Phys 0175 - Lecture 9

Phys 0175 - Lecture 9 - Reminder: Homework Assignments:...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Reminder: Homework Assignments: PHYS 0175 – LON-CAPA HW #02 Due on Sunday(@10pm), January 25 th . PHYS 0175 – LON-CAPA HW #03 Now Open: Due on Sunday(@10pm), February 1st
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Lecture 9 (Jan. 26, 2009): Chapter 23: Gauss’ Law (conclusion) Applying Gauss’ Law: planar symmetry Applying Gauss’ Law: spherical symmetry Chapter 24: Electric Potential Definition of Electric Potential Energy Definition of Electric Potential Electric Potential due to a point charge Equipotential surfaces Calculating the El. Potential from the El. Field
Background image of page 2
Applying Gauss’ Law: Planar symmetry (III) What is the electric field near two thin, infinite, parallel non-conducting sheets with uniform surface charge densities, one positive and one negative? As we saw before, each sheet sets up electric fields of magnitude E=σ/ε 0 on both sides of the sheet which point away from positive and towards negative charges. These E fields pass through the non-conducting sheets and add (as vectors) in the different regions as shown.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Applying Gauss’ Law: Spherical symmetry (I) What is the electric field produced by a solid conducting sphere with net charge q as a function of distance r from its center? E=0 inside the sphere; all charge q is spread evenly over the outer surface of the sphere. Outside the sphere E drops off as (1/r 2 ) as though all of the net charge were located at the center.
Background image of page 4
Applying Gauss’ Law: Spherical symmetry (II) What is the electric field produced by a charged spherical shell of total charge q and radius R as a function of distance r from its center? Gaussian surface S 1 encloses no charge. Hence E=0 for r<R. Gaussian surface S 2 encloses all of q. Hence for r>R 2 0 1 4 q E r πε =
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Applying Gauss’ Law: Spherical symmetry (III) What is the electric field produced by a spherically symmetric charge
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 26

Phys 0175 - Lecture 9 - Reminder: Homework Assignments:...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online