A parallel plate capacitor is charged to a potential difference of $100\,V$ and disconnected from the source of $emf$ . A slab of dielectric is then inserted between the plates. Which of the following three quantities change?
$(i)$ The potential difference
$(ii)$ The capacitance
$(iii)$ The charge on the plates
Medium
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The charge of the plates cannot change because the plates are isolated. Inserting the dielectric changes the capacitance and therefore changes the potential difference (from $\mathrm{V}=\mathrm{Q} / \mathrm{C}$ ), since the charge is fixed.
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$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled ${E_1}$
$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is ${E_2}.$
Then ${E_1}/{E_2}$ value is