Saved Bookmarks
| 1. |
Sea water at frequency v=4xx10^8Hz has permittivity epsilon~~80epsilon_0. Permeability mu~~mu_0 and resistivity rho=0.25 Omega -m. Imaging a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t)=V_0 sin (2pivt). What fraction of the conduction current density is the displacement current density? |
|
Answer» Solution :Let a be the distance between the plates. The applied voltageis given by `V(t)=V_0sin 2pivt`. The applied electric field, `E=(V(t))/d=(V_0)/d sin 2pivt` The conduction CURRENT DENSITY, `J_C=E/rho =(V_0)/(rhod) sin 2pivt=J_(0C) sin 2piv t` where `(V_0)/(rhod)=J_(0C)`=MAX. conduction current density. Displacement current density is given by `J_D=(I_D)/A=(in_0in_r d(phi_E)//dt)/A=(in_0 in_rd(EA))/(Adt)=in_0in_r (dE)/(dt) =in_0in_rd/(dt)((V_0)/d sin 2pivt)` `=in_0in_r (V_0)/d 2piv cos 2pivt =J_(0D) cos 2pivt` Where `in_0in_r (V_0)/dxx2piv=J_(0D)`=max. displacement current desity `:. (J_(0D))/(J_(0C))=(in_0 in_r V_0 2piv//d)/(V_0//(rhod))=2piv in_0in_r rho=4pi in_0 v in_r rho//2=1/(9xx10^9)xx(4xx10^8)xx80xx0.25//2=4/9` |
|