chapter24notes - Chapter 24 Gausss Law The prince of...

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

View Full Document Right Arrow Icon
Chapter 24 Gauss’s Law The prince of mathematicians: "Tell her to wait a moment 'til I'm through".
Background image of page 1

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

View Full DocumentRight Arrow Icon
Electric Flux area A E We define the electric flux Φ , of the electric field E , through the surface A, as: Φ = E . A Where A is a vector normal to the surface (magnitude A, and direction normal to the surface – outwards in a closed surface) A
Background image of page 2
Electric Flux Here the flux is Φ = E · A You can think of the flux through some surface as a measure of the number of field lines which pass through that surface. Flux depends on the strength of E , on the surface area, and on the relative orientation of the field and surface. Normal to surface, magnitude A area A E A E A
Background image of page 3

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

View Full DocumentRight Arrow Icon
Electric Flux The flux also depends on orientation area A θ A cos θ The number of field lines through the tilted surface equals the number through its projection . Hence, the flux through the tilted surface is simply given by the flux through its projection: E (A cos θ) . Φ = E . A = E A cos θ area A θ A cos θ E E A A
Background image of page 4
θ dA E What if the surface is curved, or the field varies with position ?? 1. We divide the surface into small regions with area dA 2. The flux through dA is d Φ = E dA cos θ d Φ = E . dA 3. To obtain the total flux we need to integrate over the surface A A Φ = d Φ = E . dA Φ = E . A
Background image of page 5

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

View Full DocumentRight Arrow Icon
In the case of a closed surface Φ Φ = = = d E dA q inside ε 0 The loop means the integral is over a closed surface. θ dA E
Background image of page 6
Gauss’s Law This is always true. Occasionally, it provides a very easy way to find the electric field (for highly symmetric cases). The electric flux through any closed surface equals b enclosed charge / ε 0 Φ Φ = = = d E dA q inside ε 0
Background image of page 7

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

View Full DocumentRight Arrow Icon
Calculate the electric field produced by a point charge using Gauss Law We choose for the gaussian surface a sphere of radius r , centered on the charge Q . Then, the electric field E , has the same magnitude everywhere on the surface (radial symmetry) Furthermore, at each point on the surface, the field E and the surface normal dA are parallel (both point radially outward). E . dA = E dA [cos θ = 1]
Background image of page 8
E Q Coulomb’s Law ! E = 1 4 πε 0 q r 2 E . dA = Q / ε 0 E . dA = E dA = E A A = 4 π r 2 E A = E 4 π r 2 = Q / ε 0 Electric field produced by a point charge E Q k = 1 / 4 π ε 0 ε 0 = permittivity ε 0 = 8.85x10 -12 C 2 /Nm 2
Background image of page 9

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

View Full DocumentRight Arrow Icon
Is Gauss’s Law more fundamental than Coulomb’s Law? • No! Here we derived Coulomb’s law for a point charge from Gauss’s law. • One can instead derive Gauss’s law for a general (even very nasty) charge distribution from Coulomb’s law. The two laws are equivalent. • Gauss’s law gives us an easy way to solve a few
Background image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/01/2009 for the course PHY phy2049 taught by Professor Wilfredngwa during the Fall '09 term at University of Central Florida.