The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

$A$ parallel plate capacitor having capacitance $C$ farad is connected to a battery of emf $V$ volt. Keeping the capacitor connected to the battery,a dielectric slab of dielectric constant $K$ is inserted between the plates. The dimensions of the slab are such that it fills the whole space between the capacitor plates. Then $:-$

For changing the capacitance of a given parallel plate capacitor,a dielectric material of dielectric constant $K$ is used,which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$,where $d$ is the separation between the plates of the parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $(C_{0})$ is given by the following relation:

$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?

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

$A$ parallel plate capacitor with air between the plates has a capacitance of $12 \mu F$. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $4$,what will be the new capacitance of the capacitor (in $\mu F$)?