Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

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 $(II)$):

The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. The separation between the plates is doubled and the space between the plates is then filled with wax,giving the capacitance a new value of $40 \times 10^{-12} \, F$. The dielectric constant of wax is:

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

$A$ parallel plate capacitor of capacitance $90 \ pF$ is connected to a battery of $emf$ $20 \ V$. If a dielectric material of dielectric constant $K = \frac{5}{3}$ is inserted between the plates,the magnitude of the induced charge will be.......$nC$.