A parallel plate capacitor is formed by two plates each of area $30 \pi\, cm ^{2}$ separated by $1\, mm$. A material of dielectric strength $3.6 \times 10^{7} \,Vm ^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6}\, C$, the value of dielectric constant of the material is

$\left\{ Use : \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right\}$

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

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \mu F$ are individually charged by a $100\,V$ $D.C.$ source. Capacitor $C _1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C _2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be $………V.$ (Assuming Dielectric constant $=10$ )

In the reported figure, a capacitor is formed by placing a compound dielectric between the plates of parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $=A$ )

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