. Theory of the relativity of motion . Electromagnetic Theory. 187 Let figure 15 represent a cross-section of the rotating a section of the dielectric AA which is moving perpendicularlyto the plane of the paper in the X-direction with the velocity V. Letthe magnetic field be in the F-direction parallel to the axis of problem is to calculate dielectric displacement Dz in the Z-direction. Referring to equations (248) we have V ( V \ D,+-Hy = e{Et+-Byy and, substituting the value of By given by equations (249),By+^ES = »(Hy + ^Dz\ V2\ ( V2 \ V 1 - e»~)Dz = e^l - ~)E
. Theory of the relativity of motion . Electromagnetic Theory. 187 Let figure 15 represent a cross-section of the rotating a section of the dielectric AA which is moving perpendicularlyto the plane of the paper in the X-direction with the velocity V. Letthe magnetic field be in the F-direction parallel to the axis of problem is to calculate dielectric displacement Dz in the Z-direction. Referring to equations (248) we have V ( V \ D,+-Hy = e{Et+-Byy and, substituting the value of By given by equations (249),By+^ES = »(Hy + ^Dz\ V2\ ( V2 \ V 1 - e»~)Dz = e^l - ~)E2+-(e» - 1)H„ we obtain or, neglecting terms of orders higher than —, we have Dz = + - (€M - l)Hy. (251) For a substance whose permeability is practically unity such asWilson actually used the equation reduces to D2 = eEz + ~(e-l)Hy, and this was found to fit the experimental facts, since measurementswith the electrometer show the surface charge actually to have themagnitude Dz per square centimeter in accordance with ou
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