$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

$A$ parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now,an insulating slab of same area but thickness $t = d/2$ is inserted between the plates as shown in the figure,having dielectric constant $K = 4$. The ratio of the new capacitance to its original capacitance will be:

When a dielectric slab is introduced between the plates of a parallel plate capacitor that is connected to a battery,the new charge on the plates is:

$A$ capacitor with plate separation $d$ is charged to $V$ volts. The battery is disconnected and a dielectric slab of thickness $\frac{d}{2}$ and dielectric constant $K=2$ is inserted between the plates. The potential difference across its terminals becomes

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

Two capacitors $C_1$ and $C_2$ are connected to a battery as shown in the figure. The space between the plates of $C_1$ is filled with air,and the space between the plates of $C_2$ is filled with a dielectric material. Which of the following is true regarding the charges $Q_1$ and $Q_2$ on the capacitors?