Ch32Word - Chapter 32 p. 1 CHAPTER 32 Maxwells Equations...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Chapter 32 p. 1 CHAPTER 32 Maxwells Equations and Electromagnetic Waves 1. The electric field between the plates depends on the voltage: E = V / d , so d E /d t = (1/ d ) d V /d t = (1/1.3 10 3 m)(120 V/s) = 9.2 10 4 V/m s . 2. The displacement current is I D = A (d E /d t ) = (8.85 10 12 C 2 /N m 2 )(0.038 m) 2 (2.0 10 6 V/m s) = 2.6 10 8 A . 3. The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from I D = A (d E /d t ); 1.8 A = (8.85 10 12 C 2 /N m 2 )(0.0160 m) 2 d E /d t , which gives d E /d t = 7.9 10 14 V/m s . 4. The current in the wires is the rate at which charge is accumulating on the plates and must also be the displacement current in the capacitor. Because the location is outside the capacitor, we can use the expression for the magnetic field of a long wire: B = ( /4p)2 I D / R = (10 7 T m/A)2(35.0 10 3 A)/(0.100 m) = 7.00 10 8 T . After the capacitor is fully charged, all currents will be zero, so the magnetic field will be zero . 5. The electric field between the plates depends on the voltage: E = V / d , so d E /d t = (1/ d ) d V /d t . Thus the displacement current is I D = A (d E /d t ) = ( A / d )(d V /d t ) = C d V /d t . 6. ( a ) The current in the wires is the rate at which charge is accumulating on the plates and must also be the displacement current in the capacitor: I max = I D,max = 35 A . ( b ) At any instant, the charge on the plates is Q = C , so the current is I = d Q /d t = C d /d t = C sin t , or I max = p R 2 / d ; 35 10 6 A = 2p(96.0 Hz) (8.85 10 12 C 2 /N m 2 )p(0.025 m) 2 /(2.0 10 3 m), which gives = 6.7 10 3 V . ( c ) We can find the maximum value of d E /d t from the maximum value of the displacement current: I D,max = (d E /d t ) max ; 35 10 6 A = (8.85 10 12 C 2 /N m 2 )(d E /d t ) max , which gives (d E /d t ) max = 4.0 10 6 V m/s . 7. Gausss law for electricity and Amperes law will not change. From the analogy to Gausss law for electric fields, where Q is the source, Q m would be the source of the magnetic field, so we have B d A = Q m . From the analogy to Amperes law, we have an additional current contribution to Faradays law.From the analogy to Amperes law, we have an additional current contribution to Faradays law....
View Full Document

This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.

Page1 / 9

Ch32Word - Chapter 32 p. 1 CHAPTER 32 Maxwells Equations...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online