression is spherically symmetric (it depends only on r): thus the

magnitude of the E-field is the same everywhere on the Gaussian sphere.

Furthermore, the unit normal on the sphere is simply itself, and r is

constant over the sphere. Thus:

(Coulomb‟s law)

r̂

since the area of

the sphere is 4πr2.

Doubling R quadruples the area of the

patch, but the field falls off by a factor of

4 because of the inverse square law.

Thus the flux remains unchanged.

surface area of sphere

This cancellation between the r2 coming from the surface area of the

sphere and the 1/r2 in the inverse square law is utterly crucial for

everything that follows; e.g. if Coulomb‟s law wasn‟t inverse square,

the flux would depend on the Gaussian surface chosen!

ˆ ˆ 1r r 0

2

2

0

2

0

)4(4

1

4

1

qr

r

qdA

r

qAdE

SS

11

Generalizing Gauss‟s LawClaim: IN GENERAL

We show this in two parts:

1) that the choice of Gaussian surface is arbitrary (so long as Qenc doesn‟t change)1 .

2) It holds not just for point charges, but arbitrary charge configurations.

1) Surround the charge with an arbitrary closed surface, and

consider a small area element dA on the irregular surface. This

is larger than the corresponding spherical patch at the same r. If

makes an angle relative to , two sides of the area projected onto

the spherical surface are foreshortened by a factor of . Either way,

the flux is given by . The result follows by summing over patches.

n̂ r̂cos

cosE dA

2) This follows directly from the superposition principle and the linearity of integration.

(away from the origin.)

0enc

SE

QAdE

Rigourous Proof

VVAdEE

Divergence theorem (from vector calculus) :

Compute the divergence of the E-field due to a point charge:

0313131

44

ˆ

4 5

2

35

2

35

2

3

0

333

0

2

0

r

z

rr

y

rr

x

r

Q

r

z

zr

y

yr

x

x

Q

r

rQ

where rkzjyixrzyxr /)ˆˆˆ(ˆ,222

Apply div. theorem and the result to the region between the Gaussian sphere and the arbitrary Gaussian surface:

Ssphere

sV

AdEdAnEE

ˆ0

On the sphere‟s surface, the „outward‟ unit normal to the enclosed volume points „inward‟, while the E-field points

outward, i.e.

sphere

sS

dAnEAdE ˆ

(i.e. 3D analogue of the fundamental theorem of calculus)

rnrrEE sˆˆ,ˆ)(

0

ˆ

qdArEAdE

sphereS

Thus: using the result on the previous slide.

QED 12

Gauss‟s Law

“Gauss was the greatest of all mathematicians and perhaps the

most richly gifted genius of whom there is any record.”

Equivalent statements:

In colloquial English:The electric flux through any closed surface of any charge distribution

is proportional to the charge enclosed by the surface.

One of Maxwell’s Four Equations governing all electromagnetic phenomena

Note that Qenc itself will often be an integral of a volumetric charge density: encQ dq dV that will always collapse to a single integral (at worst) for the symmetrical configurations we consider.

So it is the right hand side of Gauss‟s law that will involve calculus in our calculations, not the LHS!

While Gauss`s law can be used to compute the E-field in highly symmetrical situations (as we`ll see), it can also

be used to determine the charge configuration if the field is known, as well as to prove certain results.

0

cosˆ

enc

SSSSE

QdAEdAEdAnEAdE

13

Gauss‟s Law example of going from E-field to charge distribution…Phys 230 Fall 2009 Midterm (Short Answer 3)

3. (3 marks) In a particular region of space the E-field is given by, where z is the distance along the z-axis and A= 6.00×103

N/(C·m). How much charge is enclosed by a cylinder of circular cross section (radius 2.00 m, height 3.00 m) concentric with

the z-axis, and whose bottom face is at z = 1.00 m?

z = 4.0 m

z = 1.0 m

z

E

a) 1.50 μC b) 2.00 μC c) 2.67 μC d) 3.34 μC e) 4.50 μC

Formula(e) Used: _____________________________________________________________Tailored specifically

to the question!

14

Gauss‟s Law and Symmetry

Although Gauss‟s law is always true, it is not always useful. Unless a symmetry is present to make the flux

integral tractable/easy/trivial or if a „poor‟ choice of Gaussian surface is made, then the law is harder to use than

Chapter 21 methods. In the following examples we will always have one of the following three situations:

Spherical Symmetry: Choose Gaussian surface as a concentric sphere.

Cylindrical Symmetry: Choose Gaussian surface a coaxial cylinder

Planar Symmetry: Choose Gaussian surface as a „pillbox‟ which straddles the surface.

Although the second and third of these technically require infinite lines and planes, we will often use them in an

approximate sense (as we did in Ch 21): watch out for keywords like “very large planes” or “very long lines” or

“very close to” in the sense that the edges of the charged object are very far away.

15

The three standard examples: Uniformly Charged Spherical Shell (of radius R)

This is the shell theorem – the easy way!

In the previous chapter we worked very hard to determine to the electric field of a uniformly charged spherical shell.

Inside sphere: Choose a Gaussian sphere of radius r<R centered concentrically with the

shell. The Gaussian surface lies entirely inside the hollow shell, and so encloses no

charge. Gauss‟s law implies that E=0 everywhere on the Gaussian surface. Since r was

otherwise arbitrary, we conclude E=0 everywhere inside the shell!R

r

Outside sphere: Choose a Gaussian sphere of radius r>R centered concentrically with

the shell. Since the sphere is uniformly charged, the E-field is spherically symmetric.

This implies 1) on the Gaussian sphere the magnitude of the E-field is the same everywhere,

and 2) the E-field is directed radially outward. Thus , and comes out of

the integral over the surface:

ˆ( ) ( )E r E r r

2

0

1ˆ( )

4

QE r r

r

outside the shell.

Spherical symmetry is invariance under rotations (either of polar or azimuthal coordinate) about a center.

0

2 )4()(

Q

rEdAEAdrES

16

The three standard examples: Uniformly Charged Infinite Line

Here the symmetry is cylindrical (aka axial), so choose a cylindrical Gaussian

surface surrounding the long line charge. The symmetry implies that the E-field

must point radially outward from the line charge (otherwise there would be a

preferred direction). Thus no electric flux passes through the endcaps since .

The flux passes perpendicularly across the „side‟ of the cylinder, which has surface

area 2πrl. Thus since E must be constant on the side of the cylinder:

0 0

( )2 2

E rr

Q

rl

since the linear charge density is

uniform.

cos 0

Again, this agrees with the result we found in previously, but obtained in a much easier manner.

Cylindrical symmetry is invariance under translations along the line, and rotations about the line.

0

)2()(

Q

rlEdArEAdEsideS

17

The three standard examples: Uniformly Charged Infinite Plane

Choose the Gaussian surface to be a „pillbox‟ of cross-sectional area A

(it doesn‟t have to be circular as the diagram suggests), that symmetrically

straddles the sheet. The planar symmetry implies that the E-field is directed

perpendicular to the plane (to avoid preferred directions). Thus the flux is

directed through only the endcaps (none passes through the lateral surface).

Furthermore, translational invariance on the endcap itself implies that E is constant on

the endcaps (otherwise there are preferred points on the plane itself). Thus:

Planar symmetry is invariance under translations along either direction parallel to the plane.

0 02 2

encQ

AE

since the surface charge density is

uniform.

Thus far we haven’t even done any integrations...

0

)2()(

enc

endcapsS

QAEdArEAdE

18

YF Example 22.9: Uniformly Charged Solid Sphere (Insulated)

A slightly more nontrivial example...same spherical symmetry arguments apply

Outside Sphere: Gaussian surface sphere radius r>R

2

0

1( )

4

QE r

r

as with the point charge,

and spherical shell.

Inside Sphere: Gaussian surface sphere radius r<R. Now however, we have to

determine how much (of the total) charge is contained in the Gaussian sphere.

Uniformdq Q

constdV V

and so taking spherical shells of volume 4πr2dr

3 32

30

44

3

r

enc

r QrQ dq dV r dr

R

Gauss’s Law

32

3

0

(4 )Qr

E rR

3

0

( )4

Q rE r

R

0

2 )4()(

Q

rEAdrES

19

YF22.57 (Non-Uniform, but Spherically Symmetric Distribution)

A non-uniform but spherically symmetric distribution of charge has a charge density ρ(r) given as follows:

0 (1 / ) ,( )

0 ,

r R r Rr

r R

where 0 3

3Q

R

(a) Show that the total charge contained in the charge distribution is Q.

(b) Show that the electric field in the region r≥R is identical to that produced by a point charge Q at r=0.

(c) Obtain an expression for the electric field in the region r≤R.

(d) Graph the electric-field magnitude E as a function of r.

(e) Find the value of r at which the electric field is at its maximum, and find that maximum.

Again, we’ll need to calculate for (d) the enclosed charge from2

0( )4

r

encQ dV r r dr

region over which

ρ~const

20

9. The neutral hydrogen atom consists of a proton of charge +Q = 1.6×10-19 C and a bound electron of charge

–Q = –1.6×10-19 C. In nonrelativistic quantum mechanics we model the proton as a point charge at the origin, but the

electron’s motion is inherently non-deterministic, and we can only speak of the probability of finding it somewhere. In

fact we can think of its charge as being ‘smeared out’, and an electron in the ground state can be described by the

spherically symmetric (but non-uniform) charge distribution:

Phys 230 Fall 2009 Midterm (Long Answer 2) (See also YF 22.65)

0/2

3

0

)(ar

ea

Qr

where a0 is called the Bohr radius of the hydrogen atom.

(a) Use integration by parts to show that . [1]

(b) Find the total amount of the hydrogen atom’s charge that is enclosed within a sphere of radius r centered on the

proton. [4] (One check of your answer is that , since the atom is neutral.)

dxex

c

n

c

exdxex cxn

cxncxn 1

0)(lim

rqr

(c) Find the magnitude of the E-field as a function of r. [3]

Hand-holding. Expect to use it in part (b)!

A touch of quantum mechanics

A non-uniform but spherically symmetric charge distribution

21

Conductors and Gauss‟s Law

In the last chapter we showed that in an electrostatic situation, the E-field in the

interior of a conductor had to be everywhere zero. Consider what Gauss‟s law now

implies, by taking an arbitrary Gaussian surface entirely contained inside the conductor.

Consider a conductor with no holes, aka „cavities‟ in it.

Since the E-field everywhere on the Gaussian surface is zero, Gauss‟s law implies

the net enclosed charge is zero. Since the Gaussian surface was otherwise arbitrary,

by shrinking the surface down and moving it around within in the conductor, we can

conclude that the charge everywhere inside the conductor is zero.

In other words any excess charge carried by a conductor without cavities is carried entirely on its surface.

22

02E

for a thin sheet of charge

0

E

Warning:

Conductors continued: YF 22.51/Example YF22.8

but for a single conducting surface

(or between two oppositely charged conducting plates,

called a parallel plate capacitor.)

YF22.51

In the former case, we view the charge distribution on a charged conducting sheet as being composed of two

sheets of charge (one on each surface). In the latter case, all of the charge on each plate is drawn to one surface

(because of attraction to the other plate). Choose Gaussian cylinders with one end inside the conductor or use the

superposition principle to convince yourself of the conducting case results.

23

Conductors and Cavities

What happens if there‟s a cavity in the conductor? A Gaussian surface surrounding the cavity, Gauss‟s law, and the

fact that E=0 inside an electrostatic conductor establishes that no net charge is enclosed.

If there‟s no charge in the cavity, then there‟s no net charge on the surface of the cavity (inner surface of

conductor). In the next chapter we‟ll be able to show that in fact there is no charge anywhere on the inner surface.

If there‟s a charge +Q in the cavity then a charge of –Q must be induced in the surface of the cavity. If the

conductor was neutral, charge conservation requires a charge +Q to be form on the outside surface of the

conductor. Note this means there will be a discontinuities in the E-field across surfaces.

If the conductor has a net charge on it, then charges will arrange themselves so that the E-field inside the

conductor remains zero.

24

YF22.37 (The Coaxial Cable) A long coaxial cable consists of an inner cylindrical

conductor with radius a and an outer coaxial cylinder with inner radius b and outer

radius c. The outer cylinder is mounted on insulating supports and has no net charge.

The inner cylinder has a uniform positive charge per unit length λ. Calculate the

electric field (a) at any point between the cylinders a distance r from the axis,

(b) at any point outside the outer cylinder. (c) Graph the magnitude of the electric field

as a function of the distance r from the axis of the cable, from r=0 to r= 2c.

(d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

Remember: Conductors have their charge spread out only on their surfaces (possibly inner and outer)

in electrostatics, whereas insulators can have their charge spread throughout their volumes.

25

A hollow conducting sphere has inner and outer radii of 0.80 m and 1.20 m respectively. It carries a net charge

of –500 nC. A uniformly charged insulated ball of radius 0.20 m with a total charge of +200 nC is present at

the center of the hollow conducting sphere.

Phys 230 Fall 2009 Midterm (Short Answer 4 & 5)

Use the following information for questions 4 and 5.

0.80 m

1.2 m4. (2 marks) What is the ratio of the surface charge density on the outer surface of the conducting sphere to

the surface charge density on the inner surface of the conducting sphere? (Hint: Find the charge on the surfaces first.)

a) zero b) 3/2 c) 1 d) 2/3 e) 4/9

5. (2 marks) Qualitatively sketch the magnitude of the E-field as a function of distance r from the center of the

insulating ball.

26

Earnshaw‟s Theorem (See YF22.56)

Prove that a point charge q cannot be held in stable equilibrium by electrostatic forces alone (from external charge

distributions).Hint: Use Gauss’s law and a ‘proof by contradiction’. What must be the E-field look like

near a conjectured point of stable equilibrium?

Time & Interest permitting

YF22.48 A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by

an insulating shell with inner radius R and outer radius 2R. The insulating shell has a uniform charge density ρ.

(a) Find the value of ρ so that the net charge of the entire system is zero.

(b) If ρ has the value in (a), find the E-field (mag. and dir.) in each of the regions 0 < r < R, R < r< 2R, and r > 2R.

Graph the radial component of E as a function of r.

27

Time Permitting: Application & Superposition Example (YF22.63)

Positive charge Q is distributed uniformly over each of two spherical volumes

with radius R. One sphere of charge is centered at the origin and the other

at x= 2R. Find the magnitude and direction of the net electric field due

to these two distributions of charge at the following points on the x-axis.

(a) x= 0, (b) x = R/2, (c) x= R, (d) = 3R

28