$A$ parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now,an insulating slab of same area but thickness $t = d/2$ is inserted between the plates as shown in the figure,having dielectric constant $K = 4$. The ratio of the new capacitance to its original capacitance will be:

$A$ 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 . . . . . . .

The gap between the plates of a parallel plate capacitor of area $A$ and distance between plates $d$ is filled with a dielectric whose permittivity varies linearly from $\varepsilon_1$ at one plate to $\varepsilon_2$ at the other. The capacitance of the capacitor is

$A$ 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:

$A$ 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 plates of area $A$ separated by a distance $d$. If a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,what is the new capacitance?