If a slab of insulating material $4 \times {10^{ - 3}}\,m$ thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by $3.5 \times {10^{ - 3}}\,m$ to restore the capacity to original value. The dielectric constant of the material will be

A parallel plate capacitor has two layers of dielectrics as shown in fig. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \mathrm{~g} / \mathrm{cc}$, the dielectric constant of water will be

(Take density of water $=1 \mathrm{~g} / \mathrm{cc}$ )

A parallel plate capacitor having plates of area $S$ and plate separation $d$, has capacitance $C _1$ in air. When two dielectrics of different relative permittivities $\left(\varepsilon_1=2\right.$ and $\left.\varepsilon_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C _2$. The ratio $\frac{ C _2}{ C _1}$ is

Three capacitors $A,B$ and $C$ are connected with battery $emf\, \varepsilon $. All capacitors are identical initially. If dielectric slab is inserted between plates of capacitor $A$ slowy with help of external force then 

Putting a dielectric substance between two plates of condenser, capacity, potential and potential energy respectively