Saved Bookmarks
| 1. |
A parallel `-` plate air capacitor whose electrodes are shaped as discs of radius `R=6.0 cm` is connected to a source of an alternating sinusoidal votage with frequency `omega=1000s^(-1)`. Find the ratio of peak values of magnetic and electric energies within the capacitor. |
|
Answer» Inside the condenser the peak electrical enegry `W_(e) = (1)/(2)CV_(m)^(2)` `= (1)/(2)V_(m)^(2)(epsilon_(0)piR^(2))/(d)` `(d=` separation between the plates, `piR^(2) =` area of each plate.) `V = V_(m) sin omegat, V_(m)` is the maximum voltage Changing electric field causes a displacement current `j_(dis) = (delD)/(delt) = epsilon_(0)E_(m) cos omega t` `= (epsilon_(0)omega V_(m))/(d) cos omegat` This gives rise to a magnetic field `B(r)` (at a radial distance `r`from the centre of the plate) `B(r).2pir = mu_(0)pir^(2) j_(dis) = mu_(0)pir^(2) (epsilon_(0) omega V_(m))/(d) cos omega t` `B = (1)/(2)epsilon_(0)mu_(0) omega (r )/(d) V_(m) cos omega t` Energy associated with this field is `= intd^(3) r(B^(2))/(2mu_(0)) = (1)/(8) epsilon_(0)^(2) (omega^(2))/(d^(2)) pi underset(0)overset(R )int r^(2) rdr xx d xx V_(m)^(2) cos^(2) omegat` `=(1)/(16) pi epsilon_(0)^(2)mu_(0)(omega^(2)R^(4))/(d) V_(m)^(2)cos^(2) omegat` Thus the maximum magnetic enegry `W_(m) = (epsilon_(0)^(2)mu_(0))/(16)(omegaR)^(2) (piR^(2))/(d)V_(m)^(2)` Hence `(W_(m))/(W_(e)) = (1)/(8)epsilon_(0)mu_(0)(omegaR)^(2) = (1)/(8) ((omega R)/(c))^(2) = 5xx10^(-15)` The approximation are valid only if `omega R lt lt c`. |
|