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?

After charging a capacitor,the battery is removed. Now,by placing a dielectric slab between the plates :-

$A$ parallel plate capacitor is charged to a potential of $100 \ V$. $A$ dielectric slab of thickness $2 \ mm$ is inserted between the plates. To maintain the same potential difference,the distance between the plates is increased by $1.6 \ mm$. The dielectric constant of the slab is:

Two parallel plates with dielectric slabs placed between them are as shown in the figure. The resultant capacity of the capacitor will be: [$A$ = area of plate,$t_1, t_2, t_3$ are thicknesses of dielectric slabs,$k_1, k_2, k_3$ are dielectric constants.]

$A$ parallel plate capacitor having capacity $C_0$ is charged to $V_0$. With the battery disconnected,if the separation between the plates is doubled,then the energy stored in it is $E_1$. Instead,if the separation between the plates is doubled with the battery in connection,the energy stored in it is $E_2$. Then the value of $\frac{E_2}{E_1}$ is

Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in the figure. Capacitor $2$ contains a dielectric slab of dielectric constant $k$ as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q'_1$ and $Q'_2$. Then: