An air capacitor is connected to a battery. The effect of filling the space between the plates with a dielectric is to increase:

$A$ 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

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance:

$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced is

$A$ parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$,its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is