Between the plates of a parallel plate capacitor of plate area $A$ and capacity $0.025 \mu F$,a metal plate of area $A$ and thickness equal to $1/3$ of the separation between the plates of the capacitor is introduced. If the capacitor is charged to $100 \ V$,then the amount of work done to remove the metal plate from the capacitor is (in $\mu J$)

$A$ capacitor stores $60 \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \mu C$ flows through the battery. The dielectric constant of the material inserted is:

In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. The capacitance of the capacitor is $17.71 \, pF$. If this capacitor is connected to a $100 \, V$ supply,and a $3 \, mm$ thick mica sheet (of dielectric constant $k = 6$) is inserted between the plates,calculate the new capacitance,charge,and potential difference in the following cases:
$(a)$ While the voltage supply remains connected.
$(b)$ After the supply is disconnected.

$A$ parallel plate capacitor with plate area $A$ and separation $d$ is filled with two dielectric materials as shown in the figure. The dielectric constants are $K_1$ and $K_2$ respectively. The capacitance will be:

If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

$A$ parallel plate capacitor with square plates of side $L$ is filled with four dielectrics of dielectric constants $K_1, K_2, K_3, K_4$ arranged as shown in the figure. The effective dielectric constant $K$ will be