Chapter%203

# Chapter%203 - Chapter 3 Electric Flux Density Gauss's Law...

This preview shows pages 1–11. Sign up to view the full content.

Dr. Ray Chen© Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence

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

View Full Document
Dr. Ray Chen© 3.1 Electric flux theory D: Electric flux density, displacement, flux density, displacement density. sphere) (inner 4 2 r a r a ˆ a Q D = = π v r r b r a ˆ r Q D , b r a a ˆ b Q D = = = 2 2 4 sphere) (outer 4 v v versa vice & derived easily be can E , D know we If vacuum). (in D E 0 v v v v = ε Fig. 3.1 ab D v r
Dr. Ray Chen© Outer sphere in this case ( r>b ) does not contribute to the vector field, if only one sphere at r=a exists, the result is the same. 3.1 Electric flux theory The electric field within the sphere (hollow) is 0 everywhere, (r<a) Qa r Q r If r>a, D is the same for both cases. The sphere can be treated as a point source r r a ˆ E a ˆ r Q D v v 0 2 4 ε π = =

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

View Full Document
Dr. Ray Chen© 3.1 Electric flux theory Example 1: m 3 at D and E Find axis z the along locating nC/m 8 with charge line A = ρ v v Solution : V/m 8 143 10 854 8 2 10 8 2 12 9 0 π πε a ˆ . a ˆ . a ˆ E L = × × = = v V/m 9 47 m, 3 at a ˆ . E = = v m 3 for nC/m 424 0 C/m 10 273 1 2 10 8 2 2 2 9 9 = = × = × = = πρ . a ˆ . a ˆ a ˆ D L v
Dr. Ray Chen© 3.2 Gauss’s Law The electric flux passing through any closed surface is equal to the total charge enclosed by that surface : enclosed charge : S s S d D r v n Q Q Σ = : on distributi charge discrete For charge volume for charge surface for charge line for : on distributi charge continuous a For dv Q ds Q dl Q v s L = = = ρ

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

View Full Document
Dr. Ray Chen© 3.2 Gauss’s Law Example 2: a radius of sphere a at flux electric of amount total the Find (C/m 4 sphere the of surface the At 2 2 ) a ˆ r Q D r π = v Fig 3.3
Dr. Ray Chen© 3.2 Gauss’s Law Solution to Example 2: r a ˆ d sin r s d d d sin r ds d 2 2 φ θ = = v π d d sin Q a ˆ a ˆ d d sin a a Q s d D r r 4 4 2 2 = = v v Fig 3.3 [] ) Coulombs ( Q cos Q a ˆ a ˆ d d sin Q s d D r r 2 4 charge Total 0 2 00 = = = = ∫∫ ππ v v

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

View Full Document