ch24 - Gauss’s Law Chapter Outline ENGINEERING PHYSICS II...

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Unformatted text preview: Gauss’s Law Chapter Outline ENGINEERING PHYSICS II Electric Flux PHY 303L Gauss’s Law Applications of Gauss’s Law Chapter 24: Gauss’ Law Superposition of Electric Fields Conductors and Electric Fields Maxim Tsoi Physics Department, The University of Texas at Austin http://www.ph.utexas.edu/~tsoi 303L: Gauss’ Law (Ch.24) 303L: Gauss’ Law (Ch.24) Electric Flux Electric Flux we now treat electric field lines in a more quantitative way what if the surface is not perpendicular to the field ? • Consider an electric field that is uniform in • Normal to the surface of area A is at an both magnitude and direction angle θ to the uniform electric field • Number of lines per unit area ~ magnitude • A’ =A cosθ is a projection of A onto a of E plane oriented perpendicular to the field • Electric flux product of the E magnitude • The number of lines that cross A and A’ is and surface area A perpendicular to the field the same the flux through A equals the flux through A’ Φ = EA Φ = EA' = EA cos θ N m2/C SI units of electric flux Electric flux is proportional to the number of electric field lines penetrating some surface 303L: Gauss’ Law (Ch.24) Electric flux is proportional to the number of electric field lines penetrating some surface Example 1 303L: Gauss’ Law (Ch.24) Electric Flux Gauss’s Law what if the field is nonuniform ? relationship between flux and charge • The electric field varies over a surface • gaussian surface • Divide surface into a large number of small • Point charge q is at the center of a sphere of elements, each of area ∆A ∆Ai closed surface radius r • Electric field on the surface of the sphere: a vector with magnitude representing the area of ∆Ai surface element and direction is E = ke q r 2 perpendicular to the surface element • The electric field Ei at the location of ∆Ai makes an angle θI with ∆Ai • The electric flux through ∆Ai is: • The net flux through the sphere (ke = 1/4πε0): r r ∆Φ = E i ∆Ai cos θ i = E i ⋅ ∆Ai • The electric flux through a surface is: r r Φ = lim ∑ E i ⋅ ∆Ai = ∆Ai → 0 rr ∫ E ⋅ dA surface rr q Φ = ∫ E ⋅ dA = E ∫ dA = E 4πr 2 = ε0 The net flux through the spherical surface is proportional to the charge inside Electric flux depends both on the field pattern and on the surface 303L: Gauss’ Law (Ch.24) Example 2 303L: Gauss’ Law (Ch.24) 1 Gauss’s Law Gauss’s Law relationship between flux and charge relationship between flux and charge • S1 is spherical surface, S2 and S3 are not • The net flux through S1 is • For multiple charges or continuous distribution of charge use the superposition principle: Φ = q ε0 The electric field due to many charges is the vector sum of the electric fields produced by the individual charges • The number of electric field lines is the same through S1, S2, and S3 The net flux through any closed surface surrounding a point charge q is given by Φ= q r r r ∫ E ⋅ dA = ∫ (E ε0 1 • For a charge outside the surface, any line that enters the surface leaves it at another point • Gauss’s law r r + E 2 + ... ⋅ dA ) the net flux through any closed surface is rrq Φ = ∫ E ⋅ dA = in The net electric flux through a closed surface that surrounds no charge is zero ε0 • It can be solved for E to determine the electric field due to a system of charges or a continuous distribution of charge 303L: Gauss’ Law (Ch.24) 303L: Gauss’ Law (Ch.24) Electrostatic Equilibrium Gauss’s Law application to various charge distributions properties of conductors Gauss’s law is useful in determining electric fields when the charge distribution is characterized by a high degree of symmetry When there is no net motion of charge within a conductor, the conductor is in electrostatic equilibrium rrq Φ = ∫ E ⋅ dA = in 1) The electric field is zero everywhere inside the conductor ε0 • • 2) If an isolated conductor carries a charge, the charge resides on its surface Choose the gaussian surface that satisfies one or more of the following: • The value of E can be argued by symmetry to be constant over the surface 3) The electric field just outside a charged conductor is perpendicular to its surface and has a magnitude σ/ε0, where σ is the surface charge density at that point 4) On an irregularly shaped conductor, the surface charge density is greatest at • The dot product can be expressed as a simple algebraic product locations where the radius of curvature of the surface is smallest because E and dA are parallel • The dot product is zero because E and dA are perpendicular • The field can be argued to be zero over the surface Examples 3-7 303L: Gauss’ Law (Ch.24) 303L: Gauss’ Law (Ch.24) SUMMARY Gauss’s law • Electric flux is proportional to the number of electric field lines that penetrate a surface • Electric flux through a surface of area A is E is uniform and make an angle θ with the normal to the surface: Φ = EA cos θ • In general, the electric flux through a surface: r Φ= • Gauss’s law r ∫ E ⋅ dA surface the net electric flux through any closed gaussian surface is equal to the net charge inside the surface divided by permittivity of free space ε0 rrq Φ = ∫ E ⋅ dA = in ε0 • Properties of conductors in electrostatic equilibrium (1-4) 303L: Gauss’ Law (Ch.24) 2 ...
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