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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

  • Energy stored in capacitor is considered one the most difficult concept.

  • 43 Questions around this concept.

Solve by difficulty

If there are n capacitors in parallel connected to volt source, then the energy stored is equal to:

The work done in placing a charge of 8\times 10^{-18} 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 \Delta T, the potential difference V across the capacitance 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 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 t_{1} is the time taken for the energy stored in the capacitor to reduce to half its initial value and t_{2} is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio t_{1} / t_{2} 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

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The capacity of a condenser is 4\times 10^{-6} F and its potential is. The energy released on discharging it fully will be?

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 capacitance  C=\frac{Q}{V}.

U=\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} Q V=\frac{1}{2} C V^{2}

This energy is stored in the form of an Electric field between the plates.

The energy density per unit volume (u)

 \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}.

Note- The energy density per unit volume (u) is numerically equal to Electrostatic pressure.

 

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Energy stored in capacitor

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