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Unformatted text preview: Supplement —Phy374—Spring 2006 Prof. Ted Jacobson www.physics.umd.edu/grt/taj/374a/ Room 4115, (301)4056020 [email protected] In this supplement we discuss the relation between the integral and differential forms of Maxwell’s equations, derive the 3d wave equation for vacuum electromagnetic fields, find the general form of a plane wave solution, and discuss the field energy conservation theorem. The second section summarizes a few mathematical items from vector calculus needed for this discussion, including the continuity equation. 1 Maxwell’s equations Maxwell’s equations in differential form are the following equations: ∇ · E = ρ/ Gauss’ law (electric) (1) ∇ · B = 0 Gauss’ law (magnetic) (2) ∇ × E = ∂ t B Faraday’s law (3) ∇ × B = μ ( j + ∂ t E ) Amp` ereMaxwell law (4) The first two equations are scalar equations, while the last two are vector equations. The notation ∂ t stands for ∂/∂t . We now discuss how these differential equations are obtained from the integral equations. 1.1 Gauss’ law The integral form of Gauss’ law is R ∂V E · dS = q/ . This can be written as an equality between three dimensional volume integrals, by writing the total charge enclosed q as the integral of the charge density over the volume, and using the divergence theorem (21) to express the flux integral as the volume integral of the divergence of E . Since the two volume integrals are equal for any integration volume V , we can equate the integrands, thus obtaining the differential form of Gauss’ law (1). For the magnetic field there are no magnetic monopole charges, hence we have simply (2). 1.2 Faraday’s law Faraday’s law states that the loop integral of the induced electric field is minus the time rate of change of the magnetic flux through the loop, R ∂S E · dl = ( d/dt ) R S B · dS . We can use Stoke’s theorem (20) to write the loop integral of E as a surface integral of the curl of E . Equating integrands then yields the differential form of Faraday’s law, eqn. (3). The divergence of the left hand side of Faraday’s law, ∇ · ( ∇ × E ), vanishes identically (see (16)), so if Faraday’s law is consistent it must be true that ∇ · ∂ t B also vanishes. Since the time and space partial derivatives commute, this is the same as ∂ t ∇ · B , which vanishes thanks to the magnetic version of Gauss’ law (2). So the absense of magnetic charges is required for Faraday’s law to be selfconsistent....
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 Fall '10
 Jacobson
 Physics, Law

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