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Physics 121 - Electricity and Magnetism Lecture 04 - Gauss’ Law Y&F Chapter 22 Sec. 1 – 5 • • • • • • • • • • • • • • • Flux Definition (gravitational example) Gaussian Surfaces Flux Examples Flux of an Electric Field Gauss’ Law Gauss’ Law Near a Dipole A Charged, Isolated Conductor Spherical Symmetry: Conducting Shell with Charge Inside Cylindrical Symmetry: Infinite Line of Charge Field near an infinite Non-Conducting Sheet of Charge Field near an infinite Conducting Sheet of Charge Conducting and Non-conducting Plate Examples Proof of Shell Theorem using Gauss Law Examples Summary 1 Copyright R. Janow – Spring 2015 FLUID FLUX EXAMPLE: WATER FLOWING ALONG A STREAM Simplifying Assumptions: • • • • constant mass density r incompressible fluid constant flow velocity parallel to banks no turbulence (laminar flow) n̂' n̂ A 2 Flux measures the (current) flow: • amount/unit time flowing across an area • volume flow rate or mass flow rate past point A1 2 related fluid flow fields (currents/unit area): • velocity v represents volume flow/unit area/unit time • J = mass flow/unit area/unit time A n̂A n̂ are outward unit vectors to vector area A J rv Flux = amount of field crossing an area per unit time (field x area) V L A L vt The chunk of volume flux v A fluid moves L t t A in time t: r r mass of solid chunk m V L A v m L and \ mass flux r r A v A J A t t Through a closed surface, net flux (fluid flow) = 0 ………unless a source or drain is inside surface Copyright R. Janow – Spring 2015 Flux (symbol F): A vector field magnitude x area Applies to fluid flow, or gravitational, electric, or magnetic field Example: Gravitational flux, field ag, take component normal to area dA “unit normal” ag n̂ outward and perpendicular to surface dA dF g flux of ag through dA ag n̂dA (a scalar) Dimensions for gravitation flux: N.m Choose a closed or open surface S and integrate: - Involves doing “surface integral” of field over S FS dF a g n̂dA S S - Evaluate integrand at all points on surface S EXAMPLE : GRAVITATIONAL FLUX THROUGH A CLOSED IMAGINARY BOX (UNIFORM ACCELERATION FIELD) • No mass inside the box • F from each side = 0 since a.n = 0, F from ends cancels • TOTAL F = 0 • Example could also apply to fluid flow n̂ n̂ n̂ ag n̂ Could field still be uniform if a mass (flux source) is in the box? Could net flux be zero? Copyright R. Janow – Spring 2015 Gaussian Surfaces are closed 3D surfaces • Field lines cross a closed surface: • An odd number of times sphere for charges that are inside S • An even number of times for charges outside of S • Choose surface to match symmetry of the field or charge distribution where possible closed box cylinder with end caps no end caps not closed not a GS The flux of electric field crossing a closed surface equals the net charge inside the surface (times a constant). Example: Point charge at center of a spherical “Gaussian surface” Q 2 Flux FE Field x Surface Area E x S x 4 R Q /0 4 0R 2 Principle holds for arbitrary charges and closed surfaces? Flux is a SCALAR, Units: Nm2/C. Copyright R. Janow – Spring 2015 Electric Flux: Integrate electric field over a surface Definition: flux dE crossing (vector) area dA “unit normal” E n̂ outward and perpendicular to surface dA Divide up surface S into tiny chunks of dA each and consider one of them dA n̂dA dFE flux of E through dA E n̂dA (a scalar) Flux through surface S (closed or open): Integrate on S FE dFE S E n̂dA To do this: evaluate integrand at all points on surface S S Gauss’ Law: The flux through a closed surface S depends only on the net enclosed charge, not on the details of S or anything else. Copyright R. Janow – Spring 2015 F depends on the angle between the field and chunks of area F E n̂A EAcos() A n̂A n̂ outward unit vector F>0 F<0 WHEN E POINTS FLUX F IS from inside to outside of surface E.n > 0 POSITIVE from outside to inside of surface E.n < 0 NEGATIVE tangent to surface E.n = 0 ZERO F=0 Copyright R. Janow – Spring 2015 Evaluating flux through closed or open surfaces Special case: field is constant across pieces of the surface FE E n̂A (a scalar) Units of FE : Nm2 / C FE FE E n̂A SUM F FROM SMALL CHUNKS OF SURFACE A small areas EXAMPLE: Flux through a cube Assume: • • • • small areas Uniform E field everywhere E along +x axis Cube faces normal to axes Each side has area length2 • Field lines cut through two surface areas and are tangent to the other four surface areas • For side 1, F = -E.l 2 • For side 2, F = +E.l 2 • • For the other four sides, F = 0 Therefore, Ftotal = 0 \ FE Fi 0 i1, 6 • What if the cube is oriented obliquely?? • How would flux differ if net charge is inside?? Copyright R. Janow – Spring 2015 Flux of a uniform electric field through a cylinder • Closed Gaussian surface • Uniform E means zero enclosed charge • Calculate flux directly • Symmetry axis along E • Break into areas a, b, c F tot E dA F a Fb F c a ,b ,c Cap a: E n̂ - E, Cap c: E n̂ E, cos() - 1 \ Fa - E dA - E Acap cap a cos() 1 \ Fc E dA E Acap cap c Area b: E n̂ 0 (E A) everywhereon b \ Fb 0 \ F tot 0 What if E is not parallel to cylinder axis: • Geometry is more complicated...but... • Qinside = 0 so F = 0 still !! Copyright R. Janow – Spring 2015 Flux of an Electric Field 4-1: Which of the following figures correctly shows a positive electric flux out of a surface element? A.I. B.II. C.III. E D.IV. I. II. E.I and III. A A III. E IV. E A E A Copyright R. Janow – Spring 2015 Statement of Gauss’ Law Let Qenc be the NET charge enclosed by a (closed) Gaussian surface S. The net flux F through the surface is Qenc/0 Q F E dA enc 0 Surface Q enc 0F See Divergence Theorem, for the ambitious • Does not depend on the shape of the surface, assumed closed. • Charge outside the surface S can be ignored. • Surface integral yields 0 if E = 0 everywhere on surface Example: Derive Coulomb’s Law from Gauss’ Law Assume a point charge at center, r on spherical Gaussian surface E dA E n̂dA is always just EdA E dA E dA because E and n̂ are always radial, E is constant on S Q enc 2 F E dA E 4 r 0 S r Qenc \ E(r ) Q enc 4 0 r 2 Coulomb’s Law Copyright R. Janow – Spring 2015 Shell Theorem: Spherical Shells of Charge Spherically Symmetric Shell P Outside +Q What are the fields on Gaussian surfaces inside and outside? +Q P Inside Q F E dA enc 0 S 4-2: What is the flux through the Gaussian surface below? A. B. C. D. E. zero -6 C./0 -3 C./0 +3 C./0 not enough info Q F E dA enc 0 S Copyright R. Janow – Spring 2015 Flux through surfaces near a dipole FLUX POSITIVE from S1 • Equal positive (+Q) and negative (–Q) charges • Consider Gaussian (closed) surfaces S1...S4. • Surface S1 encloses only the positive charge. The field is everywhere outward. Positive flux. F = +Q/0 • Surface • Surface S3 encloses no charges. • + FLUX = 0 S2 encloses only the negative through S4 S1 charge. The field is everywhere inward. Negative flux. F = -Q/0 Net flux through the surface is zero. The flux is negative at the upper part, and positive at the lower part, but these cancel. F = 0 Surface S4 encloses both charges. S4 S3 FLUX = 0 through S3 S2 - Zero net charge enclosed, so equal flux enters and leaves, zero net flux through the surface. F = 0 FLUX NEGATIVE from S2 Copyright R. Janow – Spring 2015 Where does net charge reside on an isolated conductor? • Place net charge Q initially in the interior of a conductor • Charges are free to move, but can not leak off F QE • At equilibrium E = 0 everywhere inside a conductor • Charge flows until E = 0 at every interior point E = 0 • Choose Gaussian surface just below surface: E = 0 everywhere on it • Use Gauss’ Law! Q F Surface E dA 0 enc 0 Q enc 0 • The only place where un-screened charge can end up is on the outside surface of the conductor • E at the surface is everywhere normal to it; if E had a component parallel to the surface, charges would move to screen it out. Copyright R. Janow – Spring 2015 Gauss’s Law: net charge on an isolated conductor... ...moves to the outside surface F E dA Surface 1: SOLID CONDUCTOR WITH NET CHARGE ON IT Qenc 0 All net charge on a conductor moves to the outer surface E=0 S 2. HOLLOW CONDUCTOR WITH A NET CHARGE, NO CHARGE IN CAVITY E=0 S’ • Choose Gaussian surface S’ just outside the cavity • E=0 everywhere on S’, so the surface integral is zero • Gauss’ Law says: Q FS' 0 enc \ Q 0 enc 0 There is zero net charge on the inner surface: all net charge is on outer surface Does this mean that the charge density is zero at each point on the inner surface? Copyright R. Janow – Spring 2015 Hollow conductor with cavity, charge inside • Electrically neutral conducting shell, for example • Arbitrary charge distribution +Q in cavity • Choose Gaussian surface S completely within the conductor q F S E dA S enc + + - - ++ + Q + + - + + 0 • E = 0 everywhere on S, so FS = 0 and qenc= 0 • Negative charge Qinner is induced on inner surface, distributed so that E = 0 in the metal, hence… qenc 0 Q Qinner + + - - - + - - + + + Gaussian Surface Qinner -Q • The shell is neutral, so Qouter= -Qinner = +Q appears on outer surface S Whatever the inside distribution may be, outside the shell it is shielded. Qouter is uniform because the shell is spherical, so the field looks like that of a point charge at the center. OUTSIDE SPHERICAL SHELL: • Choose another spherical Gaussian surface S’ outside the spherical shell • Gauss Law: F q'enc 0 Q 4r 2E(r ) 0 E(r ) Q 4 0r 2 radially outward Copyright R. Janow – Spring 2015 Conducting spherical shell with charge inside 4-3: Place a charge inside a cavity in an isolated neutral conductor. Which statement is true? Spherical cavity ++ + Q + + + Conducting sphere (neutral) Positive charge A. E field is still zero in the cavity. B. E field is not zero in the cavity, but it is zero in the conductor. C. E field is zero outside the conducting sphere. D. E field is the same everywhere as if the conductor were not there (i.e. radial outward everywhere). E. E field is zero in the conductor, and negative (radially inward) outside the conducting sphere. Q F E dA enc 0 S Copyright R. Janow – Spring 2015 Gauss’ Law Example: Find the electric field formula at a distance r from an infinitely long (thin) line of charge z lh qenc • Cylindrical symmetry around z-axis • Uniform charge per unit length l • Every point on the infinite line has the identical surroundings, so…. • E is radial, by symmetry A • E has the same value everywhere on any concentric cylindrical surface • Flux through end caps = 0 as E is perpendicular to A • E on cylinder is parallel to unit vector for A lh \ F E dA 2rhE 0 S E(r ) l 2 0r radially outward Cylinder area = 2rh • No integration needed now due to Gauss Law • Good approximation for finite line of charge when r << L, far from the ends. Copyright R. Janow – Spring 2015 Field Lines and Conductors 4-4: The drawing shows cross-sections of three cylinders with different radii, each with the same total charge. Each cylindrical gaussian surface has the same radius (again, shown in crosssection). Rank the three according to the electric field at the gaussian surface, greatest first. A. B. C. D. E. Q F E dA enc 0 S I, II, III III, II, I All tie. II, I, III II, III, I I. II. III. Copyright R. Janow – Spring 2015 Use symmetry arguments where applicable 4-5: Outside a sphere of charge the electric field is just like that of a point charge of the same total charge, located at its center. Outside of an infinitely long, uniformly charged conducting cylinder, which statement describes the electric field? The charge per unit area is s. A. B. C. D. E. Like that of a point charge at the center of the cylinder. Like a circular ring of charge at its center. Like an infinite line of charge along the cylinder axis. Cannot tell from the information given. The field equals zero Q F E dA enc 0 S s Copyright R. Janow – Spring 2015 Use Gauss’ Law to find the Electric field near an infinite non-conducting sheet of charge qenc sA cap • Cylindrical gaussian surface penetrating sheet (rectangular OK too) • Uniform positive charge per unit area s • E has constant value as chosen point moves parallel to the surface • E points radially away from the sheet (both sides), • E is perpendicular to cylindrical part of gaussian surface. Flux through it = 0 • On both end caps E is parallel to A so F = + EAcap on each qenc sAcap F E dA 2EAcap 0 0 S s \ E 2 0 • Uniform field – independent of distance from sheet • Same as near field result for charged disk but no integration needed • Good approximation for finite sheet when r << L, far from edges. Copyright R. Janow – Spring 2015 Electric field near an infinite conducting sheet of charge – using Gauss’ Law qenc sA cap • Uniform charge per unit area s on one face • Second face is immersed in the metal • Use cylindrical or rectangular Gaussian surface • End caps just outside and just inside (E = 0) • E points radially away from sheet outside otherwise surface current would flow (!!) • Flux through the cylindrical tube = 0 E normal to surface • On left cap (inside conductor) E = 0 so F = 0 • On right cap E is parallel to A so F = EAcap \ F EAcap s r qenc sAcap 0 0 \ E s 0 F= 0 • Field is twice that for a non-conducting sheet with same s • Same enclosed charge, same total flux now “squeezed” out the right hand cap, not both • Otherwise like previous result: uniform, no r Copyright dependence, R. Janow –etc. Spring 2015 F=EA Example: Fields near parallel nonconducting sheets - 1 • • • • • Two “large” nonconducting sheets of charge are near each other, Use infinite sheet results The charge cannot move, use superposition. There is no screening, as there would be in a conductor. s Each sheet produces a uniform field of magnitude: E 0 2 0 Oppositely Charged Plates, same |s| s -s + E0 + E0 + Etot 0 • • • + Etot 2 E0 î - - E0 E0 Etot 0 Left region: Field due to the positively charged sheet is canceled by the field due to the negatively charged sheet. Etot is zero. Right region: Same argument. Etot is zero. Between plates: Fields reinforce. Etot is twice E0 and to the right. Copyright R. Janow – Spring 2015 Example: Fields near parallel nonconducting sheets - 2 Positively Charged Plates, same s + + + + E0 + + Etot -2 E0 î • • • s 2 0 s s E0 E0 E0 E0 + + Etot 0 Etot 2 E0 î Now, the fields reinforce to the left and to the right of both plates. Between plates, the fields cancel. Signs are reversed for a pair of negatively charged plates Copyright R. Janow – Spring 2015 Charge on a finite sized conducting plate in isolation L S’ S s1 E 0 or L S • E = 0 inside the conductor • Charge density s1 becomes the same on both faces, o/w charges will move to make E = 0 inside • Cylindrical Gaussian surfaces S, S’ • S charges distribute on surfaces S’ • Same field magnitude on opposite sides of each conductor, opposite directions • Same field by replacing each conductor with 2 charged non-conducting sheets alone: • same s1 • cancellation inside conductor • reinforcement outside Copyright R. Janow – Spring 2015 Proof of the Shell Theorem using Gauss’ Law (spherical symmetry) • Hollow shell of charge, net charge Q • Spherically symmetric surface charge density s, radius R • Two spherical Gaussian surfaces: - S1 is just inside shell, r1 < R - S2 is outside shell, r2 > R • qenc means charge enclosed by S1 or S2 OUTSIDE: (on S2) R r2 r1 S1 q Q F 2 E dA 4r22E2 enc 0 0 S 2 \ E2 INSIDE: (on S1) S2 Q 4 0r22 for r2 R Shell of charge acts like a point charge at the center of the sphere qenc 2 F1 E dA 4r1 E1 0 0 S 1 \ E1 0 for r1 R Shell of charge creates zero field inside (anywhere) At point r inside a spherically symmetric volume charge distribution (radius R): • only the shells of charge with radii smaller than r contribute as point charges • shells with radius between r and R produce zero fieldR. inside. Copyright Janow – Spring 2015