A) The potential difference between the plates decreases K times
B) The energy stored in the capacitor decreases K times
C) The change in energy stored is \[\frac{1}{2}C{{V}^{2}}\left( \frac{1}{K}-1 \right)\]
D) The charge on the capacitor is not conserved
Correct Answer: D
Solution :
| [d] When a parallel plate air capacitor connected to a cell of emf V, then charge stored will be |
| \[q=CV\] |
| \[\Rightarrow \] \[V=\frac{q}{C}\] |
| Also energy stored is \[U=\frac{1}{2}C{{V}^{2}}=\frac{{{q}^{2}}}{2C}\] |
| As the battery is disconnected from the capacitor the charge will not be destroyed i.e. \[q'=q\] with the introduction of dielectric in the gap of capacitor the new capacitance will be |
| \[C'=CK\] |
| \[\Rightarrow \] \[V'=\frac{q}{C'}=\frac{q}{CK}\] |
| The new energy stored will be |
| \[U'=\frac{{{q}^{2}}}{2CK}\] |
| \[\Delta U=U'-U=\frac{{{q}^{2}}}{2C}\left( \frac{1}{K}-1 \right)\] |
| \[=\frac{1}{2}C{{V}^{2}}\left( \frac{1}{K}-1 \right)\] |
| Hence, option is Incorrect. |
You need to login to perform this action.
You will be redirected in
3 sec
