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 $(II)$) 

A parallel plate condenser is connected with the terminals of a battery. The distance between the plates is $6\,mm$. If a glass plate (dielectric constant $K = 9$) of $4.5\,mm$ is introduced between them, then the capacity will become…….$times$

A parallel plate capacitor is made of two plates of length $l$, width $w$ and separated by distance $d$. A dielectric slab ( dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{{\partial U}}{{\partial x}}$ where $U$ is the energy of the capacitor when dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$ then the force on the dielectric when it is near the edge is

A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ……. $\times 10^{-4}\, m ^{2}$

A capacitor is kept connected to the battery and a dielectric slab is inserted between the plates. During this process