$A$ 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 $\text{ pF}$)

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?

An air capacitor of capacity $C = 10\,\mu F$ is connected to a constant voltage battery of $12\,V$. Now,the space between the plates is filled with a liquid of dielectric constant $K = 5$. The charge that flows from the battery to the capacitor is......$\mu C$.

$A$ parallel plate capacitor is made of two square parallel plates of area $A$,and separated by a distance $d << \sqrt{A}$. The capacitor is connected to a battery with potential $V$ and allowed to fully charge. The battery is then disconnected. $A$ square metal conducting slab also with area $A$ but thickness $d/2$ is then fully inserted between the plates,so that it is always parallel to the plates. How much work has been done on the metal slab by an external agent while it is being inserted?

The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.