$A$ parallel plate capacitor of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

$A$ parallel-plate capacitor of area $A,$ plate separation $d$ and capacitance $C$ is filled with four dielectric materials having dielectric constants $K_1, K_2, K_3$ and $K_4$ as shown in the figure. If a single dielectric material is to be used to have the same capacitance $C$ in this capacitor,then its dielectric constant $K$ is given by

The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. 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:

$A$ charged conductor produces an electric field of intensity $10^3 \ V/m$ just outside its surface in vacuum. Then,it produces an electric field of intensity $E$ just outside its surface,when it is placed in a medium of dielectric constant $4$. The value of $E$ will be (in $V/m$)

$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

Consider a parallel plate capacitor with plates in the shape of a square in the $XY$-plane. The gap between the plates is filled with a dielectric material. The dielectric constant $k$ varies along the $X$-axis as $k(x) = \left[1 + \left(\frac{x}{L}\right)^\alpha\right]$,where $\alpha$ is a constant. Let $C_d$ and $C_a$ be the capacitance in the presence of the dielectric and air,respectively. If the ratio $\frac{C_d}{C_a} = \frac{7}{6}$,then the value of $\alpha$ must be