$A$ parallel plate air capacitor has a capacitance of $100\,\mu F$. The plates are at a distance $d$ apart. If a slab of thickness $t$ $(t < d)$ and dielectric constant $K = 5$ is introduced between the parallel plates,then the new capacitance will be:

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \, pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1 = 3$ and thickness $d/3$,while the other one has dielectric constant $K_2 = 6$ and thickness $2d/3$. The capacitance of the capacitor is now.........$pF$.

$A$ parallel plate capacitor with width $4\,cm$,length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. $A$ dielectric slab of dielectric constant $5$ having length $1\,cm$,width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of the parallel plate capacitor. The electrostatic energy of this system will be ......... $\epsilon_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)

Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is:

$A$ parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. $A$ dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor,is now inserted in it. Which of the following is incorrect $?$

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant $K_1 = 6$ and thickness $\frac{d}{3}$,while the other one has a dielectric constant $K_2 = 12$ and thickness $\frac{2d}{3}$. The capacitance of the capacitor is now ......... $pF$.