The figure shows two identical parallel plate capacitors connected to a battery and a closed switch $S$. Now,the switch is opened and a dielectric material with dielectric constant $K = 3$ is inserted into the free space between the plates of both capacitors. What is the ratio of the total electrostatic energy stored in both capacitors before and after the insertion of the dielectric?

The capacity of an air-filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric of constant $K$ as shown in the figure. The new capacity becomes $C_n$. The ratio of $C_n$ to $C_0$ is:

The capacitance of a parallel plate capacitor is $5\, \mu F$. When a glass slab of thickness equal to the separation between the plates is introduced between the plates,the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is

If a slab of insulating material $4 \times 10^{-3} \ m$ thick is introduced between the plates of a parallel plate capacitor,the separation between the plates has to be increased by $3.5 \times 10^{-3} \ m$ to restore the capacity to its original value. The dielectric constant of the material will be:

$A$ parallel plate capacitor is made of two square parallel plates of area $A$,and separated by a distance $d << \sqrt{A}$. The capacitor is connected to a battery with potential $V$ and allowed to fully charge. The battery is then disconnected. $A$ square metal conducting slab also with area $A$ but thickness $d/2$ is then fully inserted between the plates,so that it is always parallel to the plates. How much work has been done on the metal slab by an external agent while it is being inserted?

$A$ 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 $(d/2 \leq x \leq d)$