$A$ parallel plate capacitor with air between the plates has a capacitance of $4 \ pF$. If the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant $K = 6$,then the value of capacitance will be . . . . . . . (in $pF$)

$A$ parallel plate capacitor is charged by connecting a $2 \ V$ battery across it. It is then disconnected from the battery and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?

$A$ 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 of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

$A$ parallel plate capacitor having capacity $C_0$ is charged to $V_0$. With the battery disconnected,if the separation between the plates is doubled,then the energy stored in it is $E_1$. Instead,if the separation between the plates is doubled with the battery in connection,the energy stored in it is $E_2$. Then the value of $\frac{E_2}{E_1}$ is

When a dielectric of constant $K = 2$ is filled in a capacitor,its capacitance becomes $C$. If the oil is removed,its capacitance will become ......