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Areas and Stuff. January 29, 2010. What’s Up?. We are finishing up the chapter on Electric Field and Gauss. Quiz Today Monday we will begin Potential .. Read first two sections in Chapter 19. WebAssign due Sunday Night Short WA is also due Tuesday Night (2 problems)
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Areas and Stuff January 29, 2010
What’s Up? • We are finishing up the chapter on Electric Field and Gauss. • Quiz Today • Monday we will begin Potential .. Read first two sections in Chapter 19. • WebAssign due Sunday Night • Short WA is also due Tuesday Night (2 problems) • Best Guess – EXAM #! – Chapters 18,19 • Wednesday February 10th. 1 hour • After this, we begin to move a bit faster through the material.
I am free • 8:30-9:30 • 9:30-10:30 Fridays • 12:30-1:30 • None of these
Results to date What the … ?? Mr. Coulomb
The Area Vector This area vector is defined as being POSITIVE because it points OUTWARD from a CLOSED surface
18.8 The Electric Field Inside a Conductor: Shielding At equilibrium under electrostatic conditions, any excess charge resides on the surface of a conductor. At equilibrium under electrostatic conditions, the electric field is zero at any point within a conducting material. The conductor shields any charge within it from electric fields created outside the condictor.
18.8 The Electric Field Inside a Conductor: Shielding The electric field just outside the surface of a conductor is perpendicular to the surface at equilibrium under electrostatic conditions.
18.8 The Electric Field Inside a Conductor: Shielding • Conceptual Example 14 A Conductor in • an Electric Field • A charge is suspended at the center of • a hollow, electrically neutral, spherical • conductor. Show that this charge induces • a charge of –q on the interior surface and • (b) a charge of +q on the exterior surface of • the conductor.
But first … QUIZ
New Topic Gauss’s Law
How about this?? • Positive point charge • Negative point charge • Large Sheet of charge • No charge • You can’t tell from this
18.9 Gauss’ Law GAUSS’ LAW The electric flux through a Gaussian surface is equal to the net charge enclosed in that surface divided by the permittivity of free space: SI Units of Electric Flux: N·m2/C
18.9 Gauss’ Law Example 15 The Electric Field of a Charged Thin Spherical Shell A positive charge is spread uniformly over the shell. Find the magnitude of the electric field at any point (a) outside the shell and (b) inside the shell.
18.9 Gauss’ Law • Outside the shell, the Gaussian • surface encloses all of the charge. (b) Inside the shell, the Gaussian surface encloses no charge.
Continuous Charge Distributions • Volume r = charge per unit volume • C/m3 • Area s = charge per unit area • C/m2 • Line m or l = charge per unit length • C/m
Which Way?? D D
Line of Charge • Gaussian Surface • It is not a REAL surface, but it is imagined. • It has multiple surfaces. In this case it has a cylindrical surface as well as two circular end-caps. • For each of the surfaces, the E field must be either normal to the surface (flux) or parallel to the surface (no flux).
Two infinite planes = capacitor!! s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 E=0 E=s/e0 E=0 The Field between two charges capacitor plates is s/e0
Charged Conductors Charge Must reside on the SURFACE - - E=0 - - E - s Very SMALL Gaussian Surface