Chapter 28. Gauss's

Home , Electric flux, Gaussian surface

Chapter 28. Gauss’s Law The nearly spherical shape Chapter 28. Gauss’s Law of the girl’s head determines Topics: the electric field that causes her hair to stream outward. • Symmetry Using Guass’s law, we can • The Concept of Flux deduce electric fields, • Calculating Electric Flux particularly those with a • Gauss’s Law high degree of symmetry, • Using Gauss’s Law simply from the shape of the charge distribution. • Conductors in Electrostatic Equilibrium Chapter Goal: To understand and apply Gauss’s law. 1 2 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

The amount of electric field passing through a surface is called Chapter 28. Reading Quizzes A. Electric flux. B. Gauss’s Law. C. Electricity. D. Charge surface density. E. None of the above.

3 4 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The amount of electric field Gauss’s law is useful for calculating passing through a surface is electric fields that are called A. symmetric. A. Electric flux. B. uniform. B. Gauss’s Law. C. due to point charges. C. Electricity. D. due to continuous charges. D. Charge surface density. E. None of the above.

Gauss’s law is useful for calculating Gauss’s law applies to electric fields that are

A. lines. A. symmetric. B. flat surfaces. B. uniform. C. spheres only. C. due to point charges. D. closed surfaces. D. due to continuous charges.

7 8 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The electric field inside a Gauss’s law applies to conductor in electrostatic equilibrium is A. lines. B. flat surfaces. C. spheres only. D. closed surfaces. A. uniform. B. zero. C. radial. D. symmetric.

The electric field inside a conductor in electrostatic equilibrium is

Chapter 28. Basic Content and Examples

A. uniform. B. zero. C. radial. D. symmetric.

11 12 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Symmetry and Electric Field The Concept of Flux We say that a charge distribution is symmetric if there are a group of geometric transformations that do not cause any physical change.

An electric field passing through a The Concept of Flux surface

15 16 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The electric Flux of a constant electric The electric Flux of a Nonuniform field Electric Field r r Φe = ∑δΦi = ∑ Ei ⋅(δ A)i i i

r r Φe = ∫ E ⋅dA surface

Electric Flux The Flux Through a Curved Surface Definition: • Electric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the field

• ΦE = EA • The field lines may make some angle θ with the perpendicular to the surface θ rrr • Then ΦE = EA cos E

normal θθθ

rrr ΦE = EA θθθ E

19 Φ = EA cos θθθ 20 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.E Electric Flux Electric Flux

Definition: • In the more general case, look at a • Electric flux is the scalar productrrr of electric small flat area element field and the vector A rrr r r rrr rrr rrr θ A E ∆Φ=∆EE ii A cos i =E i ⋅∆ A i • Φ = EA θθθ • In general, this becomes θθθ rr rr

Φ=Elimr ∑E i ⋅∆= A i EdA ⋅ rrr rrr ∆A → 0 ∫ i surface Φ==E EA EA cosθθθ > 0 or rrr E • The surface integral means the rrr θθθ A integral must be evaluated over the θθθ surface in question • The units of electric flux will be rrr rrr . 2 21 N m /C 22 Φ=E EA =− EA cosθθθ < 0 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Electric Flux: Closed Surface Electric Flux: Closed Surface rrr rrr • A positive point charge, q, is E has the same direction as A at every point. located at the center of a sphere of q E=== k e radius r r 2

• The magnitude of the electric field Spherical Then Spherical rrr rrr everywhere on the surface of the surface surface Φ=∑Ei dA i = E ∑ dA i = sphere is i i 2 E = keq / r 2 2 q ==EA0 E4 ππ r = 4 rk e 2 = • Electric field is perpendicular to the r q surface at every point, so Gauss’s Law rrr rrr =4πππ ke q = εεε E has the same direction as A 0 at every point. 1 ΦΦΦ does not depend on r BECAUSE E ∝∝∝ 2 23 24 r Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Electric Flux: Closed Surface rrr rrr EXAMPLE 28.1 The electric flux inside a E and A have opposite directions at every point. parallel-plate capacitor |q | E=== k QUESTION: e r 2 Spherical Then surface rrr rrr Φ=∑Ei dA i =− E ∑ dA i = i i |q | −−− =−EA =− Er π4π 2 =− 4 rk 2 = 0 e r 2 q Gauss’s Law = −4πππ ke | q | = εεε 0 ΦΦΦ does not depend on r ONLY BECAUSE 1 E ∝∝∝25 2 26 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. r Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

EXAMPLE 28.1 The electric flux inside a EXAMPLE 28.1 The electric flux inside a parallel-plate capacitor parallel-plate capacitor

27 28 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Gauss’s Law EXAMPLE 28.1 The electric flux inside a parallel-plate capacitor Gauss’s law states r r q Φ=E ⋅ dA = in E ∫ ε o qin is the net charge inside the surface

E is the total electric field and may have contributions from charges both inside and outside of the surface rrr

Ai rrr q E q rrr E rrr Ai q q Φ = Φ = 29 εεε 30 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education,εεε Inc.,0 publishing as Pearson Addison-Wesley. 0

What is the electric flux here? Gauss’s Law:

Gauss’s law states r r q r r in qin Φ=E ∫ E ⋅ dA = Φ=E ⋅ dA = ε Gauss’s law states E ∫ ε o o

qin is the net charge inside the surface q is the net charge inside the surface E is the total electric field and may have contributions in from charges both inside and outside of the surface E is the total electric field and may have contributions from charges both inside and outside of the surface q5 q5 q q1 q 2 2 Φ = 0 Φ = 0 q

q3 −−−q

q4 q4

q6 q6 q7 q7 31 32 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Chapter 28 Gauss’s Law: Applications

Although Gauss’s law can, in theory, be solved to find E for any charge configuration, in practice it is limited to symmetric situations r r q To use Gauss’s law, you want to choose a Gaussian Φ=E ⋅ dA = in ∫ ε surface over which the surface integral can be simplified o and the electric field determined Gauss’s Law: Applications Take advantage of symmetry Remember, the gaussian surface is a surface you choose, it does not have to coincide with a real surface

q+ q + q + q q6 q Φ = 1 2 3 4 2 q1 εεε0 q3

r r q Gauss’s Law: Point Charge Gauss’s Law: Applications Φ=E ⋅ dA = in ∫ ε rrr o E • Try to choose a surface that satisfies one or more of these conditions: +++ – The value of the electric field can be argued from symmetry to be q constant over the surface SYMMETRY: – The dot product of E.dA can be expressed as a simple algebraic product rrr Gaussian Surface – Sphere EdA because E and dA are parallel E - direction - along the radius rrr Only in this case the magnitude of – The dot product is 0 because E and dA are perpendicular E - depends only on radius , r electric field is constant on the – The field can be argued to be zero over the surface Gaussian surface and the flux can be easily evaluated q correct Gaussian surface wrong Gaussian surface Φ = - Gauss’s Law εεε0 rrr rrr 2 - definition of the Flux +++ Φ=∑EdAi i = E ∑ dA i == EA0 E4πππ r i i q Then q 2 E=== k === 4πππr E e 2 35 36 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.r Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. εεε0 Gauss’s Law: Applications Gauss’s Law: Applications

Spherically Symmetric Charge Distribution Spherically Symmetric Charge Distribution rrr The total charge is Q SYMMETRY: rrr E Q Q rrr E= = k e The electric field is the same as ∆∆∆A 4πε r2 r 2 E - direction - along the radius o for the point charge Q !!!!! rrr E - depends only on radius , r Q a For r > a Q • Select a sphere as the gaussian ≡≡≡ surface • For r >a

r r q Q Φ=⋅=E dA EdA =4πr2 E ==in E ∫ ∫ ε ε o o Q Q For r > a E= = k πε 2e 2 The electric field is the same as ≡≡≡ 4 or r for the point charge Q 37 38 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

Gauss’s Law: Applications Gauss’s Law: Applications

Spherically Symmetric Charge Distribution rrr Spherically Symmetric Charge Distribution SYMMETRY: rrr E rrr ∆∆∆A E - direction - along the radius • Inside the sphere, E varies rrr linearly with r E - depends only on radius , r E → 0 as r → 0 • Select a sphere as the gaussian • The field outside the sphere is surface, r < a equivalent to that of a point Q4 r 3 3 charge located at the center of qin =πππ rQQ =3 < 4 3 3 a πππ a the sphere 3 r r q Φ=⋅=E dA EdA =4πr2 E = in E ∫ ∫ ε o q Qr3 1 Q E=in = k = kr πε 2e 32 e 3 4 or ar a 39 40 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Gauss’s Law: Applications Gauss’s Law: Applications

Field due to a thin spherical shell Field due to a thin spherical shell • Use spheres as the gaussian surfaces • When r < a, the charge inside the surface is 0 and E = 0 • When r > a, the charge inside the surface is Q and

2 ∆∆∆A1 ∆ =σσσ ∆ ∆ =σσσ ∆ E = keQ / r q1 A 1 q2 A 2 • When r < a, the charge inside the surface is 0 and E = 0 2 2 r1 ∆A1 = r 1 ∆Ω ∆A2 = r 2 ∆Ω rrr 2 ∆∆∆E2 ∆q σσ ∆ A r ∆Ω ∆=Ek1 = k 1 = k 1 =∆Ω k σσσ the same 1 eer2 r 2 e r 2 e rrr 1 1 1 solid angle ∆∆∆E1 2 ∆q2 σσ ∆ A 2 r 2 ∆Ω ∆=Ek2 ee2 = k 2 = k e 2 =∆Ω k e σσσ r2 r 2 r 2 r2 ∆E = ∆ E 1 2 r r ∆E1 +∆ E 2 = 0

1 41 ∆∆∆A Only because in Coulomb law E ∝∝∝ 42 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc.,2 publishing as Pearson Addison-Wesley. r 2

Gauss’s Law: Applications Gauss’s Law: Applications Field from a line of charge Field from a line of charge

• Select a cylindrical Gaussian surface rrr dA The flux through this surface is 0 – The cylinder has a radius of r and a length of ℓ • Symmetry: The flux through this surface: E is constant in magnitude (depends only r r q Φ=⋅=E dA EdA = E ()2πrl = in on radius r) and perpendicular to the E ∫ ∫ ε surface at every point on the curved part o q = λl of the surface in λl E ()2πrl = ε o λ λ E= = 2 k πε e 43 2 or r 44 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The end view CopyrightThe ©end 2008 viewPearson Education, Inc., publishing as Pearson Addison-Wesley. Gauss’s Law: Applications Gauss’s Law: Applications Field due to a plane of charge rrr Field due to a plane of charge E 2 rrr • Symmetry: A2 rrr rrr E must be perpendicular to the plane and A4 A h 5 rrr must have the same magnitude at all rrr dA dA points equidistant from the plane rrr rrr h A A6 • Choose a small cylinder whose axis is 3 rrr perpendicular to the plane for the A rrr 1 gaussian surface E1

E1= E 2 = E A1= A 2 = A rr r r The flux through this surface is 0 Φ=EA + EA = EA + EA = 2 EA The flux through this surface is 0 11 22

qin σσσA σσσA σσσ Φ = = 2EA === E === does not depend on h 45 εε εεε 2εεε 46 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright0 © 2008 Pearson 0 Education, Inc., publishing0 as Pearson Addison-Wesley.0

Gauss’s Law: Applications Chapter 28

Conductors in Electric Field

σσσ E === 2εεε0

λλλ E=== 2 k e r 47 48 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Electric Charges: Conductors and Isolators Electrostatic Equilibrium

Electrical conductors are materials in which Definition: some of the electrons are free electrons when there is no net motion of charge These electrons can move relatively freely within a conductor, the conductor is said to through the material Examples of good conductors include copper, be in electrostatic equilibrium aluminum and silver Because the electrons can move freely through the Electrical insulators are materials in which all material r r F=== qE of the electrons are bound to atoms no motion means that there are no electric forces These electrons can not move relatively freely no electric forces means that the electric field through the material inside the conductor is 0 Examples of good insulators include glass, rubber and wood rrr If electric field inside the conductor is not 0, ≠ then r r E ≠≠ 0 Semiconductors are somewhere between there is an electric force F === qE and, from the second 49 50 insulatorsCopyright © 2008 and Pearson conductorsEducation, Inc., publishing as Pearson Addison-Wesley. Newton’sCopyright © 2008law, Pearson there Education, is a Inc., motion publishing of as Pearson free Addison-Wesley.electrons.

Conductor in Electrostatic Equilibrium Conductor in Electrostatic Equilibrium

• The electric field is zero everywhere inside the • If an isolated conductor carries a charge, the charge resides conductor on its surface

• Before the external field is applied, free Electric filed is 0, electrons are distributed throughout the so the net flux through conductor Gaussian surface is 0 • When the external field is applied, the r r q electrons redistribute until the magnitude Φ=E ⋅ dA = in ∫ ε of the internal field equals the magnitude o of the external field Then q = 0 • There is a net field of zero inside the in conductor

51 52 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Conductor in Electrostatic Equilibrium Conductor in Electrostatic Equilibrium

• The electric field just outside a charged conductor is σ σ > 0 E = perpendicular to the surface and has a magnitude of σ/ε ε o o

• Choose a cylinder as the gaussian surface σ < 0 • The field must be perpendicular to the surface E = 0 – If there were a parallel component to E, charges would experience a force and accelerate along the surface and it would not be σ > 0 σ < 0 in equilibrium σ < 0 • The net flux through the gaussian surface is through only the flat face outside the conductor – The field here is perpendicular to the surface • Gauss’s law: σA σ Φ==EA and E = E ε ε 53 54 Copyright © 2008 Pearson Education, Inc., publishing aso Pearson Addison-Wesley. o Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.

General Principles

Chapter 28. Summary Slides

55 56 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. General Principles Important Concepts

Important Concepts Important Concepts

Chapter 28. Questions

This box contains This box contains

A. a net positive charge. A. a net positive charge. B. a net negative charge. B. a net negative charge. C. a negative charge. C. a negative charge. D. a positive charge. D. a positive charge. E. no net charge. E. no net charge.

63 64 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. The total electric flux through this box is The total electric flux through this box is A. 6 Nm 2/C. A. 6 Nm 2/C. B. 4 Nm 2/C. B. 4 Nm 2/C. C. 2 Nm 2/C. C. 2 Nm 2/C. D. 1 Nm 2/C. D. 1 Nm 2/C. E. 0 Nm 2/C. E. 0 Nm 2/C.