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. 5 6 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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. 9 10 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 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. 13 14 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 17 18 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 E4π r = 4 π 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 =− Er4π2 =− 4 π 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 q5 q+ q + q + q q6 q Φ = 1 2 3 4 2 q1 εεε 0 q3 q4 33 34 Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley. 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 εεε 0 rrr rrr 2 - definition of the Flux +++ Φ=∑EdAi i = E ∑ dA i == EA0 E4πππ r i i q Then q 2 E=== k === 4πππ r 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.