$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K = 4$. The thickness of the dielectric material is $x$,where $x < d$.
Let $C_1$ and $C_2$ be the capacitance of the system for $x = \frac{d}{3}$ and $x = \frac{2d}{3}$,respectively. If $C_1 = 2 \mu F$,the value of $C_2$ is $........... \mu F$.

The separation between the plates of a parallel plate capacitor is $d$ and the area of each plate is $A$. When a slab of material of dielectric constant $k$ and thickness $t$ $(t < d)$ is introduced between the plates,its capacitance becomes

The plates of a parallel plate capacitor are separated by a distance $d.$ Two slabs of different dielectric constants $K_1$ and $K_2$ with thicknesses $\frac{3}{8} d$ and $\frac{d}{2}$ respectively are inserted into the capacitor. Due to this,the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25 K_2,$ find the value of $K_1.$

In a parallel plate capacitor setup,the plate area of the capacitor is $2 \, m^{2}$ and the plates are separated by $1 \, m$. If the space between the plates is filled with a dielectric material of thickness $0.5 \, m$ and area $2 \, m^{2}$ (see figure),the capacitance of the setup will be $......... \, \varepsilon_{0}$. (Dielectric constant of the material $= 3.2$) (Round off to the nearest integer).

$A$ parallel plate capacitor with square plates of side $L$ is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be

$A$ parallel plate capacitor is connected to a battery. If a dielectric slab is inserted between the plates,which of the following quantities increases?