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

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \ F$ when air is present between its plates. The space between the plates is filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \ V$. The dielectric slab is then removed. The charge passing through the wire is:

In the given figure,a capacitor is formed by placing a compound dielectric between the plates of a parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $= A$):

Write the formula for the capacitance of a capacitor having a dielectric constant $K = 2$.

Two identical capacitors $1$ and $2$ are connected in series. Capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0 \text{ volts}$. The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charges stored in the capacitors before removing the slab,and $Q'_1$ and $Q'_2$ be the values after removing the slab. Then:

$A$ capacitor, when filled with a dielectric $K = 3$, has charge $Q_0$, voltage $V_0$, and electric field $E_0$. If the dielectric is replaced with another one having $K = 9$, the new values of charge, voltage, and field will be respectively: