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Energy Stored In Capacitor - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Energy stored in capacitor is considered one the most difficult concept.
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56 Questions around this concept.
Solve by difficulty
If there are capacitors in parallel connected to V volt source, then the energy stored is equal to:
The work done in placing a charge of coulomb on a condenser of capacity 100 micro-farad is
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 , the potential difference V across the capacitance is
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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 condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process is
Let C be the capacitance of a capacitor discharging through a resistor R . Suppose is the time taken for the energy stored in the capacitor to reduce to half its initial value and
is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio
/
will be
A fully charged capacitor $C$ with initial charge $q_0$ is connected to a coil of self-inductance L at $t=0$ The time at which the energy is stored equally between the electric and the magnetic fields is
A capacitor C is fully charged with voltage $\mathrm{V}_0$. After disconnecting the voltage source, it is connected in parallel with another uncharged capacitor $\frac{C}{2}$. The energy loss in the process after the charge is distributed between the two capacitors is:
The capacity of a condenser is and its potential is. The energy released on discharging it fully will be?
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100 F
$ is connected in parallel across a 20 V supply. Then the energy stored is equal to
Concepts Covered - 1
Energy stored in capacitor
Energy stored in capacitor: When the charge is added to a capacitor then charge already present on the plate repel any new incoming charge. Hence a new charge has to be sent by applying force and doing work on it. All this work done on charges become energy stored in the capacitor. If Q is the amount of charge stored when the whole battery voltage appears across the capacitor, then the stored energy is obtained from the integral:
$$
U=\int_0^Q \frac{q}{C} d q=\frac{1}{2} \frac{Q^2}{C}
$$
This energy expression can be put in three equivalent forms by just permutations based on the definition of
$$
\begin{aligned}
& C=\frac{Q}{V} \\
& \text { capacitance } \\
& U=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} Q V=\frac{1}{2} C V^2
\end{aligned}
$$
This energy is stored in the form of an Electric field between the plates.
The energy density per unit volume (u)
$$
\begin{aligned}
& \mathrm{u}=\frac{1}{2} \mathrm{cv}^2 / \mathrm{V}=\frac{1}{2} \frac{\varepsilon_0 A E^2 d^2}{d A d} \\
& \mathrm{u}=\frac{1}{2} \varepsilon_0 E^2
\end{aligned}
$$
Note- The energy density per unit volume (u) is numerically equal to Electrostatic pressure.
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