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 :
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 (d) is Incorrect.
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