While a capacitor remains connected to a battery and a dielectric slab is inserted between the plates,then:

$A$ capacitor with air as the dielectric is charged to a potential of $100\;V$. If the space between the plates is now filled with a dielectric of dielectric constant $K = 10$,the potential difference between the plates will be: (in $;V$)

$A$ capacitor has a capacitance of $15\ \mu F$ and a plate separation of $2\ mm$. If a dielectric slab of dielectric constant $K = 2$ and thickness $t = 1\ mm$ is inserted between the plates,what will be the new capacitance (in $\mu F$)?

$A$ parallel plate capacitor has a plate area of $100\, m^{2}$ and a plate separation of $10\, m$. The space between the plates is filled up to a thickness of $5\, m$ with a material of dielectric constant $10$. The resultant capacitance of the system is $'x'\, pF$. Given $\varepsilon_{0} = 8.85 \times 10^{-12} F \cdot m^{-1}$,the value of $'x'$ to the nearest integer is:

If $q_{f}$ is the free charge on the capacitor plates and $q_{b}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates,then the bound charge $q_{b}$ can be expressed as:

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)