An air capacitor has a capacitance of $1 \mu F$. Now,the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is ($d=$ distance between two plates,$K_1=8$ and $K_2=4$ are the dielectric constants of the two dielectrics respectively).

Two identical capacitors $A$ and $B$ are connected as shown in the circuit. Initially,the switch $S$ is closed. Now,the switch is opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant $K = 3$. The ratio of total electrostatic energy stored in the capacitors before and after the introduction of the dielectric is

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants \(K _1=2, K_2=3\), and \(K _3=5\) (as shown in the figure). If new capacitance is \(\frac{ n }{3} C\) then the value of \(n\) is ___________.

The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).

$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

$A$ metal plate of thickness $2 \,mm$ and area $36 \pi \,mm^2$ is slid into a parallel plate capacitor of plate spacing $6 \,mm$ and area $36 \pi \,cm^2$. The metal plate is at a distance $3 \,mm$ from one of the plates. What is the capacitance of this arrangement (in $\,pF$)? (Let $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \,N m^2 C^{-2}$)