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Chapter 24. Review on Chapter 23 From Coulomb's Law to Gausss Law Applications of Gausss Law. Review on Chapter 23: Coulombs Law and the electric field definition. Coulombs Law: the force between two point charges The electric field is defined as - PowerPoint PPT Presentation


  • Chapter 24Review on Chapter 23From Coulomb's Law to Gausss LawApplications of Gausss Law

  • Review on Chapter 23: Coulombs Law and the electric field definitionCoulombs 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 filed

  • Two examplesExample 23.9 (page 662)Example 23.10 (page 663)

  • From Coulombs Law to Gausss 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:E = EACompare to a water flux in a tube:W = V1A1= V2A2This sign means water flows into the tube, by convention.

  • Electric Flux, plane with an angle When the field lines make an angle 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?

    Review on math: direction of a surface is defined as the (outwards) normal to that surface.Dot product of two vectors.

  • Electric Flux, GeneralIn the more general case, look at a small area element

    In general, this becomes

    The 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, E = -El 2For side 2, E = El 2For the other sides, E = 0Therefore, total = 0

  • Example 2: flux through a sphere with a charge at its center. From Coulombs Law to Gausss Law A positive point charge, q, is located at the center of a sphere of radius rAccording to Coulombs 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, so

    Combine these two equations, we have

  • The Gaussian Surface and Gausss 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.Gausss Law (Karl Friedrich Gauss, 1777 1855):The net flux through any closed surface surrounding a charge q is given by q/o and is independent of the shape of that surface

    The 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

    Gausss Law connects electric field with its source charge

  • Gausss Law SummaryGausss law states

    qin 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 Gausss law can, in theory, be solved to find for any charge configuration, in practice it is limited to a few symmetric situations

  • 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 C. The charges are fixed on the ring and cannot move. There is a bug with charge q = -0.1 C 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 bug?