Chapter 22 – Gauss Law- Charge and Electric flux- Electric Flux Calculations- Gauss’s Law and applications- Charges on ConductorsChild acquires electric charge by touching a charged metal sphere. Electrons coat each individual hair fiber and then repel each other.1. Charge and Electric Flux- A charge distribution produces an electric field (E), and E exerts a force on a test charge (q0). By moving q0around a closed box that contains the charge distribution and measuring F one can make a 3D map of E = F/q0outside the box. From that map, we can obtain the value of q inside box.- If we construct a boundary around a charge, we can think of the flow coming out from the charge like water through a screen surrounding a sprinkler.Electric Flux and Enclosed Charge:-There is a connection between sign of net charge enclosed by a closed surface and the direction of electric flux through surface (inward for -q, outward for +q).- There is a connection between magnitude of net enclosed charge and strength of net “flow” of E. - The net electric flux through the surface of a box is directly proportional to the magnitude of the net charge enclosed by the box.2q 2Eq EE~ 1/r2r1= distance of q to surface of box1.r2= 2r1= distance of qto surface of box2.In (c), E2= E1/4, since r2=2r1, but A2=4A1net flux constant.- Electric flux = (perpendicular component of E) · (area of box face)-The net electric flux due to a point charge inside a box is independent of box’s size, only depends on net amount of charge enclosed.- Charges outside the surface do not give net electric flux through surface.2. Calculating Electric FluxFlux Fluid Analogy:- If we considered flux through a rectangle, the flux will change as the rectangle changes orientation to the flow.vAdtdV=dV/dt = volume flow ratev = flow speedAvAvvAvAdtdV⋅====⊥⊥ϕcos(v┴A)Flux of a Uniform Electric Field:AEEAAEE ⊥==⋅=ΦϕcosΦE = E·AΦE = E·A·cosφΦE = 0Units: N m2/CWe can define a vector area: with n being a unit vector ┴A.nAAˆ⋅=Flux of a Non-uniform Electric Field:3. Gauss’s Law- The total electric flux through any closed surface is proportional to the total electric charge inside the surface. Point Charge Inside a Spherical Surface:- The flux is independent of the radius R of thesphere.2041RqEπε=()0220441εππεqRRqAEE==⋅=ΦE //dA at each pointPoint Charge Inside a Nonspherical Surface:- Divide irregular surface into dA elements, compute electric flux for each (E dA cosφ) and sum results by integrating.- Each dA projects onto a spherical surface element total electric flux through irregular surface = flux through sphere. 0εqAdEE=⋅=Φ∫Integral through a closed surfaceValid for + / - qIf enclosed q = 0 ΦE= 0Point charge outside a closed surface that encloses no charge. If an electric field line enters the surface at one point it must leave at another.- Electric field lines can begin or end inside a region of space only when there is a charge in that region.General form of Gauss’s law:0cosεϕenclEQdAEdAEAdE ===⋅=Φ∫∫∫⊥( )022020204444εππεπεπεqrrqdArqdArqdAEE−=−=−=−==Φ∫∫∫⊥Example: Spherical Gaussian surface around –q (negative inward flux)4. Applications of Gauss’s Law- When excess charge (charges other than ions/e-making up a neutral conductor)is placed on a solid conductor and is at rest, it resides entirely on the surface, not in the interior of the material.- Electrostatic condition (charges at rest) E = 0 inside material of conductor, otherwise excess charges will move.E = 0q = 05. Charges on Conductors- Excess charge only on surface.- Cavity inside conductor with q = 0 E = 0 inside conductor, net charge on surface of cavity = 0.- Cavity inside conductor with +q E = 0 inside conductor, -q charge on surface of cavity (drawn there by +q). Total charge inside conductor = 0 +q on outer surface (in addition to original qc).Field at the surface of a conductor:σ = q/A q = σAField outside a charged conductor is perpendicular to surface.Ex.
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