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

Two identical capacitors $1$ and $2$ are connected in series. The capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0\ volts$ . The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charge stored in the capacitors before removing the slab and $Q'_1$ , and $Q'_2$ be the values after removing the slab. Then 

A parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. A dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect $?$

The distance between the plates of a parallel plate capacitor is $d$. A metal plate of thickness $d/2$ is placed between the plates. The capacitance would then be

A parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$ its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is

Two parallel metal plates having charges $+Q$ and $- Q$ face each other at a certain distance between them. If the plates are now dipped in kerosene oil tank, the electric field between the plates will