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Clicker Question Two Finite Charged Planes are used to accelerate charged particles. A) B) C) Infinite Plane with uniform charge distribution. J.J. Thompson - 1897 1 2 New concept: Electric Flux ΦE through a surface A new technique for calculating E - fields: Gauss's Law Surface with some area A has some electric flux (E - field lines) through it. Define surface vector v A = Anˆ 3 Define surface vector v A = Anˆ 4 v v ΦE = E • A (Vector Dot Product) v v v v Φ E = E • A =| E || A | cos θ A = |A| = magnitude of the area of the surface [m2] ^n = direction of A = direction perpendicular to the surface * In special case of a single flat surface. E A A θ * Note that there are two possible directions for A 5 6 1 Clicker Question A prism-shaped closed surface is in a constant, uniform electric field E, filling all space, pointing right. The 3 rectangular faces of the prism are labeled A, B, and C. Face A is perpendicular to the E-field. The bottom face C is parallel to E. Face B is the leaning face. Which face has the largest magnitude electric flux through it? B A) A B) B C) C A D) A and B have the same C magnitude flux Think of water flowing through a net. Demonstration... E 7 Clicker Question 8 Clicker Question If I have a faucet tap turned on at the marked point, rank order which line has the most to the least water flowing through it per unit time. Assume the faucet has been on for a long time. 1 2 A) 1 > 2 > 3 > 4 B) 1 = 2 > 3 > 4 C) 1=3 > 2 > 4 D) 4 > 3 > 2 > 1 E) none of the above 1=2=3=4 3 4 If I have a faucet tap turned on at the marked point, rank order which line has the most to the least NET water flowing through it per unit time. Assume the faucet has been on for a long time. 3 2 1 4 A) 1 = 2 = 3 = 4 B) 1 > 2 > 3 > 4 C) 1 = 2 = 3 > 4 D) 1 = 3 > 2 > 4 E) none of the above 9 10 One more complication... What if our surface is not flat? 1 2 v v Φ E = ∫ E • da 3 E da Break up surface into many tiny segments of area da, which must be flat in the infinitesimal limit. Integral is over the surface. 4 11 12 2 Clicker Question Gauss's Law v v Q Φ E = ∫ E • da = inside ε0 Circle = integral over a closed surface! For a closed surface, da is always outward. The electric flux thru any closed surface S is a constant (1/ε0) times the net charge enclosed by S. 13 Clicker Question Which surface has the largest magnitude electric flux? -Q +3Q 4 1 3 2 -2Q 14 Gauss's Law can be derived from Coulomb's Law if charges are stationary. Gauss's Law is actually more general. A) 1 = 2 = 3 = 4 B) 1 > 2 = 4 > 3 C) 3 > 2 > 1 > 4 D) 3 > 2 > 1 = 4 E) None of the above 15 Example showing equivalence of two Laws. Consider single charge +q 16 Gauss's Law Coulomb's Law ∫ v kq 1 q | E |= 2 = r 4πε 0 r 2 v v Q E ⋅ da = inside surf ε0 1. Qinside = +q 2. E is always radially outward. 3. da is always radially outward. 4. Thus E.da = |E||da| everywhere! 17 18 3 ∫ v v Q E ⋅ da = inside surf v v ∫ | E || da |= surf v |E| ∫ surf How did that just work? ε0 The key is symmetry ! +q ε0 v +q | da |= ε0 v q | E | (4πr 2 ) = ε0 v | E |= q 4πε 0 r 2 1 19 Clicker Question ∫ v v Q A) E ⋅ da = inside Consider a spherical Gaussian surface ε0 surf with a source charge +q at the center. v v +q If we move the charge slightly to the B) | E || da |= ε0 right, which is the first step that is wrong? surf ∫ v C) | E | ∫ In situations with a high degree of symmetry, Gauss's Law can allow for a quick calculation of the E-field. It is only easy to compute v v if symmetric situation! ∫ E ⋅ da surf v +q | da |= ε0 surf +q 20 v q D) | E | (4πr 2 ) = E) All correct. ε0 21 Example 1: Find the E-field inside a uniform spherical shell of charge (η). + + + + + + + + + + + Exam #1 Information: By symmetry, E must be radial + + + 22 + + v v E =| E | rˆ You must show up on time to the exam. The exam will last 90 minutes. Imagine a Gaussian surface as a sphere of radius r. In this case, E and da are parallel everywhere. This is the symmetry! 23 If you have permission/disability requiring a different time/location for the exam, you will receive an email by this Wednesday letting you know the time and location. If you do not receive this email, let me know ASAP. 24 4 The exam will consist of 20 multiple choice questions. The exam will cover chapters 25, 26, 27. Coulomb's Law, E - field, E - flux, Gauss's Law This includes MP HW #1 - 4, Tutorials #1 - 3 Exam questions will look like clicker questions, parts of MP homework problems, tutorial questions. Formula sheet posted on the web will be included with your exam. Exam is closed book, closed notes, no calculators. 25 ∫ + + + surf + + + + + + + + + + + + + 26 Clicker Question v v Q E ⋅ da = inside ε0 ∫ v v Q E ⋅ da = inside ε0 surf v | E | ∫ da = 0 v | E | (4πr 2 ) = 0 v Electric Field is | E |= 0 zero everywhere + + + + + + + + + + + inside! + + + + + Using Gauss's Law we determined that E=0 inside a sphere with a uniform charge distribution. What if we change the sphere to a cube? Is E=0 inside? A) Yes, E=0 everywhere inside. B) No, E is not zero everywhere inside. 27 Example 2: Find the E-field outside a uniform spherical shell of charge (η). + + + + + + + + + + + + Clicker Question ∫ surf By symmetry, E must be radial + + 28 + + v v E =| E | rˆ + + + ε0 + + In this case, E and da are parallel everywhere. This is the symmetry! 29 v + + Imagine a Gaussian surface as a sphere of radius r. The total charge on the sphere is Q. Using Gauss's Law, find the E-field outside the sphere. v v Q E ⋅ da = inside + + + + + + + + + A) E = Q rˆ 4πε 0 r 2 B) v E = 0 C) v Q E= rˆ 4πε 0 D) None of the above. 30 5 E-Field near an infinite, uniform charged (η) plane of charge. Think about Gauss's Law and gravity. Previously we found Newton's Problem? + ++ ++ ++ ++ ++ ++ ++ + v η | E |= 2ε 0 Now try using Gauss's Law. Key Step - What shape should we use for the imaginary Gaussian surface? 31 Clicker Question A uniform, infinite plane of positive charge creates a uniform Efield. An imaginary Gaussian surface in the shape of a cylinder is shown. The flux through the surface is: 32 ∫ v v E ⋅ da = surf v v v v v v ∫ E ⋅ da + ∫ E ⋅ da + ∫ E ⋅ da endcap1 endcap 2 sideroll v v v v v v ∫ E ⋅ da =| E | | Aendcap1 | + | E || Aendcap1 | +0 surf v v v ∫ E ⋅ da = 2 | E | | πr 2 | surf E v v v Qinside πr 2η 2 π E d a E r ⋅ = 2 | | | | = = ∫ E surf 33 ε0 ε0 v η | E |= 2ε 0 34 Clicker Question Conductors in Electrostatic Equilibrium Consider a cube of side (s) of copper (conductor) in equilibrium. What is the E-field magnitude at any point inside the cube? E = 0 inside a conductor in equilibrium ! A) |E| = +infinite B) |E| = k/s2 C) |E| = 0 D) None of the above E) Cannot determine from information give. Proof: If -E field were not zero, then conduction electrons would feel a force F=qE and will move. Moving charges = not electrostatic equilibrium 35 36 6 Consider any Gaussian surface completely inside the cube. The E-field is zero everywhere inside a conductor at equilibrium, and thus the E-field is zero on all our surfaces! ∫ surf ∫ v v Q E ⋅ da = inside ε0 v v Q E ⋅ da = 0 = inside ε0 surf If we place a net charge Q on a conductor and let it settle to equilibrium, all the excess charge Q must be on the surface ! + + + + + + + + + Qinside = 0 This is a very powerful and interesting feature of equilibrium metals. + + + + + 37 This also tells us that the E-field at the surface of conductors in equilibrium must be only perpendicular to the surface. E 38 Clicker Question An excess charge +Q is put on a copper sphere. We then place a point charge +q outside the sphere. We allow the charges to settle into equilibrium. Think about what an - field parallel to the E surface would do. +q r Van de Graff demonstration. +Q What is the E-field at the center of the sphere? A) |E| = k(q+Q)/r2 B) |E| = kq/r2 C) |E| = kQ/r2 D) |E| = 0 E) None of the above. 39 Clicker Question An excess charge +Q is put uniformly on a plastic (insulating) sphere. We then place a point charge +q outside the sphere. +q r +Q What is the E-field at the center of the sphere? A) |E| = k(q+Q)/r2 B) |E| = kq/r2 C) |E| = kQ/r2 D) |E| = 0 E) None of the above. 41 40 Instead of an infinite sheet of charge, what if we have an infinite block of conductor. All the charge will be on the surface and thus have density η. The electric field is then calculated by Gauss's Law. 42 7 Clicker Question Imagine the Gaussian surface drawn below. Calculate the Efield magnitude a distance (s) away from the conductor surface. A) |E| = η/2ε0 B) |E| = η/ε0 C) |E| = 0 E) none of above D) |E| = (1/4πε0)(η/s2) Just a comment on divergence! "Surface Integrals and the Divergence" Chapter 2, Div, Grad, Curl and All That by H.M. Schey ∫ v v Q E ⋅ da = inside ε0 surf v v ρ ∇⋅E = ε0 43 Equivalent forms of the first of Maxwell's equations. 44 8