Energy Stored & Force between Plates

A parallel plate air capacitor having area of each plate '$A$' and the distance between the plates '$d$' has uniform electric field E in the space between the plates. The energy stored in the capacitor is ($\epsilon_0$ = permittivity of free space.)

A capacitor of capacitance C has charge Q and energy stored in it is W. If the charge is increased to 3Q, the energy stored in the capacitor W' will be

The electrostatic force ($F$) between the metal plates of an isolated parallel plate capacitor ($C$) having a charge $Q$ and area $A$ is ($d$ = distance between the plates)

The electric field between the plates of a parallel plate capacitor is 'E'. If the charge on the plates is Q then the force on each plate is

A capacitor of capacity $5\ \mu\text{F}$ is charged by a battery of e.m.f. $10$ V. At an instant, the potential difference across the capacitor is $4$ V and the time rate of change of potential difference across the capacitor is $0.6$ V/s. Then the time rate at which energy is stored in the capacitor at that instant is

A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed $v$. If $x$ is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to $x^{\alpha}$, where $\alpha$ is ______.

Consider a parallel plate capacitor of area A (of each plate) and separation ' $d$ ' between the plates. If $E$ is the electric field and $\varepsilon_0$ is the permittivity of free space between the plates, then potential energy stored in the capacitor is

If the charge on the capacitor is increased by 3C, the energy stored in it increases by $21\%$. The original charge on the capacitor is

A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is:

A parallel plate capacitor having plate area ' $A$ ' and separation ' $d$ ' is charged to a potential difference ' $V$ '. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is ( $\varepsilon_0 =$ permittivity of free space)

A parallel plate capacitor is charged and then isolated. If the separation between the plates is increased, which one of the following statement is NOT correct?

A $12\text{ pF}$ capacitor is connected to a $50\text{ V}$ battery, the electrostatic energy stored in the capacitor in $\text{nJ}$ is

The amount of work done in increasing the voltage across the plates of a capacitor form 5 V to 10 V is ' W '. The work done in increasing it from 10 V to 15 V will be (nearly)

If the plates of a parallel plate capacitor connected to a battery are moved close to each other, then A. the charge stored in it, increases.

B. the energy stored in it, decreases.

C. its capacitance increases.

D. the ratio of charge to its potential remains the same.

E. the product of charge and voltage increases.

Choose the most appropriate answer from the options given below:

The mean electrical energy density between plates of a charged air capacitor is (where $q =$ charge on capacitor, $A =$ Area of capacitor plate)

If the charge on the capacitor is increased by 3 coulombs, the energy stored in it increases by $44\%$. The original charge on the capacitor is

If a capacitor of capacity $900\ \mu\text{F}$ is charged to $100\text{ V}$ and its total energy is transferred to a capacitor of capacity $100\ \mu\text{F}$, then its potential will be

A parallel plate capacitor has plate area '$A$' and separation between plates is '$d$'. It is charged to a potential difference of $V_0$ volt. The charging battery is then disconnected and plates are pulled apart three times the initial distance. The work done to increase the distance between the plates is $(\varepsilon_0 =$ permittivity of free space)

A parallel plate air capacitor has a uniform electric field 'E' in the space between the plates. Area of each plate is A and the distance between the plates is '$d$'. The energy stored in the capacitor is $[\varepsilon_0 =$ permittivity of free space)

The force between the plates of a parallel plate capacitor of capacitance '$C$' and distance of separation of the plates '$d$' with a potential difference '$V$' between the plates is

A parallel plate capacitor is charged and then disconnected from the charging battery. If the plates are now moved further apart by pulling them by means of insulating handles, then

If the charge on a capacitor is increased by 2 C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

If the potential difference across a capacitor is increased from $5\text{ V}$ to $15\text{ V}$, then the ratio of final energy to initial energy stored in the capacitor is

A parallel plate capacitor has a uniform electric field $\vec{E}$ in the space between the plates. If the distance between the plates is 'd' and the area of each plate is 'A', the energy stored in the capacitor is: ($\varepsilon$0 = permittivity of free space)

In parallel plate capacitor, electric field between the plates is '$E$'. If the charge on the plates is '$Q$' then the force on each plate is

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the e.m.f. of the battery. The ratio of the energy stored int he capacitor to the work done by the battery will be

If the charge on the capacitor is increased by 2 C the energy stored in it increased by 21%. Total original charge on the capacitor is

A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2μF, how much work is done?

The electrostatic force between the metal plates of an isolated parallel plate capacitor C having a charge Q and area A, is

A parallel plate capacitor with area 200 cm2 and separation between the plates 1.5 cm, is connected across a battery of emf V. If the force of attraction between the plates is $25 \times 10^{-6}\text{ N},$ the value of V is approximately: $\left(\epsilon_0 = 8.85 \times 10^{-12}\ \frac{C^2}{N \cdot m^2}\right)$

A parallel plate air capacitor has capacity C, distance of separation between plates is d and potential difference $V$ is applied between the plates. Force of attraction between the plates of the parallel plate air capacitor is

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is

A parallel plate capacitor has a uniform electric field E in the space between the plates. If the distance between the plates is d and area of each plate is A, the energy stored in the capacitor is

The energy required to charge a parallel plate condenser of plate separation $d$ and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$, is

A battery is used to charge a parallel plate capacitor till the potential difference between the plates becomes equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work done by the battery will be

A parallel plate air capacitor is charged to a potential difference of V volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulting handle. As a result the potential difference between the plates

A fully charged capacitor has a capacitance $'C'$. It is discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $'s'$ and mass $'m'.$ If the temperature of the block is raised by $'\Delta T',$ the potential difference $'v'$ across the capacitor is

The work done in placing a charge of $8 \times 10^{-18}$ coulomb on a condenser of capacity $100$ micro-farad is

Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by

The energy stored in a capacitor of capacity $C$ and potential $V$ is given by

A $4\ \mu\text{F}$ capacitor is charged to $400\text{ V}$. If its plates are joined through a resistance of $2\text{ k}\Omega$, then heat produced in the resistance is