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24-gauss's law

# 24-gauss's law - Physics 4B Lecture Notes Chapter 24 Gauss...

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Physics 4B Lecture Notes 24-1 Chapter 24 - Gauss’ Law Problem Set #3 - due: Ch 24 - 2, 3, 6, 10, 12, 19, 25, 27, 35, 43, 53, 54 Lecture Outline 1. The Definition of Electric Flux 2. Gauss’ Law 3. The Behavior of Conductors 4. Examples Using Gauss’ Law 1. The Definition of Electric Flux Recall that the strength of the field is proportional to the density of field lines. The field can be thought of as the number of lines per unit area. The number of lines through an area is called the "flux." If a constant field is perpendicular to the surface, then the flux is given by Φ = EA . If a constant field is at an angle θ from the normal to the surface, then the flux is given by Φ = EA cos θ = r E r A . For non-constant fields, such as the field due to a point charge, the flux through the surface is the sum of the normal component of the field over each small section of the surface where the flux can be considered constant. Φ ≡ r E d r A The Definition of Electric Flux A E A E θ dA E θ

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Physics 4B Lecture Notes 24-2 Example 1: Find the flux that exits a sphere centered at the origin due to a point charge also at the origin. A small surface area on a sphere is, d r A = r 2 sin θ d θ d φ ˆ r . Using the field due to the point charge, r E = k q 2 r ˆ r , the flux can be calculated from its definition, Φ ≡ r E d r A , Φ ≡ k q 2 r ˆ r r 2 sin θ d θ d φ ˆ r Φ = k q sin θ 0 π d θ d φ 0 2 π = 4 π kq = q ε o 2. Gauss’ Law Example 2: Find the flux due to a point charge over the surface of a cube. Again start with the definition of flux, Φ ≡ r E d r A , and the field of a point charge, r E = k q 2 r ˆ r . This time however, the electric field and the area element are not always parallel and the integral is nearly impossible to complete. But, we know that the number of field lines that leave the charge is not effected by the shape of some imaginary surface that surrounds the charge. That is, the same number of field lines exit the cube as exit the sphere in example 1 so the flux must again equal Φ = q ε o . The flux through any closed surface is proportional to the charge contained inside. Gauss's Law r E d r A = q enclosed ε o Beyond the Mechanical Universe (vol. 29 frame 28270 [Ch 17]) dA θ d θ r r sin θ φ d φ r sin θ d φ r d θ x y z dA x y z q dE r
Physics 4B Lecture Notes 24-3 Example 3: Starting with Gauss’s Law derive Coulomb’s Rule. The field due to a point charge q 1 can be found by applying Gauss's Law to an imaginary gaussian sphere of radius r centered on the charge, r E d r A = q ε o = q 1 ε o . By the spherical symmetry of the problem, the field must be radial and constant on the gaussian sphere so, r E d r A = EA = E4 π r 2 = q 1 ε o E = q 1 4 πε o r 2 = k q 1 r 2 .

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24-gauss's law - Physics 4B Lecture Notes Chapter 24 Gauss...

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