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

A parallel plate capacitor $\mathrm{C}$ with plates of unit area and separation $\mathrm{d}$ is filled with a liquid of dielectric constant $\mathrm{K}=2$. The level of liquid is $\frac{\mathrm{d}}{3}$ initially. Suppose the liquid level decreases at a constant speed $V,$ the time constant as a function of time $t$ is Figure: $Image$

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

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 has plate area $40\,cm ^2$ and plates separation $2\,mm$. The space between the plates is filled with a dielectric medium of a thickness $1\,mm$ and dielectric constant $5$ . The capacitance of the system is :

If a slab of insulating material $4 \times {10^{ - 3}}\,m$ thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by $3.5 \times {10^{ - 3}}\,m$ to restore the capacity to original value. The dielectric constant of the material will be