$A$ capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

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:

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant $k_1 = 3$ and thickness $d/3$,while the other one has a dielectric constant $k_2 = 6$ and thickness $2d/3$. The capacitance of the capacitor is now . . . . . . $pF$.

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

$A$ 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?

$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel plate capacitor but has a thickness $(4/5)d$, where $d$ is the separation of the plates. The capacitance in the presence and absence of dielectric are $C$ and $C_0$ respectively. The ratio $(C/C_0)$ is