$A$ capacitor is made of a thin metal strip. Its capacitance is $2 \ \mu F$. If the space between the strips is filled with paper of thickness $0.15 \ mm$,dielectric constant $2.5$,and width $400 \ mm$,what is the length of the strip in meters?

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

The capacitance of a capacitor becomes $\frac{7}{6}$ times the original value if a dielectric slab of thickness $t = \frac{2d}{3}$ is introduced between the plates, where $d$ is the distance of separation between the plates. The dielectric constant of the slab is

Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$,then the amount of charge which will flow through the battery is (neglect internal resistance of the battery).

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 becomes $1/8^{th}$ of the original value. The value of the dielectric constant will be

In one design of a capacitor,thin sheets of metal of area $80 \ mm \times 80 \ mm$ sandwich between them a piece of paper whose thickness is $40 \ \mu m$. The relative permittivity of the paper is $4.0$ and its dielectric strength is $20 \ MVm^{-1}$. Calculate the maximum charge that can be put on the capacitor. [Permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \ Fm^{-1}$]