Ch32 - CHAPTER Maxwell's Equations and Electromagnetic...

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CHAPTER 32 Maxwell’s Equations and Electromagnetic Waves 1* A parallel-plate capacitor in air has circular plates of radius 2.3 cm separated by 1.1 mm. Charge is flowing onto the upper plate and off the lower plate at a rate of 5 A. ( a ) Find the time rate of change of the electric field between the plates. ( b ) Compute the displacement current between the plates and show that it equals 5 A. ( a ) Use Equ. 23-25: E = Q / ε 0 A ; dE / dt = ( dQ / dt )/ 0 A ( b ) Use Equ. 32-3: φ e = EA dE / dt = I / 0 A = 3.40 × 10 14 V/m . s I d = 0 A ( dE / dt ) = I = 5 A 2 In a region of space, the electric field varies according to t 2000 sin ) C / N (0.05 = E 0, where t is in seconds. Find the maximum displacement current through a 1-m 2 area perpendicular to E . Use Equs. 23-14 and 3-3 I d = (8.85 × 10 –12 × 0.05 × 2000) A = 8.85 × 10 –10 A 3 ∙∙ For Problem 1, show that at a distance r from the axis of the plates the magnetic field between the plates is given by B = (1.89 × 10 –3 T/m) r if r is less than the radius of the plates. 1. Use Equ. 32-4; I = I d ; apply cylindrical symmetry 2. Evaluate B ( r ) 2 π rB = µ 0 I d ( r 2 / R 2 ); B = 0 I d r /2 R 2 B ( r ) = (1.89 × 10 –3 T/m) r 4 ∙∙ ( a ) Show that for a parallel-plate capacitor the displacement current is given by I d = C dV / dt , where C is the capacitance and V the voltage across the capacitor. ( b ) A parallel plate capacitor C = 5 nF is connected to an emf E = E 0 cos ω t , where E 0 = 3 V and = 500 . Find the displacement current between the plates as a function of time. Neglect any resistance in the circuit. ( a ) Use Equs. 25-10 and 32-3; E = V / d ( b ) dV / dt = – E 0 sin t I d = 0 A ( dE / dt ) = ( 0 A / d )( dV / dt ) = C dV / dt I d = –(23.6 A) sin 500 t 5* ∙∙ Current of 10 A flows into a capacitor having plates with areas of 0.5 m 2 . ( a ) What is the displacement current between the plates? ( b ) What is dE / dt between the plates for this current? ( c ) What is the line integral of ! d B around a circle of radius 10 cm that lies within and parallel to the plates? ( a ) See Problem 1 ( b ) dE / dt = I d / 0 A (see Problem 1) ( c ) Use Equ. 32-4; I d enclosed = I d ( r 2 / A ) I d = 10 A dE / dt = 2.26 × 10 12 V/m . s B . d ! = 0 I d ( r 2 / A ) = 7.90 × 10 –7 T . m 6 ∙∙ A parallel-plate capacitor with circular plates is given a charge Q 0 . Between the plates is a leaky dielectric having a dielectric constant of κ and a resistivity ρ . ( a ) Find the conduction current between the plates as a function of time. ( b ) Find the displacement current between the plates as a function of time. What is the total
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Chapter 32 Maxwell’s Equations and Electromagnetic Waves (conduction plus displacement) current? ( c ) Find the magnetic field produced between the plates by the leakage discharge current as a function of time. ( d ) Find the magnetic field between the plates produced by the displacement current as a function of time. ( e ) What is the total magnetic field between the plates during discharge of the capacitor? ( a )I f Q is the charge on the capacitor plates, then the discharge current I = – dQ / dt. Also, I = V / R = VA / d ρ , where we have used Equ. 26-8. From the definition of capacitance, I = AQ / Cd . The differential equation dQ / dt + α Q = 0 has the solution Q = Q 0 e t . Here = A / Cd = 1/ ε 0 κρ . Thus, I = – dQ / dt = ( Q 0 / 0 ) e / t 0 κ .
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This note was uploaded on 04/23/2008 for the course PHYS 270 taught by Professor Drake during the Spring '08 term at Maryland.

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Ch32 - CHAPTER Maxwell's Equations and Electromagnetic...

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