$A$ 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?

When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes:

$A$ parallel plate air capacitor, with plate separation $d$, has a capacitance of $9 \text{ pF}$. The space between the plates is now filled with two dielectrics, the first having $K_1=3$ and thickness $d_1=d/3$, while the second has $K_2=6$ and thickness $d_2=2d/3$. The capacitance of the new capacitor is: (in $\text{ pF}$)

$A$ parallel plate capacitor has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)

When a parallel plate capacitor is charged up to $95 \ V$,its capacitance is $C$. If a dielectric slab of thickness $2 \ mm$ is inserted between the plates and the distance between the plates is increased by $1.6 \ mm$ such that the same potential difference is maintained,the dielectric constant of the material (slab) is: