Saved Bookmarks
| 1. |
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed . The switch is now opened and the free space between the plates of the capacitors is filled with dielectric of dielectric constant K . Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric . |
|
Answer» Solution :Here initially `C_(A) = C_(B) = C` and `V_(A) = V_(B) = V` `THEREFORE` Total initial electrostatic energy STORED in two capacitors `U= (1)/(2) C_A V_(A)^(2) + (1)/(2) C_(B) V_(B)^(2) = (1)/(2) CV^(2) + (1)/(2) CV^(2) = CV^(2) "" ….. (i)` When the space between the plates of the two capacitors is filled with a dielectric of dielectric constant K , capacitance of two capacitors changes to `C_(A) = C_(B) = KC` Now capacitor A is still connected to battery , hence `V._(A) = V_(A) = V` . However , capacitor B is disconnected from the battery , hence `V._(B) = (Q)/(C._(B)) = (CV)/(KC) = (V)/(K)` `therefore` Total final electrostatic energy stored in two capacitors `U. = 1/2 C._A V._A^2 + (1)/(2) C._B V._B^2 = (1)/(2) (KC) V^(2) + (1)/(2) (KC) ((V)/(K))^(2) = (1)/(2) KCV^(2) + (1)/(2) (CV^(2))/(K) = (1)/(2) CV^(2) (K + (1)/(K))"" .... (ii)` `IMPLIES (U)/(U.) = (CV^(2))/((1)/(2) CV^(2) (K + (1)/(K))) = (2)/(K + (1)/(K)) = (2K)/(K^(2) + 1)` |
|