$A$ parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$,the dielectric constant of the medium is

In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.

$A$ 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 has plates of area $A$ and a separation distance of $100 \ mm$. It contains two dielectric layers: one with a dielectric constant of $10$ and a thickness of $6 \ mm$,and another with a dielectric constant of $5$ and a thickness of $4 \ mm$. Find the capacitance of the capacitor.

$A$ parallel plate capacitor has a capacitance of $50\,\mu F$ in air and $110\,\mu F$ when immersed in an oil. The dielectric constant $K$ of the oil is: