$A$ 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.......$pJ$

Two parallel plate capacitors are connected in series and then connected to a $100 \ V$ battery. $A$ dielectric slab of dielectric constant $K = 4.0$ is inserted between the plates of the second capacitor. What will be the potential difference across each capacitor respectively?

$A$ parallel plate capacitor has plates of area $A$ and a separation distance of $100 \ mm$. It contains two dielectric layers: one with a dielectric constant of $10$ and a thickness of $6 \ mm$,and another with a dielectric constant of $5$ and a thickness of $4 \ mm$. Find the capacitance of the capacitor.

$A$ 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

The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)

If two charges $q_1$ and $q_2$ are separated by a distance $d$ and placed in a medium of dielectric constant $K$, what will be the equivalent distance between the charges in air for the same electrostatic force?