$A$ parallel plate capacitor with air as dielectric has a capacitance of $4 \mu F$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $K = 5$ and charged to a potential of $100 \ V$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (in $J$)

In a parallel plate capacitor,two dielectric slabs of thicknesses $t_1$ and $t_2$ and dielectric constants $K_1$ and $K_2$ respectively are inserted. What is the capacitance of this capacitor?

$A$ parallel plate capacitor is formed by two plates each of area $30 \pi \, cm^{2}$ separated by $1 \, mm$. $A$ material of dielectric strength $3.6 \times 10^{7} \, Vm^{-1}$ is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is $7 \times 10^{-6} \, C$,the value of the dielectric constant of the material is. $\{ \text{Use} : \frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \, Nm^{2}C^{-2} \}$

$A$ slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$,where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be (Given $C_{0} =$ capacitance of capacitor with air as medium between plates):

In a parallel plate capacitor,the separation between plates is $3x$. This separation is filled by two layers of dielectrics,in which one layer has thickness $x$ and dielectric constant $3k$,and the other layer has thickness $2x$ and dielectric constant $5k$. If the plates of the capacitor are connected to a battery,then the ratio of the potential difference across the dielectric layers is

$A$ parallel plate capacitor is charged by a battery and the battery remains connected. $A$ dielectric slab of constant $K$ is inserted between the plates and then taken out. What happens to the electric field between the plates?