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 $\left(\varepsilon_1=2\right.$ and $\left.\varepsilon_2=4\right)$ are introduced between the two plates as shown in the figure, the capacitance becomes $C _2$. The ratio $\frac{ C _2}{ C _1}$ is
$6 / 5$
$5 / 3$
$7 / 5$
$7 / 3$
If the distance between parallel plates of a capacitor is halved and dielectric constant is doubled then the capacitance will become
The capacity of a parallel plate condenser is $5\,\mu F$. When a glass plate is placed between the plates of the conductor, its potential becomes $1/8^{th}$ of the original value. The value of dielectric constant will be
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is introduced between the plates which results in
The capacitance of a parallel plate capacitor is $5\, \mu F$ . When a glass slab of thickness equal to the separation between the plates is introduced between the plates, the potential difference reduces to $1/8$ of the original value. The dielectric constant of glass is
A parallel plate capacitor with air between the plates has a capacitance $C$. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant $6$ , then the capacitance will become