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W02D2 Gauss ’ s Law. From Last Class Electric Field Using Coulomb and Integrating. Dipole: E falls off like 1/r 3 Spherical charge: E falls off like 1/r 2 Line of charge: E falls off like 1/r (infinite) Plane of charge: E uniform on (infinite) either side of plane.
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From Last ClassElectric Field Using Coulomb and Integrating • Dipole: E falls off like 1/r3 • Spherical charge: E falls off like 1/r2 • Line of charge: E falls off like 1/r • (infinite) • Plane of charge: E uniform on • (infinite) either side of plane
Math Review Week Three Tuesday from 9-11 pm in 26-152Vector Calculus PS 2 due Week Three Tuesday at 9 pm in boxes outside 32-082 or 26-152 W02D3 Reading Assignment Course Notes: Chapter Course Notes: Sections 3.6, 3.7, 3.10 Make sure your clicker is registered Announcements
Outline Electric Flux Gauss’s Law Calculating Electric Fields using Gauss’s Law
Gauss’s Law The first Maxwell Equation! A very useful computational technique to find the electric field when the source has ‘enough symmetry’.
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
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
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
Electric Flux Case II: E is uniform vector field directed at angle to planar surface S of area A
+q -q Concept Question: Flux The electric flux through the planar surface below (positive unit normal to left) is: • positive. • negative. • zero. • Not well defined.
+q -q Concept Question Answer: Flux Answer: 2. The flux is negative. 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
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
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 ?
Arbitrary Gaussian Surfaces True for all surfaces such as S1, S2 or S3 Why? As area gets bigger E gets smaller
Gauss’s Law Note: Integral must be over closed surface
+q +q Concept Question: Flux thru Sphere The total flux through the below spherical surface is • positive (net outward flux). • negative (net inward flux). • zero. • Not well defined.
+q +q Concept Question Answer: Flux thru Sphere Answer: 3. The total flux is zero 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, +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 Positive Negative Zero Impossible to determine without more information.
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 dA that is zero.
Virtual ExperimentGauss’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/closedSurfaces/closed.htm
Choosing Gaussian Surface In Doing Problems Desired E: Perpendicular to surface and uniform on surface. Flux is EA or -EA. Other E: Parallel to surface. Flux is zero True for all closed surfaces Useful (to calculate electric field ) for some closed surfaces for some problems with lots of symmetry.
Source Symmetry Gaussian Surface Spherical Concentric Sphere Cylindrical Coaxial Cylinder Planar Gaussian “Pillbox” 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
Using Gauss’s Law to do Problems • Based on the source, identify regions in which to calculate electric field. • Choose Gaussian surface S: Symmetry • Calculate • Calculate qenc, charge enclosed by surface S • Apply Gauss’s Law to calculate electric field:
Examples:Spherical SymmetryCylindrical SymmetryPlanar Symmetry
Group Problem Gauss: Spherical Symmetry +Q uniformly distributed throughout non-conducting solid sphere of radius a. Find everywhere.
a 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? Q • Zero • Uniform but Non-Zero • Still grows linearly • Some other functional form (use Gauss’ Law) • Can’t determine with Gauss Law
a Concept Question Answer: Flux thru Sphere Answer: 1. Zero Q Spherical symmetry Use Gauss’ Law with spherical surface. Any surface inside shell contains no charge No flux E = 0!
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.
Gauss: Planar Symmetry Symmetry is Planar Use Gaussian Pillbox Note: A is arbitrary (its size and shape) and should divide out Gaussian Pillbox
Gauss: Planar Symmetry Total charge enclosed: NOTE: No flux through side of cylinder, only endcaps + + + + + + + + + + + +
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?
Concept Question Answer: Superposition Answer 2 . The fields of each of the plates are shown in the different regions along with their sum.
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 • Dipole: E falls off like 1/r3 • Spherical charge: E falls off like 1/r2 • Line of charge: E falls off like 1/r • (infinite) • Plane of charge: E uniform on • (infinite) either side of plane