If the rate of change of electric field across the plates of a parallel plate capacitor is $E$ and the displacement current is $I$, then the area of one plate of the capacitor is ( $\varepsilon_0$ is permittivity of free space)

A parallel plate capacitor with air as dielectric has a capacitance of $4 \mu \mathrm{~F}$. The space between the plates of the capacitor is completely filled with a material of dielectric constant 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

If the rate of change in electric flux between the plates of a capacitor is $9 \pi \times 10^3 \mathrm{Vms}^{-1}$, then the displacement current inside the capacitor is

As shown in the figure, a dielectric of constant $K$ is placed between the plates of a parallel plate capacitor and is charged to a potential $V$ using a battery. If the dielectric is pulled out after disconnecting the battery from the capacitor, the final potential difference across the plates of the capacitor is

The circuit shows two capacitor $A$ and $B$ of capacitances $C$ and $2 C$ respectively.

When they are fully charged, the cell is removed and the capacitors are connected with their plates of opposite polarities touching each other. Then

(i) Charge on $A$ is $\frac{4 C E}{9}$

(ii) Charge on $B$ is zero

(iii) Loss of energy in this process is $\left(C E^2 / 3\right)$

The correct statement/s is/are

The effective capacitance between points $A$ and $B$ shown in the circuit is

The effective capacitance between points $A$ and $B$ shown in the figure is

The displacement current through the plates of a parallel plate capacitor of capacitance $30 \mu \mathrm{~F}$ is $150 \mu \mathrm{~A}$. The capacitor is charged by a source of varying potential at the rate of

If a capacitor of capacitance $100 \mu \mathrm{~F}$ is charged at a steady rate of $100 \mu \mathrm{C} \mathrm{s}^{-1}$, then the time taken to produce a potential difference of 100 V between the capacitor plates is

A parallel plate capacitor of plate area $10 \mathrm{~cm}^2$ and plate separation 3 mm is charged to a potential difference 12 V and then the battery is disconnected. A slab of dielectric constant 3 is then inserted between the plates. The work done on the system in the process of inserting the slab is $\alpha \varepsilon_0$. The value of $\alpha$ is (take $\varepsilon_0$ as the permittivity of free space)

The equivalent capacitance between points $A$ and $B$ is

The following figure shows a 9 V battery and 3 uncharged capacitors of capacitances $C_1=C_2=C_3=1 \mu \mathrm{~F}$. The switch is thrown to the right side until capacitor $C_1$ is fully charged, then the switch is thrown to the left. The final charge on capacitor $C_2$ is

Find potential difference points $A \& F$ and $F \& B$.

Assume each oil drop consists of a capacitance of $C$. If combine $n$ drops to form a bigger drop, then the capacitance of bigger drop $C^{\prime}$ would be

In $C R$-circuit the growth of charge on the capacitor is