A parallel plate capacitor having a separation between the plates $d$ , plate area $A$ and material with dielectric constant $K$ has capacitance $C_0$. Now one-third of the material is replaced by another material with dielectric constant $2K$, so that effectively there are two capacitors one with area $\frac{1}{3}\,A$ , dielectric constant $2K$ and another with area $\frac{2}{3}\,A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$ then $\frac{C}{{{C_0}}}$ is

A slab of dielectric constant $K$ has the same crosssectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}\,d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be.(Given $C _{0}=$ capacitance of capacitor with air as medium between plates.)

A frictionless dielectric plate $S$ is kept on a frictionless table $T$. A charged parallel plate capacitance $C$ (of which the plates are frictionless) is kept near it. The plate $S$ is between the plates. When the plate $S$ is left between the plates

Which one statement is correct ? A parallel plate air condenser is connected with a battery. Its charge, potential, electric field and energy are ${Q_o},\;{V_o},\;{E_o}$ and ${U_o}$ respectively. In order to fill the complete space between the plates a dielectric slab is inserted, the battery is still connected. Now the corresponding values $Q,\;V,\;E$ and $U$ are in relation with the initially stated as

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is

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