The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. To maintain the same potential difference,the distance between the capacitor plates is increased by $1.6\,mm$. The dielectric constant of the plate is:

If the distance between parallel plates of a capacitor is halved and the dielectric constant is doubled,then the capacitance will become:

$A$ parallel plate capacitor has plates of area $0.4 \pi \,m^2$ and spacing of $0.5 \,mm$. If a slab of thickness $0.5 \,mm$ and dielectric constant $4.5$ is introduced in between the plates of the capacitor, then the capacitance of the capacitor is

$A$ parallel plate capacitor with plate separation $5 \ mm$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \ mm$,while keeping the battery connections intact,the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is . . . . . . .

$A$ parallel plate capacitor with plate area $A$ and distance of separation $d$ is filled with a dielectric. What is the capacity of the capacitor when permittivity of the dielectric varies as :
$\varepsilon(x) = \varepsilon_{0} + kx$,for $(0 < x \leq d/2)$
$\varepsilon(x) = \varepsilon_{0} + k(d-x)$,for $(d/2 \leq x \leq d)$

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant $k_1 = 3$ and thickness $d/3$,while the other one has a dielectric constant $k_2 = 6$ and thickness $2d/3$. The capacitance of the capacitor is now . . . . . . $pF$.