An air capacitor is connected to a battery. The effect of filling the space between the plates with a dielectric is to increase

A parallel plate capacitor with air between the plates has 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 dielectric constant $k_1 = 3$ and thickness $\frac{d}{3}$ while the other one has dielectric constant $k_2 = 6$ and thickness $\frac{2d}{3}$ . Capacitance of the capacitor is now…….$pF$

A parallel plate capacitor with air between plates has a capacitance of $8\,\mu F$ what will be capacitance if distance between plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$ ?…..$\mu F$

A container has a base of $50 \mathrm{~cm} \times 5 \mathrm{~cm}$ and height $50 \mathrm{~cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \mathrm{~cm} \times 50 \mathrm{~cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250 \mathrm{~cm}^3 \mathrm{~s}^{-1}$. What is the value of the capacitance of the container after $10$ seconds? [Given: Permittivity of free space $\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]

The capacitance of a parallel plate capacitor with air as medium is $6\, \mu F$. With the introduction of a dielectric medium, the capacitance becomes $30\, \mu F$. The permittivity of the medium is……….$C ^{2} N ^{-1} m ^{-2}$

$\left(\varepsilon_{0}=8.85 \times 10^{-12} C ^{2} N ^{-1} m ^{-2}\right)$