Lecture05_Kim - 1/25/2012 1 Lecture 5-1 Applications of...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1/25/2012 1 Lecture 5-1 Applications of Gausss Law The net electric flux through any closed surface equals the net charge enclosed by that surface divided by . Applications to systems with symmetries: 1. Shperical Symmetry: Thin spherical shells, solid spheres 2. Cylindrial Symmetry: Infinite thin wire 3. Planar Symmetry: Infinite non-conducting sheet, Infinite conducting sheet enclosed E Q E ndA = = a i I Lecture 5-2 Spherial Symmetry Thin , uniformly charged cell (Review) E = 0 inside Discontinuity in E 2 Q E k r = outside r E = Lecture 5-3 3 Spherical Symmetry: Uniform Distribution (1) Lets start with a Gaussian surface with r 1 < R From spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. Gausss Law gives us Solving for E we find ( 29 3 1 2 1 4 3 4 r q E dA E r = = = a a d volume volume volume volume area area area area E inside = r 1 3 R Lecture 5-4 4 Spherical Symmetry: Uniform Distribution (2) 1 3 4 3 3 inside r Q E R = 1 3 inside r E = In terms of the total charge Q 1 1 3 3 4 inside Qr kQr E R R = = R 1/25/2012 2 Lecture 5-5 5 Spherical Symmetry: Uniform Distribution (3) Now consider a Gaussian surface with radius r 2 > R Again by spherical symmetry we know that the electric field will be radial and perpendicular to the Gaussian surface. Gauss Law gives us Solving for E we find E outside = kQ r 2 2 same as a point charge! same as a point charge! same as a point charge! same as a point charge! total charge total charge total charge total charge area area area area R ( 29 3 2 2 4 3 4 R Q E dA E r = = = a a d Lecture 5-6 6 Cylindrical Symmetry (Review) The electric flux through the ends of the cylinder is zero because the electric field is always parallel to the ends The electric field is always perpendicular to the wall of the cylinder so and now solve for the electric field E = 2 r = 2 k r ( 29 2 / / (Gauss) E dA EA E rL q L = =...
View Full Document

This note was uploaded on 02/05/2012 for the course PHYS 241 taught by Professor Wei during the Spring '08 term at Purdue University-West Lafayette.

Page1 / 6

Lecture05_Kim - 1/25/2012 1 Lecture 5-1 Applications of...

This preview shows document pages 1 - 3. Sign up to view the full document.

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