$A$ parallel plate capacitor has capacitance $C$. If it is equally filled with parallel layers of materials of dielectric constants $K_1$ and $K_2$,its capacity becomes $C_1$. The ratio of $C_1$ to $C$ is

Assertion: The electrostatic force between the plates of a charged isolated capacitor decreases when a dielectric fills the whole space between the plates.
Reason: The electric field between the plates of a charged isolated capacitor increases when a dielectric fills the whole space between the plates.

$A$ 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

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants \(K _1=2, K_2=3\), and \(K _3=5\) (as shown in the figure). If new capacitance is \(\frac{ n }{3} C\) then the value of \(n\) is ___________.

$A$ parallel plate capacitor has a plate area of $100\, m^{2}$ and a plate separation of $10\, m$. The space between the plates is filled up to a thickness of $5\, m$ with a material of dielectric constant $10$. The resultant capacitance of the system is $'x'\, pF$. Given $\varepsilon_{0} = 8.85 \times 10^{-12} F \cdot m^{-1}$,the value of $'x'$ to the nearest integer is:

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)