Physics 121 - Electricity and Magnetism Lecture 04 - Gauss’ Law Y&F Chapter 22 Sec. 1 – 5 Flux Definition (fluid flow example) Gaussian Surfaces Flux of an Electric Field Flux Examples 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

An important idea called Flux (symbol F): Basically a vector field magnitude x area In a fluid, flux measures volume or mass flow per second. Definition: differential amount of flux dF of field v crossing vector area dA v “unit normal” outward and perpendicular to surface dA “Phi” Flux through a closed or open surface S: calculate “surface integral” of field over S Evaluate integrand at all points on surface S EXAMPLE : FLUX THROUGH A CLOSED, EMPTY, RECTANGULAR BOX IN A UNIFORM VELOCITY FIELD No sources or sinks of fluid inside DF from each side = 0 since v.n = 0, DF from ends cancels TOTAL F = 0 Example also applies to gravitational or electric field v What do field lines look like if a flux source or sink (or mass, or charge) is in the box? Can field be uniform? Can net flux be zero.

MORE ON FLUID FLUX: WATER FLOWING ALONG A STREAM Assume: fluid with constant mass density (incompressible) constant flow velocity parallel to banks no turbulence (smooth laminar flow) Flux measures the flow (current): flow means quantity/unit time crossing area Two related fluid flow fields (currents/unit area): Velocity field v is volume flow across area/unit time Mass flow field J is mass flow across area/unit time Flux interpretation = volume of fluid crossing an area per unit time and Chunk of fluid moves distance DL in time Dt: A J v t L m lux f mass V Mass of chunk r o D = º \ Continuity: Net flux (fluid flow) through a closed surface = 0 ………unless a source or sink is inside

Flux is a SCALAR, Units: Nm2/C. Applied to Electric Field: Gaussian Surfaces are closed 3D surfaces Choose surface to match symmetry of the field or charge distribution where possible Field lines cross a closed surface: An odd number of times for charges that are inside S An even number of times for charges outside of S sphere no end caps not closed not a GS cylinder with end caps closed box The flux of electric field crossing a closed surface equals the net charge (source or sink) inside the surface (times a constant). Example: Point charge at center of a spherical “Gaussian surface” Principle holds for arbitrary charges and closed surfaces. Flux is a SCALAR, Units: Nm2/C.

Electric Flux: Integrate electric field over a surface Definition: flux dE crossing (vector) area dA E “unit normal” outward and perpendicular to surface dA Divide up surface S into tiny chunks of dA each and consider one of them 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. Flux through any surface S (closed or open): Integrate on S To do this: evaluate integrand at all points on surface S

DF depends on the angle between the field and chunks of area E.n = 0 ZERO tangent to surface E.n < 0 NEGATIVE from outside to inside of surface E.n > 0 POSITIVE from inside to outside of surface FLUX F IS WHEN E POINTS DF=0

SUM Over Small chunks of surface DA Evaluate flux through closed or open surfaces If field is constant across flat pieces of a surface, sum up A finite number of flux terms instead of integrating: SUM Over Small chunks of surface DA 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.l2 For side 2, F = +E.l2 For the other four sides, F = 0 Therefore, Ftotal = 0 EXAMPLE: Flux through a cube Assume: What if the cube is oriented obliquely?? How would flux differ if net charge is inside??

Flux through a cylinder in a uniform electric field Closed Gaussian surface UNIFORM E means zero enclosed charge Calculate flux directly Symmetry axis along E Break into areas a, b, c Cap a: Cap c: Area b: What if E is not parallel to cylinder axis: Geometry is more complicated...but... Qinside = 0 so F = 0 still !!

Flux of an Electric Field 4-1: Which of the following figures correctly shows a positive electric flux out of a surface element? I. II. III. IV. I and III. q E DA q E DA I. II. III. IV. q E DA q E DA

See Divergence Theorem, Gauss’ Law Statement Let Qenc be the NET charge enclosed by a (closed) Gaussian surface S. The net flux F through the surface is Qenc/e0 See Divergence Theorem, for the ambitious Does not depend on the shape of the surface, assumed closed. Ignore charge outside the surface S. Surface integral yields 0 if E = 0 everywhere on surface Example: Derive Point Charge Formula (Coulomb’s Law) from Gauss’ Law Point charge is at center of spherical Gaussian surface, radius r r Qenc Coulomb’s Law

Shell Theorem: Spherically Symmetric Shells of Charge What are the fields on Gaussian surfaces inside and outside shell? Use Gauss’ Law and spherical symmetry Spherical symmetry makes the flux integral trivial +Q Points Outside Above yields point charge field formula, so field of shell of charge mimics point charge field at locations outside. +Q Points Inside: Above implies E = 0 inside, since Qenc is zero. So field due to shell of charge vanishes everywhere inside the shell.

4-2: What is the flux through the Gaussian surface below? zero -6 C./e0 -3 C./e0 +3 C./e0 not enough info

+ - Example: Flux through surfaces near a dipole FLUX POSITIVE from S1 FLUX NEGATIVE from S2 FLUX = 0 through S4 through S3 + - S1 S2 S3 S4 Equal positive (+Q) and negative (–Q) charges Consider (closed) Gaussian surfaces S1...S4. Surface S1 encloses only the positive charge. The field is everywhere outward. Positive flux. F = +Q/e0 Surface S2 encloses only the negative charge. The field is everywhere inward. Negative flux. F = -Q/e0 Surface S3 encloses no charges. 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. Zero net charge enclosed, so equal flux enters and leaves, zero net flux through the surface. F = 0

Where does net charge reside on an isolated conductor? Charge flows until E = 0 at every interior point (screening) At equilibrium E = 0 everywhere inside a conductor Place a net charge q initially in the interior of a conductor…where… Charges can move, but can not leak off Choose Gaussian surface S just below surface: E = 0 everywhere on it Use Gauss’ Law: 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. Just outside surface, E = s/e0 (E for infinitely large conducting plane)

All net charge on a conductor moves to the outer surface E=0 Gauss’s Law: net charge on an isolated conductor... ...moves to the outside surface 1: SOLID CONDUCTOR WITH NET CHARGE ON IT S All net charge on a conductor moves to the outer surface E=0 2. HOLLOW CONDUCTOR WITH A NET CHARGE, NO CHARGE IN CAVITY There is zero net charge on the inner surface: all net charge is on outer surface S’ Choose Gaussian surface S’ just outside the cavity E=0 everywhere on S’, so the surface integral is zero Gauss’ Law says: E=0 Does this mean that the charge density is zero at each point on the inner surface?

Hollow conductor with cavity, charge inside Electrically neutral, conducting shell Arbitrary charge distribution +Q in cavity Choose Gaussian surface S completely within the conductor + Q - Gaussian Surface S 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… The shell is neutral, so Qouter= -Qinner = +Q appears on outer surface Logic above holds for non-spherical and spherical shells. IF SHELL IS SPHERICAL, field outside is spherically symmetric: Choose another spherical Gaussian surface S’ outside the spherical shell Gauss Law: Whatever the inside distribution may be, outside the shell it is shielded. Qouter is uniform if the shell is spherical. The field then looks like that of a point charge at the center.

Conducting spherical shell with charge inside 4-3: Place a charge inside a cavity in an isolated neutral conductor. Which statement is true? E field is still zero in the cavity. E field is not zero in the cavity, but it is zero in the conductor. E field is zero outside the conducting sphere. E field is the same everywhere as if the conductor were not there (i.e. radial outward everywhere). E field is zero in the conductor, and negative (radially inward) outside the conducting sphere. Spherical cavity + Q Positive charge Conducting sphere (neutral)

Gauss’ Law Example: Find the electric field formula at a distance r from an infinitely long (thin) line of charge z Cylindrical symmetry around z-axis Uniform charge per unit length l along line Every point on the infinite line has the identical surroundings, so…. E is radial, by symmetry E has the same magnitude everywhere on any concentric cylindrical surface Flux through end caps of surface = 0 as E is perpendicular to DA there E on cylinder is parallel to unit vectors for DA So…total flux is just cylinder area x a constant E Trivial integration due to Gauss Law Good approximation for finite line of charge when r << L, far from the ends. Cylinder area = 2prh

Field Lines and Conductors 4-4: The drawing shows cross-sections of three cylinders with different radii, each having the same total charge. Each cylindrical gaussian surface has the same radius (again, shown in cross-section). Rank the three arrangements according to the electric field at the Gaussian surface, greatest first. I, II, III III, II, I All tie. II, I, III II, III, I I. II. III.

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 on the cylinder surface is s. E is like the field of a point charge at the center of the cylinder. E is like the field of a circular ring of charge at its center. E is like the field of an infinite line of charge along the cylinder axis. Cannot tell from the information given. The E field equals zero s

Use Gauss’ Law to find the Electric field near an infinite non-conducting sheet of charge Uniform positive charge per unit area s Choose cylindrical gaussian surface penetrating sheet (rectangular is OK too) E has constant value as a chosen point moves parallel to the surface E points radially away from the sheet (both sides), E is perpendicular to cylindrical tube of gaussian surface.  Flux through it = 0 On both end caps E is parallel to A  F = + EAcap on each 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.

Electric field near an infinite conducting sheet of charge – using Gauss’ Law F=EA F= 0 r Uniform charge per unit area s on one face of sheet E points radially away from sheet outside metal otherwise surface current would flow (!!) Use cylindrical or rectangular Gaussian surface End caps just outside and just inside (E = 0) Flux through the cylindrical tube = 0 as E is normal to surface On left cap (inside conductor) E = 0 so F = 0 On right cap E is parallel to DA so F = EAcap 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 dependence, etc.

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. Each sheet produces a uniform field of magnitude: Oppositely Charged Plates, same |s| but opposite sign s -s - + 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.

Example: Fields near parallel nonconducting sheets - 2 Positively Charged Plates, same |s| same sign +s + 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

Charge on a finite sized but thin conducting plate, isolated or 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’  charges distribute on surfaces Same field on opposite sides of plate, opposite directions Same field if replace plate with 2 charged non-conducting sheets alone: same s1 same cancellation inside conductor reinforcement outside

Proof of the Shell Theorem using Gauss’ Law (Depends on Spherical Symmetry) Hollow shell, 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 S1 S2 R r1 r2 OUTSIDE: (on S2) Shell of charge acts like a point charge at the center of the sphere INSIDE: (on S1) 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 field inside.