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 distance between the plates is $d$. Now,the space between the plates is filled with two dielectrics. One dielectric has a dielectric constant $K_1 = 3$ and thickness $d/3$,while the other has a dielectric constant $K_2 = 6$ and thickness $2d/3$. Find the new capacitance of the capacitor in $pF$.

$A$ parallel plate capacitor having air as the dielectric medium is charged by a potential difference of $V$ volt. After disconnecting the battery,the distance between the plates of the capacitor is increased using an insulated handle. As a result,the potential difference between the plates . . . . . . .

$A$ parallel plate capacitor with a dielectric slab completely occupying the space between the plates is charged by a battery and then disconnected. The slab is pulled out with a constant speed. Which of the following curves represent qualitatively the variation of the capacitance $C$ of the system with time?

$A$ parallel plate capacitor has plate area $40 \ cm^2$ and plate separation $2 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and dielectric constant $5$. The capacitance of the system is ($\varepsilon_0 =$ permittivity of vacuum)

$A$ parallel plate capacitor has a dielectric slab of dielectric constant $K$ between its plates that covers $1/3$ of the area of its plates,as shown in the figure. The total capacitance of the capacitor is $C$ while that of the portion with dielectric in between is $C_1$. When the capacitor is charged,the plate area covered by the dielectric gets charge $Q_1$ and the rest of the area gets charge $Q_2$. Choose the correct option/options,ignoring edge effects.
$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{K}{2}$ $(D)$ $\frac{C}{C_1}=\frac{2+K}{K}$