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By Carl Boettcher (Auth.)

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BÖTTCHER A N D P. BORDEWIJK The electric current causes dissipation of energy. According to Joule's l a w , 3 the amount of energy dissipated during the time interval dt is given by: 2 ÖW = I-Edt dt. 60) For a harmonic field, the energy dissipation during one cycle amounts t o : 2π/ω 0 2 2 I cos (utdt AW=G(E ) J 2 = -a{E°) . 61) Hence, the average dissipation of energy per unit of time due to conduction is: 2 W=\a(E°) . 62) Comparing eqn. 62) with eqn. 55), we see that if we determine ε"(ω) from the absorption of energy in a dielectric we always obtain the sum ε" + 4πσ/ω, so that we must correct for the contribution 4πσ/ω due to the conductivity of the dielectric; this correction will be important at low frequencies.

BORDEWIJK change from an infinitely sharp loss peak to a peak with finite width, as long as the frequency region over which ε" deviates considerably from zero for each peak is small with respect to the characteristic frequency ω Λ. Therefore under these conditions eqn. 196) can be used profitably. In the frequency range corresponding with the characteristic times for the orientational polarization, one obtains again the dielectric constant of induced polarization ε^, given by eqn. 197). This also holds for bands where the integral in eqn.

Thus, the pulse-response function for the relation between the polarization and the external field in the case of a needle is identical with the pulse-response function for the relation between the polarization and the Maxwell field. The same holds for the step-response functions. This was to be expected, since for a needle in the direction of the field the Maxwell field is equal to the external field (cf. eqn. 79) with Ac = 0). For a disc-shaped dielectric we have Aa = 1, and eqn. 133) leads t o : ε [Κω) t Π R ST Ί - ^ τ - Τ - ΐ (8-135) ε- 1 { ε(ω) J In this case the external field is equal to the dielectric displacement in the dielectric (c/.

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