Answer:
Explanation:
Battery is removed so its charge stored Ā becomes fixed .
Energy of capacitor Eā = QĀ² / 2C
When dielectric is inserted , the capacity of the capacitor is increased to KC .
Charge on it still remains the same ie Q
So its new energy
Eā = QĀ² / 2 KC
Eā / Eā = 1 / K
Eā = Eā / K
Change in energy
= Eā - Eā
= Eā / K - Eā
= Eā [ (1 / K) Ā - 1 ]
Factor by which energy is changed
= [ (1 / K) Ā - 1 ] .
