If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become

A capacitor stores $60\ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric , a charge of $120\ \mu C$ flows through the battery , if the initial capacitance of the capacitor was $2\ \mu F$, the amount of heat produced when the dielectric is inserted.......$\mu J$

A medium having dielectric constant $K>1$ fills the space between the plates of a parallel plate capacitor. The plates have large area, and the distance between them is $d$. The capacitor is connected to a battery of voltage $V$. as shown in Figure ($a$). Now, both the plates are moved by a distance of $\frac{d}{2}$ from their original positions, as shown in Figure ($b$).

In the process of going from the configuration depicted in Figure ($a$) to that in Figure ($b$), which of the following statement($s$) is(are) correct?

A parallel plate capacitor has plate of length $'l',$ width $'w'$ and separation of plates is $'d'.$ It is connected to a battery of emf $V$. A dielectric slab of the same thickness '$d$' and of dielectric constant $k =4$ is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored$?$

There is an air filled $1\,pF$ parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to $2\,pF$. The dielectric constant of wax is

A parallel plate capacitor, partially filled with a dielectric slab of dielectric constant $K$ , is connected with a cell of emf $V\ volt$ , as shown in the figure. Separation between the plates is $D$ . Then