The current at the outside terminals of the capacitor is the sum of the currentused to charge the capacitor and the current through the resistor. The displacement current I d can be obtained by substitutingeq.(35.11) into eq.(35.8) The electric flux through the capacitor is therefore equal to The electricfield between the capacitor plates is therefore equal to The currentin the thin wire can be obtained using Ohm's lawī) The voltage across the capacitor is equal to the external emf. Theexterior terminals of the plates are connected to a source of alternating emfwith a voltage V = V 0 sin( t).Ī) What is the current in the thin wire ?ī) What is the displacement current through the capacitor ?Ĭ) What is the current arriving at the outside terminals of the capacitor ?ĭ) What is the magnetic field between the capacitor plates at a distance r fromthe axis ? Assume that r is less than the radius of the plates.Ī) The setup can be regarded as a parallel circuit of a resistor withresistance R and a capacitor with capacitance C (see Figure 35.2). A thin straight wire of length d lies along the axis of thecapacitor and connects the two plates. Where I d is called the displacement current and is definedasĪ parallel-plate capacitor has circular plates of area A separated by adistance d. The current I is the current intercepted by whatever surface is used in thecalculation, and is not necessarily the same as the current in the wires.Equation (35.6) is frequently written as In such acase, the effects of the electric flux and the electric current must becombined, and Ampere's law becomes Where E is the electric flux through the surface indicated inFigure 35.1 In the most general case, the surface spanned by the integrationpath of the magnetic field can intercept current and electric flux. The magnetic field around the wire can now be found by modifying Ampere's law The electric flux will therefore also be timedependent, and the rate of change of electric flux is equal to If a current I is flowing through the wire, then the charge on the capacitorplates will be time dependent. The electric flux, E, intercepted by the surface shown in Figure 35.1 is equalto The electric field outside the capacitor is equal to zero. If the plates have a surface area A then the electricfield between the plates is equal to At a certain time t the charge on thecapacitor plates is Q. Suppose the capacitor is an idealcapacitor, with a homogeneous electric field E between the plates and noelectric field outside the plates. Ampere's law in a capacitor circuit.Īlthough the surface shown in Figure 35.1 does not interceptany current, it intercepts electric flux. However, thecurrent intercepted by an arbitrary surface now depends on the surface chosen.For example, the surface shown in Figure 35.1 does not intercept any current.Clearly, Ampere's law can not be applied in this case to find the magneticfield generated by the current.įigure 35.1. This current will generate a magnetic field and ifwe are far away from the capacitor, this field should be very similar to themagnetic field produced by an infinitely long, continuous, wire. A current will flow through the wire during the chargingprocess of the capacitor. However, consider the casein which the current wire is broken and connected to a parallel-plate capacitor(see Figure 35.1). The calculation of the magnetic field of a current distribution can, inprinciple, be carried out using Ampere's law which relates the path integral ofthe magnetic field around a closed path to the current intercepted by anarbitrary surface that spans this path:Īmpere's law is independent of the shape of the surface chosen as long as thecurrent flows along a continuous, unbroken circuit. THE DISPLACEMENT CURRENT AND MAXWELLS EQUATIONS The magnetic field of an accelerated chargeģ5. The Electric Field of an Accelerated Charge THE DISPLACEMENT CURRENT AND MAXWELLS EQUATIONS.THE DISPLACEMENT CURRENT AND MAXWELLS EQUATIONS CHAPTER 35
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