If half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $4$,the capacitance is $C_1$. If one third of the space between the plates of the capacitor is filled with the medium of dielectric constant $4$,the capacitance is $C_2$. If in both cases,the dielectric is placed parallel to the plates of the capacitor,then $C_1: C_2=$

$A$ parallel plate capacitor with square plates of side $L$ is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \ pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has a dielectric constant $K_1 = 6$ and thickness $\frac{d}{3}$,while the other one has a dielectric constant $K_2 = 12$ and thickness $\frac{2d}{3}$. The capacitance of the capacitor is now ......... $pF$.

If a capacitor with air,a capacitor with a dielectric,and a capacitor with a conducting slab have capacitances $C_1$,$C_2$,and $C_3$ respectively,then:

An air-filled parallel plate capacitor has a capacity of $2 \ pF$. If the separation between the plates is doubled and the interspace between the plates is filled with a dielectric material,the capacity increases to $6 \ pF$. The dielectric constant of the material is:

$A$ parallel plate capacitor has a capacitance of $5 \ \mu F$. When a glass plate is inserted between the plates of the capacitor,its potential becomes $1/8$ of its original value. What is the value of the dielectric constant?