The space between the plates of a parallel plate capacitor is halved and a dielectric medium of relative permittivity $10$ is introduced between the plates. The ratio of the final and initial capacitances of the capacitor is

The capacity of a parallel plate capacitor is $10\,\mu F$ without a dielectric. If a dielectric of constant $K = 2$ is used to fill half the distance between the plates,the new capacitance in $\mu F$ is:

Three parallel plate capacitors $C_1, C_2$ and $C_3$ each of capacitance $5 \mu F$ are connected as shown in the figure. The effective capacitance between points $A$ and $B$,when the space between the parallel plates of $C_1$ capacitor is filled with a dielectric medium having a dielectric constant of $4$,is (in $\mu F$)

$A$ 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:

$A$ 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?

$A$ parallel plate capacitor is constructed with two plates. Each plate has an area of $100 \ cm^2$ and they are separated by a distance of $1 \ mm$. $A$ dielectric with a dielectric constant of $5.0$ and a dielectric strength of $1.9 \times 10^7 \ V/m$ is filled between the plates. Find the maximum charge that can be stored on the capacitor without causing dielectric breakdown.