Lecture 3 - University of Birminghamepweb2.ph.bham.ac.uk/user/lazzeroni/EM2_2017/Lecture3_EM... ·...

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Electromagnetism II Cristina Lazzeroni [email protected] Lecture 3

Transcript of Lecture 3 - University of Birminghamepweb2.ph.bham.ac.uk/user/lazzeroni/EM2_2017/Lecture3_EM... ·...

  • Electromagnetism II

    Cristina [email protected]

    Lecture 3

  • Bet your shirt on it “If you have bought one of those T-shirts with Maxwell’s equations on the front, you may have to worry about itsgoing out of style, not about its becoming false. We willgo on teaching Maxwellian electrodynamics as long asthere are scientists.”

    Steven Weinberg, Nobel prizewinning physicist

  • Maxwell’s equations for free space

    Contain all information about EM fields in vacuum at all length scales, down to QM scales.All M fields in vacuum are solutions of Maxwell’s eqs.

  • Lecture 3:

    - Electrostatic solutions to Maxwell’s equations- Write Laplace’s and Poisson’s eqs. for electrostatic potential- Uniqueness theorem- Use method of images in simple situations- Applications of numerical methods to find numericalsolutions to Laplace’s eq in simple cases

  • Static fields (no time dependence)

    V = scalar field

    If charge-free region Laplace’s equation

    Poisson’s equation in free space

    Electrostatic

  • Uniqueness theorem

  • Example:

    Consider a conducting enclosure (Faraday cage).Inside the enclosure the electrostatic potential satisfiesLaplace’s equation. On the surface of the enclosurethe electrostatic potential is a constant,V0

    Write a solution for the potential inside the enclosure

  • Example:

    Consider a conducting enclosure (Faraday cage).Inside the enclosure the electrostatic potential satisfiesLaplace’s equation. On the surface of the enclosurethe electrostatic potential is a constant,V0

    Write a solution for the potential inside the enclosure

    V0 This satisfies Laplace’s equation and boundary conditions

    How do we know this is the only correct solution ?

    Uniqueness theorem: there is only one possible solution(N.B. V constant - no E field)

  • Enclosure of sensitive electronics from stray electric fields

    In any conducting enclosure, V is constant and E is zero

  • Solutions to Poisson’s equation:

    a. Direct integration (in simple cases)

    b. Method of images (requires symmetry)

    c. Numerical methods (general use)

  • a. Direct integrationExample:Solid sphere of radius R, uniformly distributed charge density ρ.Calculate potential V (and field E) inside and outside the sphere

    r ≥ R: externally same as if the charge were all at its centre

  • r ≤ R: internally use Poisson’s equation

    Symmetry: only r-coordinateis relevant

  • b. Method of images

  • Example:

  • Example:+q is near two semi-infinite perpendicular conductive planesat zero potential. Determine an expression for the electrostaticpotential in the region between the plates.

  • c. Numerical solution - finite differences method

  • Assign know potentials at the boundaries and make a guess elsewhere.Guess potential at C is VCCalculate A,B,D,E using Taylor

    Apply to every point on grid, untilprocedure converges (negligiblechange in the map from one cycleto the next)

  • In practice finer grids are used. Electrostatic voltages in complexsystems.

    Electrostatic lens, designed to focus beam of positive lead ionsfor the LHC at CERN.

  • Summary:

    Uniqueness theorem: any solution of Poisson / Laplace equation that satisfies the boundary conditions for a particular problem is unique.This means that if a solution is found, it is the only solution.

    A solution can be found via integration or method of imagesin particular cases, or via numerical techniques in the general case.

  • Method of Images

    Replace conducting surfaces by a simple equivalent set of virtualsource charges (image charges) such that boundary conditionsare satisfied.The image charges are located so that the surface where theconductor had been is equipotential, its potential being that ofthe conductor, and all other boundary conditions are matched.With the image charges in place the electric field can be calculated.

    In the case where the conductor surface is flat, each image chargecoincides with the optical mirror image of an existing charge and has the opposite value.

  • Next lecture:

    Dielectric materials

    Recommended readings:Grant+Phillips: 3.1, 3.1.1, 3.1.2, 3.2, 3.3, 3.3.1