1 of 10 Gauss's Law Gauss's Law is one of the 4 fundamental laws of electricity and magnetism called Maxwell's Equations. Gauss's law relates charges and electric fields in a subtle and powerful way, but before we can write down Gauss's Law, we need to introduce a new concept: the electric flux Φ through a surface. E surface with area A Consider an imaginary surface which cuts across some E-field lines. We say that there is some electric flux through this surface. To make the notion of flux precise, we must first define a surface vector. Definition: surface vector A = , associated with a flat surface of area A. ˆAAn=KMagnitude of vector A = area A of surface. A area A Direction of vector A = direction perpendicular (normal) to surface = direction of unit normal . ˆn Notice that there is an ambiguity in the direction . Every flat surface has ˆntwo perpendicular directions. A smaller area, shorter A The electric flux Φ through a surface A is defined as AE θ (for E = constant, surface flat ) EA EAcosΦ= ⋅ = θKK The flux Φ has a the following geometrical interpretation: flux ∝ the number of electric field lines crossing the surface. Think of the E-field lines as rain flowing threw an open window of area A. The flux is a measure of the amount of rain flowing through the window. To get a big flux, you need a large E, a large A, and you need the area perpendicular to the E-field vector, which means the area vector A is parallel to E. (In the rain analogy, you need the window to be facing the rain direction.) A EE A θ = 90o , cosθ = 0 , Φ = 0 θ = 0, cosθ =1 , Φ max PHYS1120 Lecture Notes, Dubson, 9/10/2009 ©University of Colorado at Boulder2 of 10 (A cosθ) is the projection of the area A onto the plane perpendicular to E. It is the area which "faces the rain". Only the area facing the rain gives flux. E The formula is a special case formula: it only works if the surface is flat and the E-field is constant. If the E-field varies with position and/or the surface is not flat, we need a more general definition of flux: E AΦ= ⋅KK E da "surface integral of "Φ= ⋅ =∫EKK To understand a surface integral, do this: in your imagination, break the total surface up into many little segments, labeled with an index i. The surface vector of segment i is idaK. If the segment is very, very tiny, it is effectively flat and the E-field is constant over that tiny surface, so we can use our special case formula E AΦ=⋅KK . The flux through segment i is therefore iiEdaiΦ=⋅KK. (Ei is the field at the segment i) The total flux is the sum: Φ=iiiEda Eda⋅ → ⋅∑∫KKKK (In the limit that the segments become infinitesimal, there are an infinite number of segments and the sum becomes an integral.) In general, computing surface integral Eda⋅∫KKcan be extremely messy. So why do we care about this thing called the electric flux? The electric flux is related to charge by Gauss's Law. E Edaθ A Eθ θ area = A cosθ PHYS1120 Lecture Notes, Dubson, 9/10/2009 ©University of Colorado at Boulder3 of 10 Gauss's Law (the 1st of 4 Maxwell's Equations) enclosed0qEda⋅=ε∫KKv In words, the electric flux through any closed surface S is a constant (1/ε0) times the total charge inside S. E da surface integral⋅=∫KKvclosed "closed" imaginary spherical surface S, radius r r + da A surface is closed if it has no edges, like a sphere. For a closed surface, the direction of da is always the outward normal. The constant ε0 is related to k by 01k4=πε. da12 12coul220kq q q q1Fr4==πε r1208.85 10−ε= × (SI units) Gauss' Law can be derived from Coulomb's Law if the charges are stationary, but Gauss's Law is more general than Coulomb's Law. Coulomb's Law is only true if the charges are stationary. Gauss's Law is always true, whether or not the charges are moving. It is easy to show that Gauss's Law is consistent with Coulomb's Law. From Coulomb's Law, the E-field of a point charge is20kQ 1 QEr4==πε2r(si . We get the same result by applying Gauss's Law: nce E is parallel to da on S) (since E is constant on S) (says Mr. Gauss) ()S20Eda EdaEdaEAQE4 r⋅====π=ε∫∫∫KKvvvSolving for E, we have 201QE4r=πε . Done. PHYS1120 Lecture Notes, Dubson, 9/10/2009 ©University of Colorado at Boulder4 of 10 When viewed in terms of field lines, Gauss's Law is almost obvious (after a while). Recall that flux is proportional to the number of field lines passing through the surface. Notice also that flux can be positive or negative depending on the angle θ between the E-field vector and the area vector. Where the field lines exit a closed surface, the flux there is positive; where the field lines enter a closed surface, the flux there is negative. So the total flux through a closed surface is proportional to da E (exiting S)θ θ da [(# field lines exiting) – (# field lines entering)] E (entering S) If a closed surface S encloses no charges, then the number of lines entering must equal the number of lines exiting, since there are no charges inside for the field lines to stop or start on. imaginary surface S E SEda 0←⋅=∫KKv S So only charges inside the surface can contribute to the flux through the surface. Positive charges inside produce positive flux; negative charges produce negative flux. The net flux is due only to the net charge inside: enclosed0qEda⋅=ε∫KKv. Using Gauss's Law to solve for the E-field Gauss's Law is always true (it's a LAW). But it is not always useful. Only in situations with very high symmetry is it easy to compute the flux integral Eda⋅∫KKv. In these few cases of high symmetry, we can use Gauss's Law to compute the E-field. + – PHYS1120 Lecture Notes, Dubson, 9/10/2009 ©University of Colorado at Boulder5 of 10 Example of Spherical Symmetry: Compute E-field everywhere inside a uniformly-charged spherical shell. By symmetry, E must be radial (along a radius), so E = E(r). We choose an imaginary surface S concentric with and inside the charged sphere. + + + + + +Since the E-field is radial and the surface vector da on S is also radial, we have EKK. (The dot product of the parallel vectors is just the product of the magnitudes.) So we have da Eda⋅=++ S Q SEda Eda E da EA,⋅= = =∫∫ ∫KKvvv+ + + where A is the area of surface S. We are able to take E outside the integral only because E = E(r) and so E = constant
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