24.1

Chapter 24: Gauss’ Law

In Chapter 23, the tool symmetry was very important in simplifying problems. Gauss’ law takes these symmetry arguments and maximizes their efficiency in simplifying E-field calculations. • Coulomb’s Law Method is barbaric and uses pure “brute force” when solving for the

E-field. We derived some equations using some fairly strenuous integration. This is the hard way of doing things. Coulomb's method for solving electrostatics can be formidable - much of the rest of electrostatics is devoted to assembling a bag of tools and tricks for avoiding these integrals

There is a saying that goes something like this: there is a hard way and easy way to do the same job – the easy way usually involves using the right tools. Gauss’ law is the right tool.

• Gauss’ Law Method is a more elegant approach for solving E-fields when there is high degree of symmetry. Consequently, we will be restricted to a limited number of charge distributions. As you will see, we will be able to derive some E-fields in only a few lines!

The basic idea of Gauss' law is 1. Enclose a charge distribution with an imaginary or Gaussian surface. We then

mathematically count the number of E-field lines poking through this Gaussian surface. This counting is the electric flux.

2. If this Gaussian has the right symmetry, we will then be able to relate the Gaussian surface area to the E-field function.

This may sound rather like an indirect method of expressing things, but it turns out to be a tremendously useful relationship.

What is a Gaussian A Gaussian is an imaginargy closed surface with a high degree of symmetry. If it has the proper symmetry, then a Gaussian is a constant E-field surface. The key point being that if you cannot do this, then it is easier to use Coulomb’s Method for determining the E-field.

Gaussian Surface constant E-field surface⇔ There are only 3 types of Gaussians’: spherical, cylindrical, and cubic (“pill box”) shell.

The connection between Electric Flux and Electric Charge

Definition: The number of E-field lines poking out or in through the Gaussian is measured by the electric flux.

EElectric Flux number of E-field lines passing through a GaussianE dA≡ Φ = ⋅ =∫

An enclosed charge creates a flux through a Gaussian tells us what is going on inside. In other words, we will be able to determine the magnitude and sign of the enclosed charge. Here are three key points about Gaussians and flux: i. The electric flux is directly proportional to the enclosed charge. ii. Charges outside a Gaussian will not contribute to the flux of the enclosed charge. iii. Flux is independent of size and shape of the Gaussian.

Part 1: Electric Flux is proportional to the Enclosed Electric Charge (Φ ∝ q) Suppose an electric charge lines within the Gaussian, (i) how much and what sign does the enclose charge have?

24.2

HWR applet Gauss law Suppose a positive charge is enclosed by a spherical Gaussian. The electric charge will produce an electric flux that pokes through the Gaussian. • Measuring the amount of flux through the Gaussian will tell us the number of field

lines coming through the Gaussian. That is, if four lines poking through is for one electric charge, then 8 lines is twice the charge enclosed.

• If the flux is pointing outwards, then the enclosed charge is positively charged whereas inwards, its negatively charged

In summary,

( ) ( )

enclosed encE

Electric Flux Enclosed Electric Charge

q q

∝

Φ ∝ ± ≡ ±

ii. Electric charges outside a surface will not contribute to the electric flux of qenc Suppose a Gaussian encloses some electric charges AND there are different, external electric charges outside the Gaussian. The net flux produced by these external electric charges not affect the flux reading of the enclosed charge.

enc encenc net q q

cancel out

2qq + −= → Φ + Φ + Φ = ΦΦ =

iii. The Electric Flux is Independent of the Size and Shape of the Gaussian If the enclosed charge is fixed, then the flux is

enc Eq A Econstant constant E AΦ ⋅ == → = = ⋅ Consider two spherical Gaussians where one of the spheres has twice the radius as the other. Since the magnitude of the field decreases as 1/r2, the field at the surface of a sphere twice as large has its E-field reduced by ¼ when compared to the other sphere:

2 2 21 21 1 1 1r (2r) 4 r

E vs. E∝ ∝ =

On the other hand, a sphere is twice as large, has four times the area: 1 2

2 2 2A 4 r vs. A 4 (2r) 4 4 r= π = π = ⋅ π Comparing the fluxes through both spheres tells us

1 21

1 1 1 2 2 2 1 14

The Electric Flux is Independent of the size of the sphere

E A and E A E 4A Φ = ΦΦ = ⋅ Φ = ⋅ = ⋅ →

Calculating the Electric Flux The electric flux depends on the E-field and the area. The area vector is a vector that is normal to the surface: A = An = (A⊥, A∥). It is customary to pick the outward direction to be the positive normal direction. There are 2 area vector components:

24.3

Let’s first start by viewing this applet.

Applet Davidson’s Flux

The amount of flux (i.e., number of field lines poking through the enclosed area) is the dot product between E and dA.

EFlux E dA Ecos dA≡ Φ = =⋅ θ∫ ∫

where out of the surface

ˆ ˆA An n normal into the surface

+= → = =

−

Units: [ ] [ ][ ] 2 2E A mN/C N m /CΦ = = ⋅ = ⋅ However, since the Gaussian is independent of area and not going into the details of the situation, when the “dust settles”, the important detail for now is the cosine angle between E and dA. The angle dependent of the flux shows that there are three possible angles but only two will be important for our purposes: θ = 0° and 90°.

Question: A rubber balloon has a single point charge in its interior. Does the electric flux through the balloon depend on whether or not it is fully inflated? Explain your reasoning.

Gauss’ Law (Stated without Proof) The flux through any kind of surface is equal to the enclosed charge divided by ϵ0. The E-field is the total electric field at each point on the Gaussian Surface.

A Deeper look at the vector E- field From vector calculus, there is a mathematical theorem known as Gauss law that relates an electric field vector on a surface area to the divergence in a volume:

E dA EdV⋅ = ∇ ⋅∫ ∫

where the divergence is

number of field lines coming

through a Gaussian surfaceE ∝

∇ ⋅

Here is a much deeper inside view of what an electric field is and its physical meaning. An ordinary vector can be multiplied 3 different ways:

aA, A B, A B⋅ ×

The same applies to three-dimensional derivatives or Del Operators:

curlgradient divergence

A , A , A∇ ∇ ⋅ ∇ ×

encE surface

0

Gauss' Law: q E dAΦ = ⋅ =∫

24.4

What do these mathematical operations measure?

1. Gradient is the standard derivative which measures slope 2. The divergence mathematically looks like

( ) ( )x y z x y z x x y y z zA , , A ,A ,A A A A∇ ⋅ = ∂ ∂ ∂ ⋅ = ∂ + ∂ + ∂

Geometrical interpretation: A∇ ⋅

is measure of how much the vector A

spreads out (diverges) from a point in question.

3. Will save this one for when we get to magnetism but here is a “little bite”: The curl

A∇ ×

is measure of how much the vector A

curls (rotates) about a point.

Example 24.1 A charged particle is held at the center of two concentric conducting spherical shells. The figure on the right gives the net flux Φ through a Gaussian sphere centered on the particle, as a function of the radius r of the sphere. What are (a) the charge of the central particle and the net charges of (b) shell A and (c) shell B?

Solution The point charge at the center has no dimensions but the shell A has a width being r2 − r1 and while shell B is r4 − r3. a. To determine qenc inside of shell A, a Gaussian (of radius ra) is placed inside r1 (ra <

r1). From the plot, we read that Φ = –9.0×105 in SI units. Substituting this into Gauss’ law, we get

( )( )12 2 2 5 2enc 0

central

enc 0 8.85 10 C / N mq 9.0 10 N m / C

8 C q

q /

−× ⋅Φ = = ε Φ = − × ⋅

= − µ =

ε →

b. To determine qenc outside of shell A, a Gaussian (of radius rb) is placed between r3 > rb > r2. The value of the flux is +4.0×105, which implies that

( )( )12 2 2 5 2

0enc b 3 8.85 10 C / N m 4.0 10 N m / C 3.5 Cq (r r ) −× ⋅ × ⋅ == ε Φ = µ<

But we have already accounted for some of that charge in part (a), so the result is

enc b 3 A central A enc b 3 central

A

q (r r ) q q q q (r r ) q 3.5 C ( 8 C)

11.5 C 12 C q

< = + → = < −

= µ − − µ

= µ ≈ µ =

This makes physical sense because to go from a large negative to a positive flux requires a lot of positive charge.

c. To determine qenc outside of shell B, a Gaussian (of radius rc) is placed outside of r4 (rc > r4). The flux value is – 2.0×105, which implies

( )( )12 2 2 5 2

0enc c 4 8.85 10 C / N m 2.0 10 N m / C 1.77 Cq (r r ) −× ⋅ − × ⋅ == ε Φ = − µ<

Putting it all together the result is

24.5

enc c 4 A B central B enc c 4 A centralq (r r ) q q q q q (r r ) q q 1.77 C 12 C ( 8 C)

5.7 C 6 C q

< = + + → = < − −

= − µ − µ − − µ

= − µ ≈ − µ = B

Applications of Gauss’ Law (Spoken Remarks) 1. Gauss’ Law is most useful and should only be applied to cases where there is a high

degree of symmetry. When there is not a high degree of symmetry, it may be easier to use Coulomb’s Method to get the electric field.

2. Gauss' law is almost of magical power and it all hinges on the fact that Coulomb's law has a 1/r2-character. Without this character, Gauss' law is not possible. Gauss’ law is most commonly used if the enclosed electric charge is known, what is the electric field produced by this enclosed charge?

3. Once a gaussian has been chosen, the E-field function is mathematically pulled out of the integral because it is a constant everywhere on the gaussian surface:

enc

enc 00Step 2Step 1

q1E dA Ecos dA q / EdA

Φ = ⋅ = θ = → =∫ ∫ ∫

PROBLEM SOLVING STRATEGIES There is a 3-step process for applying Gauss’ law to derive the form of the electric field.

Step 1: Pick a good Gaussian Surface • Symmetry condition (pulls E = constant out of the integral sign) • Dot-product condition evaluates the cosine function either as ±1 or 0.

Step 2: Evaluate the enclosed charge qenc Step 3: Apply Gauss’ law to derive the E-field expression

Step 1: Picking a good Gaussian The goal is to pull out of the flux integral the electric field function and the cosine term. A good Gaussian satisfies the (i) Symmetry and (ii) Dot-Product conditions.

Symmetry Condition: The symmetry of the charge distribution must also be the symmetry of the Gaussian surface: • Spherically symmetric charge distribution choose→ Spherically Gaussian shell

• Cylindrical symmetric charge distribution choose→ Cylindrically Gaussian shell

• Planar symmetric charge distribution choose→ Cubically Gaussian shell The job of the symmetry condition is to pull the E-field variable out of the flux integral sign:

surface surfaceE dA Ecos dA ⋅ = θ∫ ∫

That means that the electric field will have the same value for most, if not all, everywhere on the gaussian. Picture wise, we get

Dot-Product Condition: The Dot Product between E and dA yields a cosine function: E dA (E dA)cos⋅ = ⋅ θ

. We look at the direction of the area and E-field vectors, where are three possibilities:

• E ∥ dA → θ=0°→ cos(0°) = +1 → positive flux • E (antiparallel) dA → θ=180°→ cos(180°) = −1 → negative flux • E ⊥ dA → θ=90°→ cos(90°) = 0 → negative flux The result of step 1 is

24.6

surface surface

Step 1

E dA Ecos dA ⋅ = θ∫ ∫

Step 2: Evaluating the Enclosed Charge qenc There are two possible situations for the charge distribution through the volume: (i) constant or (ii) variable.

Constant charge distribution A constant charge distribution is described by its charge distribution function defined by the symbol rho ≡ ρ. If rho is constant () then the enclosed charge is defined via the enclosed charge equation:

3enc encenc

total enc totalV Vq VQ q Q where [ ] C/m

Vρ ≡ = → = ρ =

There are two situations that will be encountered in problem sets. • Point P outside the charge distribution

If the Gaussian encloses the total charge, then the enclosed charge is the total charge:

enc total(point outside the surface)q Q V= = ρ

• Point P inside the charge distribution If the Gaussian encloses only part of the total charge, then the enclosed charge is

encenc

total

V(point inside the surface)

Vq Q=

Step 3: Applying Gauss’ Law Substituting Steps 1 & 2 into Gauss’ Law, solving for the E-field function gives

enc encsurface surface

0 0Gaussian surface

q q1E dA E dA EdA

Φ = ⋅ = = ⇒ =∫ ∫ ∫

One is left with an integral over the area of the Gaussian surface:

2cylinder spherecylinder sphere

2 34cylinder sphere 3

A dA 2 rL A dA 4 rCylinder Sphere

V r L V r

= = π = = π= =

= π = π

∫ ∫

I will now apply this 3-step process for several examples; insulated solid sphere, spherical shell, a thin wire (or cylinder), and an infinite plane sheet.

Insulating Solid Sphere An insulating solid sphere of radius R has a uniform volume charge density ρ and carries a positive total charge of qtotal = Q. (a) Derive the E-field using Gauss’s law (show the details) for 0 < r < ∞. (b) Plot E(r) vs. r for 0 < r < ∞.

Solution a. When ever the charge distribution is said to be uniformly charged, it means that the

charges are in the inside as well as on the surface. There are two possibilities: outside and inside the sphere.

i. E (r > R), outside the solid insulating sphere Step 1: Pick a good Gaussian

24.7

The charge distribution has spherical symmetry, so the Gaussian is a spherical shell. One draws a spherical Gaussian with a dashed line at a distance r that encloses the total charge. The flux at the Gaussian surface has E and dA parallel, so the cos θ function is cos (0) = +1. The flux integral is therefore

E spheresphericalshell

2E

E dA Ecos A

E 4 r (spherical shell)

Φ = ⋅ = θ ⋅

= + ⋅ π = Φ

∫

Step 2: Determine the enclosed charge Since the Gaussian encloses all of the charge of the solid sphere, the magnitude of the enclosed charge is

encq Q Q= =

Remember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian.

Step 3: Apply Gauss’ law to determine E (R1 < r < R2) Gauss’ law gives

2enc 0 0 2sphere

0Step 2Step 1

QE dA q / E 4 r Q/ E(r R)4 r

⋅ = →+ ⋅ π = → > =π∫

ii. E (0 < r < R), inside the solid insulating sphere Step 1: Pick a good Gaussian The charge distribution inside the insulating sphere still has spherical symmetry, and we have already solved this for a spherical Gaussian. Since it has already been done, we just take the result:

2E E 4 rΦ = + ⋅ π

Step 2: Determine the enclosed charge Is there an E-field inside a charge insulator? Absolutely! Would one expect a larger E-field inside the charged insulating solid? NO – why? If the Gaussian encloses less charge, then the flux is smaller and this leads to a weaker E-field.

Inside an insulating sphere, one must use the enclosed charge equation to determine qenc.

enc total encenc

enc total total

Vconstant

V Vq q q QV

ρ = = = → =

There are two different volumes in this equation: 34 3 34

total 3 enc 33 3434

total 3enc 3

V VVV

(sphere) R r r 1 for r RR R(Gaussian) r

= π π → = = < < π= π

That is, the enclosed charge qenc is less than the total charge Q since the ratio of volumes is less than one. Solving for the enclosed charge gives

3enc

enc enc3total

VV

rq Q Q qR

= = =

Remember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian.

Step 3: Apply Gauss’ law to determine E (0 < r < R) Gauss’ law gives

24.8

32

enc 0 03 3sphere0Step 2Step 1

r QE dA q / E 4 r Q/ E(0 r R) rR 4 R

⋅ = →+ ⋅ π = → < < =π∫

In summary,

30

20

insulatingsphere

(inside)

(outside)

Qr, 0 r R

4 R1 Q

, r R 4 R

E⋅ < <

π

>π

=

Interpretation of Results 1. A plot of E(r) versus r is shown on the right. Note that Einside is different

from Eoutside − this is a key signature of E-field across boundaries (Ein ≠ Eout). What would happen if the field was not discontinuous? The field would continue to behave as E ∝ 1/r2 for r < R, so that the field would then be infinite at r = 0, which is physically impossible. However, Gauss' law shows that in the limit as the radius of the Gaussian goes to zero, there is a smaller and smaller enclosed charge until there is no enclosed charge. So, the E-field goes to zero:

insider 0limE 0

→→

This inside result eliminates the problem that would exist at r = 0 if E(r) varied as 1/r2 inside the sphere as it goes outside the sphere.

Side Note: This is a problem with classical electromagnetic theory and modern particle physics. Point particle have infinite fields at their origin. How does one eliminate this type of problem? One extends a point particle into a “string” and these infinities vanish! Superstrings – however, it has its own problems.

2. These two expressions for the electric field match at the boundary (surface):

total totalinside outside3 2

0 0r R

q q1E r E4 ε R 4 ε R

=

= ⋅ = = π π

3. For those of you who have had Phys 4C or remember Snell’s law from optics, the index of refraction occurs because of the electrical properties of a dielectric. That is, at the boundary of a dielectric, the E-fields are not the same just inside and outside a dielectric ( in outE E⊥ ⊥≠ ). This is the main reason why light is refracted across the boundary and therefore, why a change in the index of refraction occurs.

My conundrum with Gauss’ law in electrostatics If I use Gauss’ law to calculate the electric field outside of a charged (conducting or insulating) sphere or a point charge, the fields are the same. However, as a test approaches a point charge, the field becomes infinite whereas the test charge approaching the surface of the sphere reaches a finite value of Q/4πϵ0R2, where R and Q are the radius and charge of the sphere. Why wouldn’t the field also go to infinity as I get infinitesimally close to the surface of sphere? Those are just “point charges” on the surface. What ultimately led me to this question was trying to physically understand why Gauss’ law predicts the same field for a charge point and charged sphere (obviously r >R).

Answer: there are two ways to see out of this conundrum. 1. A physical way to see the answer is to ask what the test charge sees as it

approaches the surface of the conductor. At first, with a test charge far from the

24.9

charged sphere, it sees the sphere as point-like with a field value of Q/4πϵ0r2, where r is the radius of the Gaussian spherical shell. However, as the test charge gets really close to the surface of a conducting sphere, the surface of the sphere appears as an infinite plane whose field value is

2

20 00

Q Q/4 RE(r R)4 R

σπ→ = = =

π

In other words, as the test charge approaches the conductor’s surface, it’s “field-of-view” is reduced and “sees” less field lines. This compensates for the diminishing influence of any particular piece, giving the electric field a fixed and constant value at the surface. In other words, at really small distances from the surface, the conductor is an infinite plane, as shown above.

2. The "point charge" you approach has an infinitesimal charge (thus cancelling out and

giving a net zero field). There are infinite such points, true, but you don't approach all of them – just one. The rest combine to give a finite value, Q/4πϵ0R2. Gauss law predicts the same field because the net enclosed charge is the same and has the same symmetries for r >R.

You might the follow question: How does the rest of the charges combine to give the finite value of the sphere's surface? Answer: We do not approach the rest of the charges, so the r for them is not zero. Each one is a finite distance away, and as each one is infinitesimal charge the force by each one is infinitesimal. In this case, the infinite sum of infinitesimals converges.

Conducting Spheres

Thin Spherical Shell A thin spherical shell of radius R has a total electric charge Q distributed uniformly over its surface. (a) Derive the E-field using Gauss’s law (show the details) for 0 < r < ∞. (b) Plot E(r) vs. r for 0 < r < ∞.

Solution i. E (r > R), outside the conducting spherical shell The calculation for the electric field outside the shell is identical to the previous one for the insulating solid sphere:

20

(outside)1 Q

r R , r R 4 R

E( )> >π

=

ii. E (0 < r < R), inside the solid insulating sphere Since there is no enclosed charge, there cannot be any flux produced within the shell, and therefore, no electric field exists within the spherical shell.:

enc in insideq 0 0 E 0, r R ( )= →Φ = → = <

One can obtain the same result using Coulomb’s method; however, this type of calculation is rather complicated. Gauss’ law allows us to determine these results considerably easier – no doubt.

Conductors in Electrostatic Equilibrium Good conductors have charges that are NOT bound to any one atom and therefore, charges are free to move within a material. When there is no net motion of charge within a conductor, the conductor is said to be in electrostatic equilibrium. A conductor in such a state has the following properties:

Basic Properties of Conductors

24.10

(i) E = 0 inside a conductor. Why? Because if there were any field, those free charges would move, and it wouldn't be electrostatics anymore (If this was not so, picking up a piece of steel (say, a fork) would have a current flowing through it and could possibility shock you if you somehow tap into this current.). Well ... that's hardly a satisfactory explanation; maybe all it proves is that you can't have electrostatics when conductors are present. We had better examine what happens when you put a conductor into an external E-field E0. Initially, this will drive any free negative charges to the left, but when they depart the right side is left with a net positive charge (due to the stationary nuclei). When they come to the edge of the material, the charges pile up: plus on the right side, minus on the left. Now, these induced charges produce a field of their own, Einduced, which, as you can see from the figure, is in the opposite direction to Eexternal. That's the crucial point, for it means that the field of the induced charges tends to cancel off the original field. Charge will continue to flow until this cancellation is complete, and the resultant field inside the conductor is precisely zero. The whole process is practically instantaneous (10−16 s). This is called a Faraday Cage – a hollow conductor

(ii) ρ = q = 0 inside a conductor. This follows from gauss's law (∇∙E = ρ/ε0). If E = 0, so also is ρ. There is still charge around, but exactly as much plus charge as minus, so the net charge density in the interior is zero.

(iii) If an isolated conductor carries a charge, it resides on its surface. That's the only other place it can be.

I think it is strange that the charge on a conductor flows to the surface. Because of their mutual repulsion, the charges naturally spread out as much as possible, but for all of them to go the surface seems like a waste of the interior space. Surely we could do better, from the point of view of making each charge as far as possible from its neighbors, to sprinkle some of them through the volume ... Well, it simply is not so. You do best to put all of the charge on the surface, and this is true regardless of the size or shape of the conductor. A better way to phrase this problem is in terms of energy. Like any other free dynamical system, the charge on a conductor will seek the configuration that minimizes its potential energy. What property (iii) asserts is that the electrostatic energy of a solid object (with specified shape and total charge) is a minimum when the charge is spread over the surface. For instance, the energy of a sphere if the charge is uniformly distributed over the surface is lower compared to that of a sphere with charge uniformly distributed throughout the volume.

(iv) E is perpendicular to the surface, just outside a conductor. Otherwise, as in (i), charge will immediately flow around the surface until it kills off the tangential component. (Perpendicular to the surface, charge cannot flow, of course, since it is confined to the conducting object.)

The electric field very close to the surface of a charged conductor is perpendicular to the surface and has a magnitude of Esurface = η/ϵ0. Why? Because all of the field lines are pointing in one direction for a conductor and with the infinite plane, they are pointing both directions. So the conductor must have twice the line density as an infinite plane.

24.11

(v) On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is smallest, i.e., sharp points (stated this without proof).

Applications of a Faraday Cage (Electric shielding)

DEMO radio in Faraday cage and telephones in aluminum foil

Youtube (i) Partial vs. Full Faraday cage, (ii) Tesla coil with cage, (iii) Faraday Suit

Examples: MRI, cellphones on trains, cars struck by lightning, and Theater of Electricity in Boston Example 24.2 A solid conducting sphere of radius R1 is concentric with a spherical conducting shell of inner radius R2 and outer R3. The sphere has a charge q1; the shell has a net charge q2 = −3q1. (a) What is the net charge on the inner and outer surface of the shell? (b) Derive the E-field using Gauss’s law (show the details) for 0 < r < ∞. (c) Plot E(r) vs. r for 0 < r < ∞.

Solution a. The inside sphere has charge q1 while the spherical shell

has charge −3q1. The inside solid sphere must have all of its charge q1 on its surface (as already discussed). Charge q1 induces (or pulls in) the oppose charge of −q1 on the inner surface (at R2) of the spherical shell. This then leaves charge −2q1 on the outer surface (at R3).

b. The charge distributions have spherical symmetry and therefore, the electric flux calculations are all identical. Let’s calculate the fluxes starting from the inside (R1 < r < R2) and then move outwards.

iii. E (R1 < r < R2), between the solid sphere and inside surface of the shell Step 1: Pick a good Gaussian The charge distribution of the inside charged sphere has spherical symmetry, so the Gaussian is a spherical shell. Drawing (red circle) a Gaussian of radius r between the sphere (at R1) and the inner surface (at R2). The flux through the Gaussian has E and dA parallel, where cos (0) = +1. The flux integral is therefore

2E sphere Espherical

shellE dA Ecos A E 4 r (spherical shell)Φ = ⋅ = θ ⋅ = + ⋅ π = Φ∫

Step 2: Determine the enclosed charge Since the Gaussian encloses all of the charge of the solid sphere, the magnitude of the enclosed charge is

enc 1 1q q q= =

Remember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian.

24.12

Step 3: Apply Gauss’ law to determine E (R1 < r < R2) Gauss’ law gives

2 1enc 0 1 0 1 2 2sphere

0Step 2Step 1

qE dA q / E 4 r q / E(R r R )4 r

⋅ = →+ ⋅ π = → < < =π∫

iv. E (R2 < r < R3), inside the shell Step 1: Pick a good Gaussian To determine the flux for a Gaussian inside the shell (R2 < r < R3), one draws a Gaussian a distance r (which is different from the previous calculation. The charge distribution is still spherically symmetric; therefore, we can use the previous result for the flux through the Gaussian and write

2sphere

E dA E 4 r⋅ = + ⋅ π∫

Step 2: Determine the enclosed charge Since the Gaussian encloses the negative charge of the inside surface of the shell AND the positive charge of the solid sphere, the magnitude of the enclosed charge is

enc inside solid 1 1 encsurface sphere

q q q q q 0 q= + = − + = =

Once again, the charge on the outside surface of the shell will not contribute to the net flux because it is outside this Gaussian. One immediately concludes that

2 3E(R r R ) 0< < =

Step 3: Apply Gauss’ law to determine E (R1 < r < R2) Gauss’ law gives

2 1enc 0 1 0 1 2 2sphere

0Step 2Step 1

qE dA q / E 4 r q / E(R r R )4 r

⋅ = →+ ⋅ π = → < < =π∫

v. E (r > R3), outside the shell Step 1: Pick a good Gaussian In this case, one draws a Gaussian outside the shell (r > R3), where it encloses everything. Although the calculation of the flux is the same as in the two previous cases, the cosine is now negative since E and dA point in opposite directions: cos(180) = −1. The flux is

2sphere

E dA E 4 r⋅ = − ⋅ π∫

Step 2: Determine the enclosed charge Since the Gaussian encloses everything, the magnitude of the enclosed charge is

enc 1 1 1 1 encq 2q q q 2q q= − − + = = Remember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian.

Step 3: Apply Gauss’ law to determine E (R1 < r < R2) Gauss’ law gives

2 1enc 0 1 0 2 2sphere

0Step 2Step 1

2qE dA q / E 4 r 2q / E(r R )4 r−

⋅ = →− ⋅ π = → > =π∫

24.13

In summary, the E-fields for this configuration are

1

11 2 2

0

2 3

13 2

0

E(r R ) 0qE(R r R )

4 rE(r)

E(R r R ) 02qE(r R )

4 r

< = + < < = π= < < = −

> =π

Example 24.3 A thick spherical shell has uniform volume charge density ρ, inner radius R1 and outer radius R2 = 2R1. (a) Derive the E-field using Gauss’s law (show the details) for 0 < r < ∞. (b) Plot E(r) vs. r for 0 < r < ∞. (c) Numerically calculate the E-field at (i) r = R1/2, (ii) 3R1/2, and (iii) 2R2 where R1 = 10.0 cm and ρ = 1.84 nC/m3.

Solution a. The description of the charge density in the problem statement says that the spherical shell is uniformly charged, which implies it is an insulator. Furthermore, since it is a sphere, the flux part of Gauss’ law will always be the same, and is given by

2sphere

E dA E 4 r⋅ = + ⋅ π∫

The electric field calculation will only depend on the nature of the enclosed charge that is

enc2

0

qE(r)4 r

=π

The focus now is to determine the enclosed charge for each of the three regions.

i. E (r < R1), inside of inner surface Since there is no charge inside of R1, the enclosed charge and E(r < R1) are zero:

enc 1q 0 E(r R ) 0= → < =

ii. E (R1 < r < R2), inside the shell To determine the enclosed charge, we start off with the charge density equation:

total enc encenc total

total enc total

Q q Vconstant q Q V V V

ρ = = = → =

The enclose charge depends on two terms, Vtotal and Venc where Qtotal is fixed. The total volume is calculated by subtracting two volumes:

3 3total 2 1 2 1

43V V(R ) V(R ) )R R(= − = −π

The same applies to the enclosed volume but the two radii are r and R1: 3 3

enc 1 143V V(r) V(R ) )r R(= − = −π

The enclosed charge is then

24.14

encenc total

total

43Vq Q

V= =

π 3 31

43

)r R( −

π

3 31

enc3 33 32 12 1

)Q Q q))

r RR RR R(((

−= =

−−

Now, we can write the electric field for in-between the shells:

3 3enc 1

1 2 2 2 3 30 0 2 1

3 31

1 22 3 30 2 1

q )1E(R r R ) Q4 r 4 r )

)Q E(R r R )4 r )

r RR R

r RR R

((

((

−< < = =

π π −

−= = < <

π −

iii. E (r > R2), outside the shell When the Gaussian is drawn outside the shell, the total enclosed charge is Q, and so the electric field is

22

0

QE(r R )4 r

> =π

In summary, the E-fields for this configuration are

13 3

11 2 2 3 3

0 2 1

2

E(r R ) 0)QE(r) E(R r R )

4 r )E(r R ) 0

r RR R((

< =

−= < < = π − > =

Infinitely Long, Thin Wire A common type of coaxial long thin wires is (i) current carrying wires and (ii) coaxial wires:

Example 24.4 An insulating rod of radius R1 and length L has a uniform volume charge density is inside a thin-walled coaxial conducting cylindrical shell of radius R2 > R1. The net charge on the rod is −4q and on the shell is +2q. (a) Derive the electric field using Gauss’s law (show the details) for 0 < r < ∞. (b) Plot E(r) vs. r for 0 < r < ∞.

Solution How are the charges distributed on the cylinder? If the insulating rod has a net charge of −4q and the thin-walled conducting cylindrical shell is +2q1, the net charge outside the coaxial cable is q(r > R2) = −4q +2q = −2q whereas q(0 < r < R2) is variable because it is insulating.

i. E (r > R2), outside the cylindrical shell Step 1: Picking the Gaussian Surface Symmetry Condition: the charge distribution is cylindrically symmetric, so one chooses a cylindrical Gaussian of arbitrary radius r and arbitrary length L. From the previous chapter (23), a line charge

24.15

has an electric field that is perpendicular to the surface of the wire. A fixed distance r perpendicular to the wire is where the Gaussian is a constant electric field surface. However, there are two area parts to this cylindrical Gaussian, the side walls and the end caps. The electric flux integral is then broken into these two area parts:

E end pieces curved curvedend piecesside walls side walls2 E dA E dAΦ = Φ + Φ = ⋅ ⋅ + ⋅∫ ∫

Dot-Product Condition: looking at the image above, the end pieces have E perpendicular to dA (cos 90° = 0), and therefore, there is NO flux through the end pieces:

end pieces90E dA EdA cos 0

θ= °⋅ = θ = = Φ

The curved side walls have E and dA parallel to each other (cos 180 = −1), so that

cylinder side walls cylinderside wallsE dA E 2 rLΦ = Φ = + − Φ= ⋅ π =∫

Now remember, all cylinders have this very relationship regardless of their charge density.

Step 2: Evaluating the Enclosed Charge qenc Since the Gaussian encloses the whole wire, the enclosed charge is

enc encq 4q 2q 2q q = − + = = Step 3: Applying Gauss’ Law Applying Gauss’ Law gives us

enc 0 0side wallsStep 2Step 1

infinite 2 infinite 2cylinder cylinder0 0

EE dA q / 2 rL 2q/

2q/L 2 E (r R ) E (r R )2 r 2 r

−Φ = ⋅ = → ⋅ π =

− − λ> = = = >

π π

∫

ii. E (R1 < r < R2), in-between the cylinder and cylindrical shell Step 1: Picking the Gaussian Surface The charge distribution inside the insulating cylinder still has cylindrical symmetry, and we have already solved this for a cylindrical Gaussian. Since it has already been done, we have

E E 2 rLΦ = − ⋅ π

Step 2: Determine the enclosed charge Since the Gaussian encloses the whole wire, the enclosed charge is

enc encq 4q 4q q = − = =

Step 3: Apply Gauss’ law to determine E (0 < r < R1) Gauss’ law gives

enc 0 0 1 2sphere0Step 2Step 1

4E dA q / E 2 rL 4q/ E(R r R )2 r− λ

⋅ = →− ⋅ π = → < < =π∫

iii. E (0 < r < R1), outside the cylindrical shell Step 1: Picking the Gaussian Surface The charge distribution inside the insulating cylinder still has cylindrical symmetry, and we have already solved this for a cylindrical Gaussian. Since it has already been done, we just take the result:

24.16

E E 2 rLΦ = − ⋅ π

Step 2: Determine the enclosed charge Inside an insulating cylinder, one must use the enclosed charge equation to determine qenc.

enc total encenc

enc total total

Vconstant

V Vq q q QV

ρ = = = → =

There are two different volumes in this equation: 2 2 2

total enc2 22

totalenc

V VVV

(cylinder) R L r L r 1 for r RR L R(Gaussian) r L

= π π → = = < < π= π

That is, the enclosed charge qenc is less than the total charge Q since the ratio of volumes is less than one. Solving for the enclosed charge gives

2enc

enc enc2total

VV

rq Q Q qR

= = =

Remember that the flux due to the charge on the shell will not contribute to the net flux because this charge is outside this Gaussian.

Step 3: Apply Gauss’ law to determine E (0 < r < R1) Gauss’ law gives

2

enc 0 02 2sphere0Step 2Step 1

r 4E dA q / E 2 rL (4q/ ) E(0 r R) rR 2 R

− λ⋅ = →− ⋅ π = → < < =

π∫

In summary,

20

0

0

1

concentric 1 2cylinders

2 )

4, 0 r R

2 R4

, R r R2 r

4, r R

2 r

r

E

− λ< <

π

− λ< <

π

− λ>

π

=

Application: Geiger counter and concentric cylinders

DEMO Geiger tube

A Geiger tube consists of a tube filled with a low-pressure inert gas. The tube contains two electrodes, between which there is a voltage of 400–600V. The walls of the tube are either metal or have their inside surface coated with a conductor to form the cathode while the anode is a wire passing up the center of the tube. When ionizing radiation strikes the tube, some of the molecules of the fill gas are ionized, either directly by the incident radiation or indirectly by means of secondary electrons produced in the walls of the tube. This creates positively charged ions and electrons, known as ion pairs, in the fill gas. The strong electric field created by the tube's electrodes accelerates the positive ions towards the cathode and the electrons towards the anode. Close to the anode the electrons gain sufficient energy to ionize additional gas molecules, creating an avalanche which is collected by the anode. This is the “gas multiplication” effect.