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

Flux itself is a SCALAR, Units: Nm2/C. Electric Flux (F): Integrate electric field over a surface Basically, flux is a vector field x Area Definition: dF is flux of electric field E crossing vector area dA E “unit normal” outward and perpendicular to surface dA Divide up a finite sized surface S into tiny chunks of dA each and consider flux through one of them: Flux itself is a SCALAR, Units: Nm2/C. Gauss’ Law: The flux of E through a closed surface S depends only on the net enclosed charge, not on the details of S or anything else. To find the flux through a finite surface integrate over closed or open surface S to find Flux FE : To do this evaluate integrand at all points on surface S

Example: Point charge at center of a spherical “Gaussian surface” Definition: 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.

Total Flux: SUM over chunks of surface DA EXAMPLE: flux through a cube If field is constant across flat surface sections, sum up a finite number of flux terms instead of integrating: Total Flux: SUM over chunks of surface DA Uniform E field everywhere, along x Cube faces are normal to axes Each side has area L2 Field lines cut through surface areas 1 & 2 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 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 Symmetry axis along E Uniform E means zero enclosed charge 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

Statement of Gauss’ Law See Divergence Theorem if you are mathematically advanced Qenc is the NET charge enclosed by a (closed) Gaussian surface S. The net flux F through the surface is Qenc/e0 F 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

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: Derive point charge formula (Coulomb’s Law) from Gauss’ Law Place point charge Q at center of spherical Gaussian surface, radius r. r Qenc Symmetry allows simplification: Coulomb’s Law

Shell Theorem: spherically symmetric shell of charge Find the field on Gaussian surfaces inside and outside shell. Use Gauss’ Law and spherical symmetry Spherical symmetry makes the flux integral trivial, as for point charge +Q Points Outside Above yields point charge field formula, so field of a shell of charge mimics a point charge field at locations outside shell. +Q Points Inside: Above implies E = 0 inside, since Qenc is zero. So field due to shell of charge vanishes everywhere inside the shell.

Where does net charge stay 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 somewhere inside or on a conductor…where charges can move, but can not leak off Pick Gaussian surface S just below surface: E = 0 everywhere on it Gauss’ Law says: The only place where un-screened charge can end up is on the outer 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 another 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 every point on the inner surface?

Demonstration Source: Pearson Study Area - VTD Chapter 22 Electroscope in Conducting Shell https://mediaplayer.pearsoncmg.com/assets/secs-vtd33_electroscope Discussion: What made the needle depart from the neutral position after charging? What changed when the second half of the sphere was put in place? Where did the charge go?

Example: Hollow conductor with cavity & charge inside Conducting shell (may not be spherical) is neutral Arbitrary charge +Q in cavity (may not be spherical) Choose Gaussian surface S completely within the conductor + Q - Gaussian Surface S E = 0 everywhere on S, so FS = 0  qenc= 0 Q induces charge Qinner on inner surface, distributed so that E = 0 in the metal, hence… For the shell, Qnet = Qinner + Qouter = 0 The shell is neutral, so Qouter= -Qinner = +Q appears on outer surface Logic above holds for non-spherical as well as spherical shells. BUT ONLY 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 Q may be, outside the shell it is shielded. Qouter is uniform & spherical if the shell is spherical. The field outside then looks like that of a point charge at the center.

Conducting Shell with charge inside 4-3: Place a charge Q 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. Conducting Shell (Neutral) Positive Charge Q + Q Cavity

Gauss’ Law Example: Find a formula for the electric field 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 identical surroundings, so…. E is radial, by symmetry (perpendicular to line charge) E has the same magnitude everywhere on any concentric cylindrical surface Flux through end caps of surface = 0 since E is perpendicular to DA there E on cylinder is radial as are 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 ends. Cylinder area = 2prh

Field Lines and Conductors 4-4: The drawing shows cross-sections of three charged cylinders with different radii, but each has the same total charge. Each cylindrical gaussian surface has the same radius. 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.

Example: Cylindrical Analog of Shell Theorem 4-5: Outside a sphere of charge the electric field mimics 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 below describes the electric field? The charge per unit area on the cylinder surface and is uniform. 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

Gauss’ Law example: find the Electric field near an infinite non-conducting sheet of charge s is a uniform positive charge per unit area Choose cylindrical gaussian surface penetrating sheet (rectangular is OK too) E remains constant 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 fewer steps 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.

SUPPLEMENTARY MATERIAL

Evaluate integrand at all points on surface S Gauss’ Law depends on the notion of Flux (symbol F): Basically vector field x area In fluids: Flux measures volume flow or mass flow per second. Definition: dF is flux of velocity field v crossing vector area dA v “unit normal” outward and perpendicular to surface dA “Phi” Dimensions: Volume/unit time Flux through a closed or open surface S: INTEGRATE field over surface S Evaluate integrand at all points on surface S EXAMPLE : FLUID FLUX THROUGH A CLOSED, POROUS 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 fields. v What do field lines look like if a flux source or sink is in the box? Can field be uniform? Can net flux be zero?

MORE DETAIL ON FLUID FLUX: WATER FLOWING IN A STREAM Assume: Incompressible fluid (constant mass density) Constant flow velocity parallel to banks No turbulence (smooth laminar flow) Mass density is r 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 at velocity v 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

from outside to inside of surface from inside to outside of surface DF depends on the angle between the field and vectorized chunks of area DF>0 DF=0 DF<0 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

Example: Fields near parallel nonconducting sheets - 1 Two infinite nonconducting sheets of charge are near each other, Equal and opposite surface charge densities. Charge cannot move. Use superposition. There is no screening, as there would be in a conductor. Each sheet produces a uniform field of magnitude: Etot for oppositely Charged Plates, same |s| but opposite sign s -s - + Left region: Field of the positively charged sheet is canceled by the field of 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 non-conducting sheets - 2 Etot for 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

Field near a thin but finite sized conducting plate, isolated OPTIONAL TOPIC S S’ L 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

Details of Shell Theorem proof using Gauss’ Law (depends on spherical symmetry) Hollow shell, net charge Q, radius R Surface charge density s , becomes spherically symmetric (i.e. s = Q/4pr2). Consider 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 shells of charge with radii smaller than r contribute to E as point charges. shells with radius between r and R produce zero field inside.