8.022pset11solution

8.022pset11solution - Massachusetts Institute of Technology...

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Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution to Assignment 11: Poynting vector practice 1. Poynting vector and capacitor. (a) E = 4 πQ A = 4 Q a 2 . This feld points From the top plate to the bottom plate. (b) Generalized Ampere’s law: I ~ B · d~s = 1 c d Φ E dt . By right hand rule, iF we look at the top plate From above, we see that the displace- ment current induces a magnetic feld that circulates in a clockwise sense. To get the magnitude, we need to do some integrals: choosing a contour oF radius r , 2 πrB ( r ) = 1 c d dt µ πr 2 4 Q a 2 -→ B ( r ) = 2 r ca 2 dQ dt . (c) Magnetic field cycles clockwise Electric field points into page Looking at capacitor from above Cross product points toward center of circle. (d) The surFace we want to choose is the cylindrical side oF radius a and height s between the top and the bottom plate. The magnitude oF the Poynting vector is | ~ S | = c 4 π | ~ E × ~ B | = c 4 π ˆ 8 Q ca 3 dQ dt ! = 2 Q πa 3 dQ dt .
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Then, P = Z ~ S · d~a = S A cyl where A cyl = 2 πas is the area of the cylinder connecting the top and bottom plates: P = ˆ 2 Q
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8.022pset11solution - Massachusetts Institute of Technology...

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