In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

$A$ 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

$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

$A$ sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

When a dielectric material is introduced between the plates of a charged capacitor,the electric field between the plates:

Which one statement is correct? $A$ parallel plate air condenser is connected to a battery. Its charge,potential,electric field,and energy are ${Q_0}$,${V_0}$,${E_0}$,and ${U_0}$ respectively. $A$ dielectric slab is inserted to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as: