$A$ parallel plate air capacitor, with plate separation $d$, has a capacitance of $9 \text{ pF}$. The space between the plates is now filled with two dielectrics, the first having $K_1=3$ and thickness $d_1=d/3$, while the second has $K_2=6$ and thickness $d_2=2d/3$. The capacitance of the new capacitor is: (in $\text{ pF}$)

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \,F$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \,V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

$A$ parallel plate capacitor of capacitance $200 \,\mu F$ is connected to a battery of $200 \,V$. $A$ dielectric slab of dielectric constant $2$ is now inserted into the space between the plates of the capacitor while the battery remains connected. The change in the electrostatic energy in the capacitor will be ...... $J$.

$A$ parallel plate capacitor is connected to a battery. The quantities charge,voltage,electric field,and energy associated with the capacitor are given by $Q_0, V_0, E_0$,and $U_0$ respectively. $A$ dielectric slab is introduced between the plates of the capacitor,but the battery remains connected. The corresponding quantities now given by $Q, V, E$,and $U$ related to the previous ones are:

The capacitance of a parallel plate capacitor with air as the medium is $3 \mu F$. With the introduction of a dielectric medium between the plates,the capacitance becomes $15 \mu F$. The permittivity of the medium in $SI$ units is: [Given: $\epsilon_{0} = 8.85 \times 10^{-12} \text{ SI units}$]

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled as shown in the figure with a dielectric constant $K = 5$, the percentage increase in the capacitance is . . . . . . .