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Chapter 22 Gauss’s Law PowerPoint® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Copyright © 2012 Pearson Education Inc – Modified by Scott Hildreth, Chabot College 2016. Goals for Chapter 22 • Use electric field at a surface to determine charge within the surface • Learn meaning of electric flux & how to calculate • Learn relationship between electric flux through a surface & charge within the surface Goals for Chapter 22 • Use Gauss’s law to calculate electric fields • Recognizing symmetry • Setting up two-dimensional surface integrals • Learn where charge on a conductor is located Charge and electric flux • Positive charge within the box produces outward electric flux through the surface of the box. Charge and electric flux • Positive charge within the box produces outward electric flux through the surface of the box. More charge = more flux! Charge and electric flux • Negative charge produces inward flux. • Flux can be negative!! Charge and electric flux • More negative charge – more inward flux! Zero net charge inside a box • Three cases of zero net charge inside a box • No net electric flux through surface of box! 1 Zero net charge inside a box • Three cases of zero net charge inside a box • No net electric flux through surface of box! 2 Zero net charge inside a box • Three cases of zero net charge inside a box • No net electric flux through surface of box! 3 Zero net charge inside a box • Three cases of zero net charge inside a box • No net electric flux through surface of box! 1 2 3 What affects the flux through a box? • Doubling charge within box doubles flux. What affects the flux through a box? • Doubling charge within box doubles flux. • Doubling size of box does NOT change flux. Uniform E fields and Units of Electric Flux For a UNIFORM E field (in space) F= E·A = EA cos(q°) [F] = N/C · m2 = Nm2/C Calculating electric flux in uniform fields Calculating electric flux in uniform fields Calculating electric flux in uniform fields Example 22.1 - Electric flux through a disk Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 103 N/C. What is the flux? Example 22.1 - Electric flux through a disk Disk of radius 0.10 m with n at 30 degrees to E, with a magnitude of 2.0 x 103 N/C. What is the flux? F = E·A = EA cos(30°) A = pr2 = 0.0314 m2 F =54 Nm2/C Example 22.1 - Electric flux through a disk Disk of radius 0.10 m, magnitude E of 2.0 x 103 N/C. What is the flux if n is perpendicular to E? Example 22.1 - Electric flux through a disk Disk of radius 0.10 m, magnitude E of 2.0 x 103 N/C. What is the flux if n is Parallel to E? Electric flux through a cube • An imaginary cube of side L is in a region of uniform E. Find the flux through each side… Electric flux through a cube • An imaginary cube of side L is in a region of uniform E. Find the flux through each side… • Start with easy ones! • f3 & f4 = ? • f5 & f6 = ? • F3,4,5,6 = 0! Electric flux through a cube • An imaginary cube of side L is in a region of uniform E. Find the flux through each side… • Which will be positive? • f2 = POSITIVE + EL2 • F1 = NEGATIVE - EL2 Electric flux through a cube • An imaginary cube of side L is in a region of uniform E. Find the flux through each side… • Which will be positive? • f2 = POSITIVE + EL2 • F1 = NEGATIVE - EL2 • NET flux = 0 • No charge inside!! Electric flux through a cube • An imaginary cube of side L is in a region of uniform E. Find the flux through each side… Electric flux through a cube • An imaginary cube of side L is in a region of uniform E. Find the flux through each side… • Which will be positive? • f2,4 = POSITIVE + • F2 = +EL2 cosq • F4 = +EL2 cos(90-q) • F1,3 = NEGATIVE – Electric flux through a sphere Consider flux through a sphere of radius r around a charge of +q…. r + q Electric flux through a sphere E varies in direction everywhere on the surface, but not magnitude Normal to surface varies in direction everywhere on the surface r + q Electric flux through a sphere Imagine a small segment of area dA; what is flux through that? Electric flux through a sphere Now what is flux across entire surface? F = ∫ E·dA Electric flux through a sphere F= ∫ E·dA E = kq/r2 = 1/(4pe ) 0 q/r2 and is parallel to dA everywhere on the surface F= ∫ E·dA = E ∫dA = EA Electric flux through a sphere F = ∫ E·dA E = kq/r2 and is parallel to dA everywhere on the surface F = ∫ E·dA = E ∫dA = EA For q = +3.0nC, flux through sphere of radius r=.20 m? Gauss’ Law F= ∫ E·dA =qenc S e0 Gauss’ Law F= ∫ E·dA =qenc S Electric Flux is produced by charge in space e0 Gauss’ Law F= ∫ E·dA =qenc S You integrate over a CLOSED surface (two dimensions!) e0 Gauss’ Law F= ∫ E·dA =qenc S E field is a VECTOR e0 Gauss’ Law F= ∫ E·dA =qenc S Infinitesimal area element dA is also a vector; this is what you sum e0 Gauss’ Law F= ∫ E·dA =qenc S Dot product tells you to find the part of E that is PARALLEL to dA at that point (perpendicular to the surface) e0 Gauss’ Law F= ∫ E·dA =qenc S Dot product is a scalar: E·dA = ExdAx + EydAy + EzdAz = |E||dA|cosq e0 Gauss’ Law F= ∫ E·dA =qenc S The TOTAL amount of charge… e0 Gauss’ Law … but ONLY charge inside S counts! F= ∫ E·dA =qenc S e0 Gauss’ Law F= ∫ E·dA =qenc S The electrical permittivity of free space, through which the field is acting. e0 Why is Gauss’ Law Useful? • Given info about a distribution of electric charge, find the flux through a surface enclosing that charge. •Given info about the flux through a closed surface, find the total charge enclosed by that surface. •For highly symmetric distributions, find the E field itself rather than just the flux. Gauss’ Law for Spherical Surface… • Flux through sphere is independent of size of sphere • Flux depends only on charge inside. F= ∫ E·dA = +q/e0 Point charge inside a nonspherical surface As before, flux is independent of surface & depends only on charge inside. Positive and negative flux • Flux is positive if enclosed charge is positive, & negative if charge is negative. Conceptual Example 22.4 • What is the flux through the surfaces A, B, C, and D? Conceptual Example 22.4 • What is the flux through the surfaces A, B, C, and D? FA = +q/e0 Conceptual Example 22.4 • What is the flux through the surfaces A, B, C, and D? FB = -q/e0 FA = +q/e0 Conceptual Example 22.4 • What is the flux through the surfaces A, B, C, and D? FB = -q/e0 FA = +q/e0 FC = 0 ! Conceptual Example 22.4 • What is the flux through the surfaces A, B, C, and D? FB = -q/e0 FA = +q/e0 FC = 0 ! FD = 0 !! Applications of Gauss’s law • Recall from Chapter 21… Under electrostatic conditions, E field inside a conductor is 0! WHY????? E=0 inside! Applications of Gauss’s law • Under electrostatic conditions, E field inside a conductor is 0! • Assume the opposite! IF E field inside a conductor is not zero, then … – E field WILL act on free charges in conductor – Those charges WILL move in response to the force of the field – They STOP moving when net force = 0 – Which means IF static, THEN no field inside conductor! Applications of Gauss’s law • Under electrostatic conditions, any excess charge on a conductor resides entirely on its surface. WHY ??? Applications of Gauss’s law • Consider Gaussian surface inside conductor! Applications of Gauss’s law • Under electrostatic conditions, field outside ANY spherical conductor looks just like a point charge! Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? ·E = ? ·E = ? Charge/meter = l ·E = ? ·E = ? Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? Charge/meter = l ·E = ? • You know charge, you WANT E field (Gauss’ Law!) • Choose Gaussian Surface with symmetry to match charge distribution to make calculating ∫ E·dA easy! Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? Imagine closed cylindrical Gaussian Surface around the wire a distance r away… Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? • Look at ∫ E·dA; remember CLOSED surface means you sum flux over ALL sides! • Three components: the cylindrical side, and the two ends. Need flux across each! Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? •E is orthogonal to dA at the end caps. •E is parallel (radially outwards) to dA on cylinder Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? •E is constant in value everywhere on the cylinder at a distance r from the wire! Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? •E is parallel to dA everywhere on the cylinder, so E ◦ dA = EdA F = ∫ E·dA= ∫ EdA Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? • Integration over curved cylindrical side is twodimensional |dA| = (rdq) dl Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? •E is constant in value everywhere on the cylinder at a distance r from the wire! F = ∫ E·dA = ∫ ∫ E(rdq)dl Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? •E is constant in value everywhere on the cylinder at a distance r from the wire! F = ∫ E·dA = ∫ ∫ (E) · (rdq)dl = E ∫ ∫ rdqdl Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? •Limits of integration? • dq goes from 0 to 2p • dl goes from 0 to l (length of cylinder) F = ∫ E·dA = E ∫ ∫ rdqdl Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? Surface Area of cylinder (but not end caps, since net flux there = 0 Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? •E is constant in value everywhere on the cylinder at a distance r from the wire! F = ∫ E·dA = (E) x (Surface Area!) = E(2pr)l Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? How much charge enclosed in the closed surface? Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? How much charge enclosed in the closed surface? Q(enclosed) = (charge density) x (length) = l l Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? • So… q/e0 = (l l) /e0 • Gauss’ Law gives us the flux • F = E(2pr) l = q/e0 = (l l) /e0 Example 22.6: Field of a line charge • E around an infinite positive wire of charge density l? And… E = (l l) /e0 r = (l / 2pr e0) r Don’t forget E is a vector! (2pr) l Unit vector radially out from line Field of a sheet of charge • Example 22.7 for an infinite plane sheet of charge? Field of a sheet of charge • Example 22.7: What is E field a distance x away for an infinite plane sheet of charge with Q coulombs/sq. meter ? Find Q enclosed to start! Qenc = sA Where s is the charge/area of the sheet Field of a sheet of charge • Example 22.7: What is E field a distance x away for an infinite plane sheet of charge with Q coulombs/sq. meter ? Flux through entire closed surface of the cylinder? F = ∫ E·dA = E1A (left) + E2A (right) (only) Why no flux through the cylinder? Field of a sheet of charge • Example 22.7: What is E field a distance x away for an infinite plane sheet of charge with Q coulombs/sq. meter ? Flux through entire closed surface of the cylinder? F = ∫ E·dA = Qenc / e0 = 2EA = Qenc / e0 = sA / e0 So E = s / 2e0 for infinite sheet Ex. 22.8 Field between two parallel conducting plates • E field between oppositely charged parallel conducting plates. Field between two parallel conducting plates • Superposition of TWO E fields, from each plate… Field between two parallel conducting plates • Superposition of 2 fields from infinite sheets with same charge! For EACH sheet, E = s / 2e0 And s is the same in magnitude, directions of E field from both sheets is the same, so Enet = 2(s / 2e0) = s / e0 Ex. 22.9 - A uniformly charged insulating sphere • E field both inside and outside a sphere uniformly filled with charge. Ex. 22.9 - A uniformly charged insulating sphere • E field both inside and outside a sphere uniformly filled with charge. • E field outside a sphere uniformly filled with charge? EASY • Looks like a point=charge field! A uniformly charged insulating sphere • E field inside a sphere uniformly filled with charge. Find Qenclosed to start! Start with charge density function r r = Qtotal/Volume (if uniform) Nota Bene! NB Density function could be non-uniform! r(r) = kr2 A uniformly charged insulating sphere • E field inside a sphere uniformly filled with charge. Find Qenclosed in sphere of radius r <R? Qenclosed = (4/3 pr3 ) x r A uniformly charged insulating sphere • E field inside a sphere uniformly filled with charge. Find Flux through surface? • E field uniform, constant at any r • E is parallel to dA at surface so… F= ∫ E·dA = E(4pr2) A uniformly charged insulating sphere • E field inside a sphere uniformly filled with charge. F= ∫ E·dA = E(4pr2) = (4/3 pr3 ) x r And E = rr/3e0 (for r < R) or E = Qr/4pR3e0 (for r < R) Ex. 22.9 - A uniformly charged insulating sphere • E field both inside and outside a sphere uniformly filled with charge. Charges on conductors with cavities • E = 0 inside conductor… Charges on conductors with cavities • Empty cavity inside conductor has no field, nor charge on inner surface Charges on conductors with cavities • Isolated charges inside cavity “induce” opposite charge, canceling field inside the conductor! A conductor with a cavity • Conceptual Example 22.11 Electrostatic shielding • A conducting box (a Faraday cage) in an electric field shields the interior from the field. Electrostatic shielding • A conducting box (a Faraday cage) in an electric field shields the interior from the field. See http://www.youtube.com/watch?v=FvtfE-ha8dE