$A$ parallel plate capacitor has plates of area $A$ separated by a distance $d$. If a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,what is the new capacitance?

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

Assertion: The electrostatic force between the plates of a charged isolated capacitor decreases when a dielectric fills the whole space between the plates.
Reason: The electric field between the plates of a charged isolated capacitor increases when a dielectric fills the whole space between the plates.

$A$ parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. $A$ dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$

The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in the figure,will be (area of plate $= A$)

Two metal plates each of area $A$ form a parallel plate capacitor with air in between the plates. The distance between the plates is $d$. $A$ metal plate of thickness $\frac{d}{2}$ and of same area $A$ is inserted between the plates to form two capacitors of capacitances $C_1$ and $C_2$ as shown in the figure. If the effective capacitance of the two capacitors is $C^{\prime}$ and the capacitance of the capacitor initially is $C$,then $\frac{C^{\prime}}{C}$ is