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

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Chapter 24 Gauss’s Law The prince of mathematicians: "Tell her to wait a moment 'til I'm through".

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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
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

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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
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

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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
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

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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]
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

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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
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