Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant of a material,which when fully filled between the plates produces the same capacitance?

The figure shows two identical parallel plate capacitors connected to a battery and a closed switch $S$. Now,the switch is opened and a dielectric material with dielectric constant $K = 3$ is inserted into the free space between the plates of both capacitors. What is the ratio of the total electrostatic energy stored in both capacitors before and after the insertion of the dielectric?

$A$ parallel plate capacitor is formed by two plates each of area $30 \pi \, cm^{2}$ separated by $1 \, mm$. $A$ material of dielectric strength $3.6 \times 10^{7} \, Vm^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6} \, C$,the value of the dielectric constant of the material is. $\{ \text{Use} : \frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, Nm^{2}C^{-2} \}$

Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes: