$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

$A$ sheet of aluminium foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor

$A$ 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:

$A$ parallel plate capacitor with air as dielectric has a capacitance of $4 \mu F$. The space between the plates of the capacitor is completely filled with a material of dielectric constant $K = 5$ and charged to a potential of $100 \ V$. The work done to completely remove the dielectric material after the capacitor is disconnected from the battery is (in $J$)

$A$ 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. If the initial capacitance of the capacitor was $2 \ \mu F$,the amount of heat produced when the dielectric is inserted is ... $\mu J$.

$A$ slab of material with a dielectric constant of $3$ has the same area as the plates of a parallel plate capacitor but has a thickness of $\left(\frac{3}{4}\right) d$,where $d$ is the separation between the plates. What is the electrical potential difference between the plates when the slab is inserted? The initial electrical potential difference is $V_0$.