The capacitance of an air capacitor is $15\,\mu F$ and the separation between the parallel plates is $6\,mm$. $A$ copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$.

Two parallel plate capacitors of capacity $C$ and $3C$ are connected in parallel and charged to a potential difference of $18V$. The battery is then disconnected,and the space between the plates of the capacitor of capacity $C$ is completely filled with a material of dielectric constant $K=9$. The final potential difference across the combination of capacitors will be $V^{\prime}$. Find $V^{\prime}$. (in $V$)

$A$ slab of dielectric constant $K$ has the same cross-sectional 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):

$A$ parallel plate capacitor has a capacity $80 \times 10^{-6} \,F$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $20$. The capacitor is now connected to a battery of $30 \,V$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is

$A$ parallel plate air capacitor is charged up to $100 \,V$. $A$ plate $2 \,mm$ thick is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by $1.6 \,mm$. The dielectric constant of the thick plate is

The energy and capacity of a charged parallel plate capacitor are $U$ and $C$ respectively. Now a dielectric slab of dielectric constant $K = 6$ is inserted in it. Then,the new energy and capacity become (Assume the charge on the plates remains constant).