Chapter 24Review on Chapter 23: Coulomb’s Law and the electric field definitionTwo examplesFrom Coulomb’s Law to Gauss’s lawSome preparation: Electric Flux through a perpendicular planeElectric Flux, plane with an angle θElectric Flux, GeneralExample 1: flux through a cube of a uniform electric fieldExample 2: flux through a sphere with a charge at its center. From Coulomb’s Law to Gauss’s LawThe Gaussian Surface and Gauss’s LawGauss’s Law – SummaryPreview sections and homework 1/27, due 2/3Chapter 241. Review on Chapter 232. From Coulomb's Law to Gauss’s Law3. Applications of Gauss’s LawReview on Chapter 23: Coulomb’s Law and the electric field definitionCoulomb’s Law: the force between two point charges The electric field is defined as and is represented through field lines.The force a charge experiences in an electric filed1 212 122ˆeq qkr=F rroq�FErreq=F Er rTwo examplesExample 23.9 (page 662)Example 23.10 (page 663)From Coulomb’s Law to Gauss’s lawTry to calculate the electric field ofA point chargeAn infinitely long straight wire with evenly distributed chargeA wire loopA round diskAn infinitely large planeA solid sphere with evenly distributed chargeAre there other ways to calculate electric field generated from a charge distribution? Electric field is generated by source charges, are there ways to connect electric field directly with these source charges?The answer is YES!Some preparation:Electric Flux through a perpendicular planeElectric flux is the product of the magnitude of the electric field and the surface area, A, perpendicular to the electric field:ΦE = EACompare to a water flux in a tube:ΦW = –V1A1= V2A2This sign means water flows into the tube, by convention.Electric Flux, plane with an angle θ When the field lines make an angle θ with the direction (i.e., the normal) of the surface, the flux is calculated as:And the electric field E has to be a constant all over the area A. Question: when this is not the case, what do you do the get the flux?AEcosEAEAEReview on math: 1. direction of a surface is defined as the (outwards) normal to that surface.2. Dot product of two vectors.Electric Flux, GeneralIn the more general case, look at a small area elementIn general, this becomescosE i i i i iE AθDF = D = �DE Ar r0surfacelimiE i iAEE AdD �F = �DF = ���E Ar rThe surface integral means the integral must be evaluated over the surface in questionIn general, the value of the flux will depend both on the field pattern and on the surfaceWhen the surface is closed, the direction of the surface (i.e. the normal of it) points outwards.The unit of electric flux is N.m2/CReview on math: Integral over a surface.Example 1: flux through a cube of a uniform electric fieldThe field lines pass through two surfaces perpendicularly and are parallel to the other four surfacesFor side 1, ΦE = -El 2For side 2, ΦE = El 2For the other sides, ΦE = 0Therefore, Φtotal = 0Example 2: flux through a sphere with a charge at its center. From Coulomb’s Law to Gauss’s Law A positive point charge, q, is located at the center of a sphere of radius rAccording to Coulomb’s Law, the magnitude of the electric field everywhere on the surface of the sphere is The field lines are directed radially outwards and are perpendicular to the surface at every point, soCombine these two equations, we have 24AE rEd AEEdAdAEdnE2rqkEe0222444qqkrrqkrEeeEThe Gaussian Surface and Gauss’s LawClosed surfaces of various shapes can surround the chargeOnly S1 is sphericalThe flux through all other surfaces (S2 and S3) are the same. These surfaces are all called the Gaussian Surface.Gauss’s Law (Karl Friedrich Gauss, 1777 – 1855):The net flux through any closed surface surrounding a charge q is given by q/εo and is independent of the shape of that surfaceThe net electric flux through a closed surface that surrounds no charge is zeroSince the electric field due to many charges is the vector sum of the electric fields produced by the individual charges, the flux through any closed surface can be expressed as Gauss’s Law connects electric field with its source charge0AEqdE02121A)EE(AE...qqd...dEGauss’s Law – SummaryGauss’s law statesqin is the net charge inside the Gaussian surface represents the electric field at any point on the surface is the total electric field at a point in space and may have contributions from charges both inside and outside of the surfaceAlthough Gauss’s law can, in theory, be solved to find for any charge configuration, in practice it is limited to a few symmetric situationsErErEr0AEinEqd Preview sections and homework 1/27, due 2/3Preview sections:Section 24.3Section 24.4Homework:Problem 4, page 687.Problem 9, page 687.(optional = do it if you find it fun, or would like to challenge yourself) Problem 11, page 687. (optional): On an insulating ring of radius R there evenly distributed 73 point charges, each with a charge Q =+1 μC. The charges are fixed on the ring and cannot move. There is a bug with charge q = -0.1 μC sits at the center of the ring, and enjoys zero net force on it. When one of the charge Q is removed from the ring, what is the net force of the remaining charges exert on the poor
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