When air in a capacitor is replaced by a medium of dielectric constant $K$, the capacity

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 parallel plate capacitor with width $4\,cm$, length $8\,cm$ and separation between the plates of $4\,mm$ is connected to a battery of $20\,V$. A dielectric slab of dielectric constant $5$ having length $1\,cm$, width $4\,cm$ and thickness $4\,mm$ is inserted between the plates of parallel plate capacitor. The electrostatic energy of this system will be......... $\in_{0}\,J$. (Where $\epsilon_{0}$ is the permittivity of free space)

A parallel plate capacitor with plate area $A$ and plate separation $d =2 \,m$ has a capacitance of $4 \,\mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K =3$ (as shown in figure) will be .........$ \mu \,F$

There are two identical capacitors, the first one is uncharged and filled with a dielectric of constant $K$ while the other one is charged to potential $V$ having air between its plates. If two capacitors are joined end to end, the common potential will be

Voltage rating of a parallel plate capacitor is $500\,V$. Its dielectric can withstand a maximum electric field of ${10^6}\,\frac{V}{m}$. The plate area is $10^{-4}\, m^2$ . What is the dielectric constant if the capacitance is $15\, pF$ ? (given ${ \in _0} = 8.86 \times {10^{ - 12}}\,{C^2}\,/N{m^2}$)