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Conquer Radio Frequency

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1.4 V & I or E & H ? 7 If we now calculate the ratio of voltage and current from (1.4-2), (1.4-3), and (1.4-1)(a) we obtain: Also, substituting (1.3-2) into (1.4-4) we obtain Equation (1.4-5) shows that relates and at a specific point in space around the resistor (identified by radial distance from its axis) in a linear fashion. This derivation is not entirely accurate as one would need to take into account material properties to be fully rigorous. Nonetheless it demonstrates how, in principle, the value of the resistance gives an indication of the relative magnitudes of vector quantities and , not just of the scalar quantities and which may be used to represent and in low frequency circuits. 1.4.2 Capacitors Equation (1.4-1)(b) shows that the capacitance C is a constant of proportionality between the current through a capacitor and the rate of change of the voltage across its terminals ( ). Now let us consider the field inside a parallel plate capacitor (figure 1.4-2). This may be easily calculated from Gauss's law and works out to be where is the sheet charge density on each plate, is the absolute value of the charge on either plate and is the surface area of each plate. Now let us take a look at the magnetic field between the plates of the capacitor. It would appear that, since no current, in its classical form 2 , can flow between the plates, we are unable to use equation (1.3-1) to obtain the magnetic field from the current. This in fact is not the case since a displacement current exists between the plates which is equal in magnitude to the current flowing in and out of the terminals of the capacitor. This comes from Maxwell's generalisation of Ampere's law and its proof is beyond the scope of this treatment 3 . 2 conduction current 3 For the reader who wishes to research this further, the forth Maxwell equation is used to come to this result ∮ ⁄ (1.4-6) (1.4-7) (1.4-4) (1.4-5) Conquer Radio Frequency 7 www.cadence.com/go/awr

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