Saved Bookmarks
| 1. |
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1//2) QE, where Q is the charge on the capacitor , and E is the magnitude of electric field between the plates . Explain the origin of the factor 1//2. |
|
Answer» Solution :If F is the force on each plates of parallel plate CAPACITOR, then work done in increasing the seperation between the plates by `Delta x = f Delta x` This must be the increase in potential energy of the capacitor Now the increase the volume of capacitor is `= A Delta x` If U = energy DENSITY = energy stroed/volume then the increase in potential energy `= U.A Delta X` `:. f Delta X = U. A Delta X` `f = U.A = ((1)/(2) epsilon_(0) epsilon^(2)) A = (1)/(2) (epsilon_(0) A epsilon) epsilon` `F = (1)/(2) (epsilon_(0) A (V)/(d)) epsilon [ :' C = (epsilon_(0) A)/(d)]` `F = (1)/(2) (CV) E = (1)/(2) QE` The origin of factor 1/2 in force can be explained by the fact that inside the conductor field is zero and outside the conductor, the field is E. Therefore the AVERAGE value of the field (i.e E/2) contributes to the force. |
|