$A$ parallel plate capacitor has two layers of dielectrics as shown in the figure. This capacitor is connected across a battery. The ratio of the potential difference across the dielectric layers $(V_1/V_2)$ is:

$A$ slab of material with a dielectric constant of $3$ has the same area as the plates of a parallel plate capacitor but has a thickness of $\left(\frac{3}{4}\right) d$,where $d$ is the separation between the plates. What is the electrical potential difference between the plates when the slab is inserted? The initial electrical potential difference is $V_0$.

$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$. $A$ metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_1$ and $C_2$ as shown in the figure. If the effective capacitance of the two capacitors is $C^{\prime}$ and the capacitance of the capacitor initially is $C$,then $\frac{C^{\prime}}{C}$ is

Two capacitors $C$ and $2C$ are connected in parallel and charged to a potential $V$ by a battery. The battery is then removed. The capacitor $C$ is filled with a dielectric of constant $K$. What is the new potential difference across the capacitors?

$A$ parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates,then: