$A$ capacitor, when filled with a dielectric $K = 3$, has charge $Q_0$, voltage $V_0$, and electric field $E_0$. If the dielectric is replaced with another one having $K = 9$, the new values of charge, voltage, and field will be respectively:

An air capacitor has capacitance $C_1$. The space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The new capacitance of the capacitor is $C_2$. The ratio $\frac{C_1}{C_2}$ is ($d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the two dielectrics respectively).

Two parallel plates have equal and opposite charges. When the space between them is evacuated,the electric field between the plates is $2 \times 10^5 \ V/m$. When the space is filled with a dielectric,the electric field becomes $1 \times 10^5 \ V/m$. Find the dielectric constant of the dielectric material.

The capacity of a parallel plate capacitor is $5\,\mu F$. When a glass plate is placed between the plates of the capacitor,its potential becomes $1/8^{th}$ of the original value. The value of the dielectric constant will be

$A$ capacitor is charged by using a battery which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of $V_0 \, V$. If one of the capacitors is completely immersed in a liquid with dielectric constant $K$,the potential difference between the plates of the other capacitor will change to: