An air capacitor has a capacitance of $1 \mu F$. Now the space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The capacitance of the capacitor is: [$d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the first and second dielectric respectively] (in $\mu F$)

$A$ 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 capacitor is made of two square plates of side $a$,separated by a distance $d$ $(d \ll a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$,as shown in the figure. The capacitance of this capacitor is

$A$ parallel plate capacitor having capacitance $12 \, pF$ is charged by a battery to a potential difference of $10 \, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is.......$pJ$

$A$ capacitor is made of a thin metal strip. Its capacitance is $2 \ \mu F$. If the space between the strips is filled with paper of thickness $0.15 \ mm$,dielectric constant $2.5$,and width $400 \ mm$,what is the length of the strip in meters?

$A$ parallel plate capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to: