Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be

$A$ container has a base of $50 \text{ cm} \times 5 \text{ cm}$ and height $50 \text{ cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \text{ cm} \times 50 \text{ cm}$. 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 $250 \text{ cm}^3 \text{ s}^{-1}$. What is the value of the capacitance of the container after $10 \text{ s}$ (in $\text{ pF}$)? [Given: Permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]

The parallel combination of two air-filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage $V$. When the capacitors are fully charged,the battery is removed,and after that,a dielectric material of dielectric constant $K$ is placed between the two plates of the first capacitor. The new potential difference of the combined system is

Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant of a material,which when fully filled between the plates produces the same capacitance?

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 ___________.

In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be: