When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?

Two capacitors $C_1$ and $C_2$ are connected to a battery as shown in the figure. The space between the plates of $C_1$ is filled with air,and the space between the plates of $C_2$ is filled with a dielectric material. Which of the following is true regarding the charges $Q_1$ and $Q_2$ on the capacitors?

The plates of a parallel plate capacitor are separated by a distance $d$ with air as the medium between them. $A$ dielectric slab of dielectric constant $K = 3$ is introduced between the plates so as to increase the capacity by $50 \%$. The thickness of the dielectric slab is

$A$ 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

In the given circuit,two identical capacitors $C_1$ and $C_2$ are connected to a battery. The space between the plates of $C_1$ is filled with air,while the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K$. 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