In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

$A$ parallel plate air capacitor has a capacitance of $100\,\mu F$. The plates are at a distance $d$ apart. If a slab of thickness $t$ $(t < d)$ and dielectric constant $K = 5$ is introduced between the parallel plates,then the new capacitance will be:

$A$ parallel plate capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to:

$A$ parallel plate capacitor is made of two plates of length $l$,width $w$ and separated by distance $d$. $A$ dielectric slab (dielectric constant $K$) that fits exactly between the plates is held near the edge of the plates. It is pulled into the capacitor by a force $F = -\frac{\partial U}{\partial x}$ where $U$ is the energy of the capacitor when the dielectric is inside the capacitor up to distance $x$ (See figure). If the charge on the capacitor is $Q$,then the force on the dielectric when it is near the edge is

When a dielectric material is introduced between the plates of a charged capacitor,the electric field between the plates:

The capacity of a parallel plate capacitor is $5 \mu F$. When a glass plate is placed between the plates of the capacitor,its potential difference reduces to $1/8$ of the original value. The magnitude of the relative dielectric constant of the glass is