A capacitor when filled with a dielectric $K = 3$ has charge ${Q_0}$, voltage ${V_0}$ and field ${E_0}$. If the dielectric is replaced with another one having $K = 9$ the new values of charge, voltage and field will be respectively

Two identical parallel plate capacitors, of capacitance $C$ each, have plates of area $A$, separated by a distance $d$. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants $K_1$ , $K_2$ and $K_3$ . The first capacitor is filled as shown in fig. $I$, and the second one is filled as shown in fig. $II$. If these two modified capacitors are charged by the same potential $V$, the ratio of the energy stored in the two, would be ( $E_1$ refers to capacitor $(I)$ and $E_2$ to capacitor $(II)$) 

There is an air filled $1\,pF$ parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to $2\,pF$. The dielectric constant of wax is

The capacity of an air condenser is $2.0\, \,\mu F$. If a medium is placed between its plates. The capacity becomes $ 12\, \,\mu F$. The dielectric constant of the medium will be

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in

Following operations can be performed on a capacitor : $X$ - connect the capacitor to a battery of $emf$ $E.$ $Y$ - disconnect the battery $Z$ - reconnect the battery with polarity reversed. $W$ - insert a dielectric slab in the capacitor