A parallel plate capacitor filled with a medium of dielectric constant $10$ , is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant $15$ . Then the energy of capacitor will ......................

The plates of a parallel plate capacitor are charged up to $100 \,volt$ . A $2 \,mm$ thick plate is  inserted between the plates, then to maintain the same potential difference, the distance  between the capacitor plates is increased by $1.6\, mm$. The dielectric constant of the plate  is :-

In a parallel plate condenser, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. There is a glass slab of $3\,mm$ thick and of radius $12\,cm$ with dielectric constant $6$ between its plates. The capacity of the condenser will be

How does the polarised dielectric modify the original external field inside it ?

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 $(II)$) 

Two capacitors, each having capacitance $40\,\mu F$ are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant $K$ such that the equivalence capacitance of the system became $24\,\mu F$. The value of $K$ will be.