The distance between the plates of a parallel plate capacitor is $0.05\, m$. An electric field of $3 \times 10^4\, V/m$ is established between the plates. The capacitor is disconnected from the battery,and an uncharged metal plate of thickness $0.01\, m$ is inserted. If a dielectric slab of dielectric constant $K = 2$ is inserted instead of the metal plate,what will be the potential difference in $kV$?

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

$A$ parallel plate capacitor having a separation between the plates $d$,plate area $A$,and material with dielectric constant $K$ has capacitance $C_0$. Now,one-third of the material is replaced by another material with dielectric constant $2K$,such that effectively there are two capacitors: one with area $\frac{1}{3}A$,dielectric constant $2K$,and another with area $\frac{2}{3}A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$,then $\frac{C}{C_0}$ is:

When a dielectric slab is introduced between the plates of a parallel plate capacitor that is connected to a battery,the new charge on the plates is:

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

$A$ parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $1/3$ of the area of its plates,as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_1$. When the capacitor is charged,the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. Choose the correct option/options,ignoring edge effects.
$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{K}{2}$ $(D)$ $\frac{C}{C_1}=\frac{2+K}{K}$