Energy Stored in a Capacitor

A  $6\mu F$ capacitor is charged from $10\mathrm{V}$ to $20\mathrm{V}$. Increase in energy will be

The plates of a parallel plate condenser are pulled apart with a velocity $v.$ If at any instant mutual distance of separation is $d,$ then the magnitude of the time of rate of change of electrostatic energy of the capacity depends on $d$ as follows: (potential difference between plates is kept constant)-

If the charge on a body is increased by $2 \mathrm{\mu C},$ the energy stored in it increases by $21\%.$ The original charge on the body in micro-coulombs is

Two identical conducting very large plates $P_1$ and $P_2$ having charges $+4Q$ and $+6 Q$ are placed very closed to each other at separation $d$. The plate area of either face of the plate is $A$. The potential difference between plates $P_1$ and $P_2$ is

In the above question, if plates $P_1$ and $P_2$ are connected by a thin conducting wire, then the amount of heat produced will be

A parallel plate capacitor is connected to a battery. The plates are pulled apart with uniform speed. If $x$ is the separation between the plates, then the rate of change of electrostatic energy of the capacitor is proportional to

On increasing the plate separation of charged condenser, the energy

Three plates $A,B,C$ each of area $50 {\mathrm{cm}}^2$ have separation $3 \mathrm{mm}$ between $A$ and $B$ and $3 \mathrm{mm}$ between $B$ and $C$. The energy stored when the plates are fully charged is

In the circuit given below, the capacitor $C$ is charged by closing the switch $S_1$ and opening the switch $S_2$. After charging, the switch $S_1$ is opened and $S_2$ is closed, then the maximum current in the circuit

The potential differences that must be applied across the parallel and series combination of $3$ identical capacitors is such that the energy stored in them becomes the same. The ratio of potential difference in parallel to series combination is

The capacity of a capacitor is $4\times {10}^{-6} \mathrm{F}$ and its potential is $100 \mathrm{V}$. The energy released on discharging it fully will be

If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to,

A $10 \mathrm{\mu F}$ capacitor is fully charged to a potential difference of $50 \mathrm{V}$ After removing the source voltage it is connected to an uncharged capacitor in parallel. Now the potential difference across them becomes $20 \mathrm{V}$. The capacitance of the second capacitor is :

Electric field at a distance '$r$' from an infinitely large conducting sheet is proportional to:

The plate separation in a parallel plate condenser is $d$ and plate area is $A$. If it is charged to $V$ volt & battery is disconnected then the work done in increasing the plate separation to $2d$ will be

Two capacitors of equal capacitance $\left(C_1=C_2\right)$ are shown in the figure. Initially, while the switch $S$ is open, one of the capacitors is uncharged and the other carries charge $Q_0.$ The energy stored in the charged capacitor is $U_0$. Sometimes after the switch is closed, the capacitors $C_1$ and $C_2$ carry charges $Q_1$ and $Q_2$, respectively, the voltage across the capacitors are $V_1$ and $V_2,$ and the energies stored in the capacitors are $U_1$ and $U_2.$ Which of the following statements is incorrect?

In a parallel plate capacitor, the distance between the plates is $d$ and the potential difference across the plates is $V$. The energy stored per unit volume between the plates of the capacitor is,

The capacities of two conductors are $C_1$ and $C_2$ and their respective potentials are $V_1$ and $V_2.$ If they are connected by a thin wire, then the loss of energy will be given by

A parallel plate air capacitor has capacity $C$ farad, potential $V$ volt and energy $E$ joule. When the gap between the plates is completely filled with dielectric