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By J. E. Parton, S. J. T. Owen, M. S. Raven (auth.)

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Extra resources for Applied Electromagnetics

Sample text

D =p v Maxwell's first equation (differential form) If we integrate the equations we obtain fD. D)d(vol) S vol Maxwell's first equation (integral form) vol This is also one form of the divergence theorem by means of which the surface integral of Dis replaced by the volume integral of div D, or vice versa. 6 Energy and potential We have seen that, to obtain the distribution of the electric field strength E, either Coulomb's law or Gauss's theorem may be used. The first requires a vector summation of the contributions due to each charge; the second is really only applicable to symmetrical charge distributions where a suitable gaussian surface can easily be seen.

The equipotential V surfaces are rectangular hyperbolae and a few values are shown in the figure. E=-grad V=-VV = -((avjax)ax + (avjay)ay) = -lOO(yax + xay) V/m The flow lines of the electric field can be sketched in orthogonally to the potential V surfaces. Since the potential off S rises, the conductor lies below A, as shown. The surface of the conductor is an equipotential. The electric field E must be normal to the surface S through A. 5 Dielectrics Previously in this chapter we have considered conductors in which the charges move freely in response to an electric field so that there is no electric field within the material.

5, where surface Sis a conductor and which passes through A ( 1, -1, 0). Just off the surface the potential rises. Sketch the field and fmd the surface density at A. 5 Two-dimensional field sketch; V =lOOxy. The field has z axis symmetry. The equipotential V surfaces are rectangular hyperbolae and a few values are shown in the figure. E=-grad V=-VV = -((avjax)ax + (avjay)ay) = -lOO(yax + xay) V/m The flow lines of the electric field can be sketched in orthogonally to the potential V surfaces. Since the potential off S rises, the conductor lies below A, as shown.

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