$A$ parallel plate capacitor has plates of area $A$ separated by a distance $d$. If a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,what is the new capacitance?

$A$ parallel plate capacitor with air between the plates has a capacitance of $9 \, pF$. The separation between its plates is $d$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1 = 3$ and thickness $d/3$,while the other one has dielectric constant $K_2 = 6$ and thickness $2d/3$. The capacitance of the capacitor is now.........$pF$.

$A$ combination of parallel plate capacitors is maintained at a certain potential difference. When a $3 \, mm$ thick slab is introduced between all the plates,in order to maintain the same potential difference,the distance between the plates is increased by $2.4 \, mm$. Find the dielectric constant of the slab.

$A$ parallel plate capacitor with plate separation $5 \ mm$ is charged up by a battery. It is found that on introducing a dielectric sheet of thickness $2 \ mm$,while keeping the battery connections intact,the capacitor draws $25 \%$ more charge from the battery than before. The dielectric constant of the sheet is . . . . . . .

$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used,is $C$. If the dielectric is replaced by another with dielectric constant $20$,the capacity will become