W 02 D 2 Gausss Law 1 From

W 02 D 2 Gauss’s Law 1

From Last Class Electric Field Using Coulomb and Integrating 1) Dipole: E falls off like 1/r 3 1) Spherical charge: E falls off like 1/r 2 1) Line of charge: (infinite) E falls off like 1/r 4) Plane of charge: (infinite) E uniform on either side of plane 2

Announcements Math Review Week Three Tuesday from 9 -11 pm in 26 -152 Vector Calculus PS 2 due Week Three Tuesday at 9 pm in boxes outside 32 -082 or 26 -152 W 02 D 3 Reading Assignment Course Notes: Chapter Course Notes: Sections 3. 6, 3. 7, 3. 10 Make sure your clicker is registered 3

Outline Electric Flux Gauss’s Law Calculating Electric Fields using Gauss’s Law 4

Gauss’s Law The first Maxwell Equation! A very useful computational technique to find the electric field when the source has ‘enough symmetry’. 5

Gauss’s Law – The Idea The total “flux” of field lines penetrating any of these closed surfaces is the same and depends only on the amount of charge inside 6

Gauss’s Law – The Equation Electric flux (the surface integral of E over closed surface S) is proportional to charge enclosed by the volume enclosed by S 7

Electric Flux Case I: E is a uniform vector field perpendicular to planar surface S of area A Our Goal: Always reduce problem to finding a surface where we can take E out of integral and get simply E*Area 8

Electric Flux Case II: E is uniform vector field directed at angle to planar surface S of area A 9

Concept Question: Flux The electric flux through the planar surface below (positive unit normal to left) is: +q 1. 2. 3. 4. positive. negative. zero. Not well defined. -q 10

Concept Question Answer: Flux Answer: 2. The flux is negative. +q -q The field lines go from left to right, opposite the assigned normal direction. Hence the flux is negative.

Open and Closed Surfaces A rectangle is an open surface — it does NOT contain a volume A sphere is a closed surface — it DOES contain a volume 12

Area Element: Closed Surface Case III: not uniform, surface curved For closed surface, is normal to surface and points outward ( from inside to outside) if points out if points in 13

Group Problem Electric Flux: Sphere Consider a point-like charged object with charge Q located at the origin. What is the electric flux on a spherical surface (Gaussian surface) of radius r ? 14

Arbitrary Gaussian Surfaces True for all surfaces such as S 1, S 2 or S 3 Why? As area gets bigger E gets smaller 15

Gauss’s Law Note: Integral must be over closed surface 16

Concept Question: Flux thru Sphere The total flux through the below spherical surface is +q 1. 2. 3. 4. positive (net outward flux). negative (net inward flux). zero. Not well defined. 17

Concept Question Answer: Flux thru Sphere Answer: 3. The total flux is zero + q We know this from Gauss’s Law: No enclosed charge no net flux. Flux in on left cancelled by flux out on right

Concept Question: Gauss’s Law The grass seeds figure shows the electric field of three charges with charges +1, and -1, The Gaussian surface in the figure is a sphere containing two of the charges. The electric flux through the spherical Gaussian surface is 1. Positive 2. Negative 3. Zero 4. Impossible to determine without more information. 19

Concept Question Answer: Gauss’s Law Answer 3: Zero. The field lines around the two charged objects inside the Gaussian surface are the field lines associated with a dipole, so the charge enclosed in the Gaussian surface is zero. Therefore the electric flux on the surface is zero. Note that the electric field E is clearly NOT zero on the surface of the sphere. It is only the INTEGRAL over the spherical surface of E dotted into d. A that is zero. 20

Virtual Experiment Gauss’s Law Applet Bring up the Gauss’s Law Applet and answer the experiment survey questions http: //web. mit. edu/viz/EM/visualizations/electrostatics/flux/closed. Surfaces/closed. htm 21

Choosing Gaussian Surface In Doing Problems True for all closed surfaces Useful (to calculate electric field ) for some closed surfaces for some problems with lots of symmetry. Desired E: Perpendicular to surface and uniform on surface. Flux is EA or -EA. Other E: Parallel to surface. Flux is zero 22

Symmetry & Gaussian Surfaces Desired E: perpendicular to surface and constant on surface. So Gauss’s Law useful to calculate electric field from highly symmetric sources Source Symmetry Gaussian Surface Spherical Concentric Sphere Cylindrical Coaxial Cylinder Planar Gaussian “Pillbox” 23

Using Gauss’s Law to do Problems 1. Based on the source, identify regions in which to calculate electric field. 2. Choose Gaussian surface S: Symmetry 3. Calculate 4. Calculate qenc, charge enclosed by surface S 5. Apply Gauss’s Law to calculate electric field: 24

Examples: Spherical Symmetry Cylindrical Symmetry Planar Symmetry 25

Group Problem Gauss: Spherical Symmetry +Q uniformly distributed throughout non-conducting solid sphere of radius a. Find everywhere. 26

Concept Question: Spherical Shell We just saw that in a solid sphere of charge the electric field grows linearly with distance. Inside the charged spherical shell at right (r<a) what does the electric field do? 1. 2. 3. 4. 5. a Q Zero Uniform but Non-Zero Still grows linearly Some other functional form (use Gauss’ Law) Can’t determine with Gauss Law 27

Concept Question Answer: Flux thru Sphere Answer: 1. Zero Q a Spherical symmetry Use Gauss’ Law with spherical surface. Any surface inside shell contains no charge No flux E = 0! P 04 -

Demonstration Field Inside Spherical Shell (Grass Seeds): 29

Worked Example: Planar Symmetry Consider an infinite thin slab with uniform positive charge density. Find a vector expression for the direction and magnitude of the electric field outside the slab. Make sure you show your Gaussian closed surface. 30

Gauss: Planar Symmetry is Planar Use Gaussian Pillbox Note: A is arbitrary (its size and shape) and should divide out Gaussian Pillbox 31

Gauss: Planar Symmetry Total charge enclosed: NOTE: No flux through side of cylinder, only endcaps + + + 32

Concept Question: Superposition Three infinite sheets of charge are shown above. The sheet in the middle is negatively charged with charge per unit area , and the other two sheets are positively charged with charge per unit area. Which set of arrows (and zeros) best describes the electric field? 33

Concept Question Answer: Superposition Answer 2. The fields of each of the plates are shown in the different regions along with their sum. 34

Group Problem: Cylindrical Symmetry An infinitely long rod has a uniform positive linear charge density. Find the direction and magnitude of the electric field outside the rod. Clearly show your choice of Gaussian closed surface. 35

Electric Fields 1) Dipole: E falls off like 1/r 3 1) Spherical charge: E falls off like 1/r 2 1) Line of charge: (infinite) E falls off like 1/r 4) Plane of charge: (infinite) E uniform on either side of plane 36