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The spacebetween the electrodes of a parallel-plate capacitor is filled witha uniformpoorlyconducting mediumof conductivitysigma and permittivity epsilon. The capacitorplates shaped as rounddiscs are separtedby a distance d. Neglectingthe edgeeffects, find the magnetic fieldstrength betweenthe plates at a distance r fromtheiraxis ifan ac voltage V = V_(m) cos omega tis appliedto the capacitor. |
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Answer» Solution :The electricfield between the plates can be written as, `E = Re (V_(m))/(d) e^(I omega t)`, insteadof `(V_(m))/(d) cos omega t`. The gives RISE to a conductioncurrent, `j_(e) = sigmaE = Re (sigma)/(d) V_(m) e^(I omega t)` and a displacementcurrent, `j_(d) = (DEL D)/(del t) = Re epsilon_(0) epsilon i omega (V_(m))/(d) e^(i omega t)` The totalcurrent is, `j_(T) = (V_(m))/(d) sqrt(sigma^(2) + (epsilon_(0) epsilon omega)^(2)) cos (omega t + alpha)` where, `tan alpha = (sigma)/(epsilon_(0) epsilon omega)` on takingthe real PART ofthe resultant. The correspondingmagneticfieldis obtainedby usingcircularion THEOREM, `H.2pi r= pi r^(2) j_(r)` or, `H = H_(m) cos(omega t + alpha)`, where, `H_(m) = (r V_(m))/(2d) sqrt(sigma^(2) + (epsilon_(0) epsilon omega)^(2))` |
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