Capacitor · Physics · TS EAMCET

A parallel plate capacitor of capacitance $10 \mu \mathrm{~F}$ is charged by a 220 V supply. The capacitor is then disconnected from the supply and is connected to another uncharged parallel plate capacitor of capacitance $12 \mu \mathrm{~F}$. The loss of electrostatic energy in this process is

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 $F \& B$.

A capacitor is fully charged with a battery and then disconnected. A dielectric is then inserted into the capacitor. How do charges on surface of the dielectric and outer surface of the plates of the capacitor would change respectively?

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