A parallel plate capacitor is charged to a potential difference of $100\,V$ and disconnected from the source of $emf$ . A slab of dielectric is then inserted between the plates. Which of the following three quantities change?

$(i)$  The potential difference

$(ii)$ The capacitance

$(iii)$ The charge on the plates

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, igonoring edge effects.

$(A)$ $\frac{E_1}{E_2}=1$ $(B)$ $\frac{E_1}{E_2}=\frac{1}{K}$ $(C)$ $\frac{Q_1}{Q_2}=\frac{3}{K}$ $(D)$ $\frac{ C }{ C _1}=\frac{2+ K }{ K }$

If the dielectric constant and dielectric strength be denoted by $k$ and $x$ respectively, then a material suitable for use as a dielectric in a capacitor must have

Between the plates of a parallel plate condenser, a plate of thickness ${t_1}$ and dielectric constant ${k_1}$ is placed. In the rest of the space, there is another plate of thickness ${t_2}$ and dielectric constant ${k_2}$. The potential difference across the condenser will be

The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be