$A$ parallel plate capacitor is immersed in an oil of dielectric constant $K = 2$. The electric field between the plates is:

$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced,is

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

$A$ capacitor has capacitance $C_0$ when there is no dielectric between its plates. Two slabs of dielectric constant $K_1$ and $K_2$ respectively,with area equal to the area of the plates but thickness half of the distance between the plates,are placed in between the plates. Then the new capacitance is

In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

Between the plates of a parallel plate capacitor, a dielectric plate is introduced to completely fill the space between the plates. The capacitor is charged and then disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference $V$ across the plates versus the length $x$ of the dielectric plate drawn out is: