Initially,the circuit is in a steady state. Now,one of the capacitors is filled with a dielectric of dielectric constant $K = 2$. Find the heat loss in the circuit due to the insertion of the dielectric.

$A$ parallel-plate capacitor with plate area $A$ has a separation $d$ between the plates. Two dielectric slabs of dielectric constants $K_{1}$ and $K_{2}$,each having an area of $A/2$ and thickness $d/2$,are inserted in the space between the plates as shown in the figure. The equivalent capacitance of the capacitor will be:

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery,then the charge on the plates relative to the earlier charge:

$A$ parallel plate capacitor of capacitance $5\,\mu F$ and plate separation $6\, cm$ is connected to a $1\, V$ battery and charged. $A$ dielectric of dielectric constant $4$ and thickness $4\, cm$ is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is........$\mu C$

$A$ 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 has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)