The force between two point charges kept with a separation of $9 \ cm$ in air is $98 \ N$. If a dielectric slab of constant $4$,thickness $6 \ cm$ and another dielectric slab of constant $9$,thickness $3 \ cm$ are introduced between the two charges,then the new force becomes (in $N$)

$A$ parallel plate capacitor having plates of area $S$ and plate separation $d$,has capacitance $C_1$ in air. When two dielectrics of different relative permittivities $(\varepsilon_1=2$ and $\varepsilon_2=4)$ are introduced between the two plates as shown in the figure,the capacitance becomes $C_2$. The ratio $\frac{C_2}{C_1}$ is

$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

The plates of a parallel plate capacitor are separated by a distance $d.$ Two slabs of different dielectric constants $K_1$ and $K_2$ with thicknesses $\frac{3}{8} d$ and $\frac{d}{2}$ respectively are inserted into the capacitor. Due to this,the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25 K_2,$ find the value of $K_1.$

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:

$A$ parallel plate capacitor of capacitance $90 \ pF$ is connected to a battery of $emf$ $20 \ V$. If a dielectric material of dielectric constant $K = \frac{5}{3}$ is inserted between the plates,the magnitude of the induced charge will be.......$nC$.