The distance between plates of a parallel plate capacitor is $5d$. Let the positively charged plate is at $ x=0$ and negatively charged plate is at $x=5d$. Two slabs one of conductor and other of a dielectric of equal thickness $d$ are inserted between the plates as shown in figure. Potential versus distance graph will look like :

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

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 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

Dielectric constant for metal is

A dielectric slab of dielectric constant $K$ is placed between the plates of a parallel plate capacitor carrying charge $q$. The induced charge $q^{\prime}$ on the surface of slab is given by