Assertion : If the distance between parallel plates of a capacitor is halved and the dielectric constant is increased to three times its original value,then the capacitance becomes $6$ times.
Reason : The capacity of a capacitor does not depend upon the nature of the material between the plates.

The distance between the plates of a parallel plate capacitor is $0.05\, m$. An electric field of $3 \times 10^4\, V/m$ is established between the plates. The capacitor is disconnected from the battery,and an uncharged metal plate of thickness $0.01\, m$ is inserted. If a dielectric slab of dielectric constant $K = 2$ is inserted instead of the metal plate,what will be the potential difference in $kV$?

$A$ parallel plate capacitor is charged and then disconnected from the source of potential difference. If the plates of the capacitor are then moved farther apart by the use of an insulated handle,which one of the following is true?

$A$ 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.
$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{K}{2}$ $(D)$ $\frac{C}{C_1}=\frac{2+K}{K}$

$A$ parallel plate capacitor with air between the plates has a capacitance $C$. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant $6$,then the capacitance will become

$A$ parallel plate capacitor is constructed with two plates. Each plate has an area of $100 \ cm^2$ and they are separated by a distance of $1 \ mm$. $A$ dielectric with a dielectric constant of $5.0$ and a dielectric strength of $1.9 \times 10^7 \ V/m$ is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing dielectric breakdown.