PreambleFluxFlux Through a Closed SurfaceExampleFlux of a Point ChargeGauss’ LawApplications of Gauss’ LawInfinite Sheet of ChargeInfinite Line of ChargeSpheres of ChargeElectric Fields inside ConductorsElectric Fields on the Surface of ConductorsConducting sheet of ChargeMethod of Image ChargesPHY2061 Enriched Physics 2 Lecture Notes Gauss Gauss’ Law Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor. Preamble In this lecture we learn a simple and powerful technique for calculating the electric field for situations involving a high degree of symmetry. Flux Flux, Φ, is a measure of the amount of a vector field, , passing through a surface with area (direction defined perpendicular to the plane of the surface): FA cosFAθΦ= ⋅ =FA where θ is the angle between the field direction and the normal to the area surface. As one example, consider the flow of air or water through a surface. The voulume of ar passing through the surface per unit time is given by: cosvAθΦ= ⋅ =vA Obviously this flow rate depends on the orientation of the surface. As another example, consider the light from the Sun impinging onto the surface of the Earth. Obviously land near the equator receives a large light flux than that at the poles. In this course, we will consider the flux of electric and magnetic fields (essentially light!) passing through a surface. D. Acosta Page 1 9/2/2005PHY2061 Enriched Physics 2 Lecture Notes Gauss Flux Through a Closed Surface Fig. from HRW 7/eConsider the flux of electric field through a closed surface. Let’s break the surface into a large number (N) of small elements of area ∆Athat are nearly flat. By convention, let’s take the direction of the area vector to be perpendicular to the surface and pointing out. The flux of electric field passing through this closed surface is then: 1Ni=Φ= ⋅∆∑EA If we let and , then the small elements become infinitesimal and we can write: N →∞ 0∆→A dΦ= ⋅∫EAv This is a closed surface integral. It is a 2-dimensional integral over a 3-dimensional surface. The “circle” in the integral sign denotes an integral over an entire closed surface (versus just a single face). D. Acosta Page 2 9/2/2005PHY2061 Enriched Physics 2 Lecture Notes Gauss Example y topAKEK Consider a square cube immersed in a constant electric field: 0ˆE=ExK. Calculate the flux through each face. iAK is the area vector pointing out from cube face i, which has magnitude , where s is the side length of the cube. 2||is=AK Note that top top0 since ⋅= ⊥EA E AKKKKAlso: bottomfrontback000⋅=⋅=⋅=EAEAEAKKKKKK What is left is: 2right right 0 right2left left 0 left||| | since | || | since opposite EsEs⋅= =⋅=− =−EA E A E AEA E A E AKK KKK K&KKKK K 6220010 sum over all 6 facesiiEs Es=⇒Φ= ⋅ = − =∑EA Thus, the total flux into the cube is balanced by the flux going out. This will hold for any arbitrary shape of the surface, provided the electric field is constant x z rightAKbottomAKD. Acosta Page 3 9/2/2005PHY2061 Enriched Physics 2 Lecture Notes Gauss Flux of a Point Charge Consider a spherical shell surface of radius r enclosing a point charge at the center. dAErq Note that points out from the closed surface, and that since E points radially as well, dAK so that dd⋅=EA EAEAKKdKK&KK By Coulomb’s Law, 2qKr=E So the flux through the spherical surface is: since is constant at fixed radiusSSSdddΦ= ⋅ ==∫∫∫EA EAEA Evvv 24 surface area of sphereSdrπ=∫Av To see this explicitly, let’s do the surface integral: 2220sin 4Sodrddππrθθφπ==∫∫ ∫Av So, the flux is: 2222200144441qqrK rrrqrππππεεΦ= = =⇒Φ=E Note that this is a constant, independent of the size of the sphere, and only dependent on the amount of charge enclosed. In fact, it turns out that it does not even depend on the shape of the surface! D. Acosta Page 4 9/2/2005PHY2061 Enriched Physics 2 Lecture Notes Gauss Gauss’ Law Gauss’ Law generalizes the previous example, and states that: enc0SqdεΦ= ⋅ =∫EAv So the amount of flux passing through any closed surface S depends only on the net amount of electric charge enclosed. Why is this general case true? Well, and arbitrary surface can always be inscribed by a sphere within it. And the flux through that sphere, calculated in the previous example, must be equal to the flux intercepted by the outer surface S. S q Gauss’ Law relates the electric flux through a closed surface to the net charge enclosed. It was derived as a consequence of Coulomb’s Law, but in fact is completely equivalent to Coulomb’s Law. It can form an alternative starting point for calculating electric fields. D. Acosta Page 5 9/2/2005PHY2061 Enriched Physics 2 Lecture Notes Gauss Applications of Gauss’ Law Gauss’ Law is a powerful technique to calculate the electric field for situations exhibiting a high degree of symmetry. Infinite Sheet of Charge Let’s calculate the electric field from an infinite sheet of charge, with a charge density of σ (measured in C/m2). EK EKrightAleftA By symmetry, we expect E to point perpendicular to the surface for a very large sheet (and far from the edges). We would not expect any other angle, because why would any particular direction be preferred over any other for such a symmetric situation? Now consider a closed surface (which we will call a Gaussian surface) that extends through the sheet of charge. The sides, here assumed cylindrical, are chosen to be perpendicular to the sheet. So. The caps are parallel to the sheet, so since both vectors always point in the same direction. The total flux through the surface is thus: side0⋅=EAleft right⋅=⋅EA EA left right2EAΦ= ⋅ + ⋅ =EA EA By Gauss’ Law, enc0qεΦ= , where encqAσ=. So: 0022AEAEσεσεΦ= =⇒= This is exactly the same solution for the magnitude of the electric field from an infinite sheet that we obtained by painstaking integration of rings of charge! And as noted before, the electric field is a constant and does not depend on the distance from the sheet. D. Acosta Page 6 9/2/2005PHY2061 Enriched Physics 2 Lecture Notes Gauss Infinite Line of Charge Let’s calculate the electric field
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