Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,what will be the new (final) capacitance?

$A$ parallel plate capacitor of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$,then the potential difference across the other capacitor will become:

$A$ parallel plate capacitor is charged and then disconnected from the source of potential difference. If the plates of the capacitor are then moved farther apart by the use of an insulated handle,which one of the following is true?

Two condensers of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is then disconnected and the condenser of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors in the second case is

Two capacitors $C$ and $2C$ are connected in parallel and charged to a potential $V$ by a battery. The battery is then removed. The capacitor $C$ is filled with a dielectric of constant $K$. What is the new potential difference across the capacitors?