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Energy Stored in a Capacitor

Energy stored in a charged capacitor is the work done in charging it and is stored in the electric field between its plates.

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Student-friendly explanation

As charge is added to a capacitor, its potential difference increases, so later charges require more work. The total work done becomes electrostatic energy stored in the capacitor. The same energy can be written in three equivalent forms: U = 1/2 CV^2, U = 1/2 QV, and U = Q^2/(2C). The correct form depends on which quantities are known or held constant.

How to write this in exams

  1. 1

    Start with the exact idea

    Energy stored in a charged capacitor is the work done in charging it and is stored in the electric field between its plates.

  2. 2

    Then show how to use it

    List given Q, V, and C; convert microfarad to farad; choose U = 1/2 CV^2, 1/2 QV, or Q^2/(2C); substitute with SI units; write energy in joule; interpret changes using constant V or constant Q when conditions change.

  3. 3

    Add one concrete example

    If a capacitor remains connected to a battery, voltage remains constant when a dielectric is inserted. If it is disconnected, charge remains constant. This changes how stored energy is calculated.

  4. 4

    Avoid this incomplete answer

    Using Q^2/(2C) when voltage is fixed after capacitance changes can lead to the wrong comparison unless Q is recalculated.

Definition

Energy stored in a charged capacitor is the work done in charging it and is stored in the electric field between its plates.

Example

If a capacitor remains connected to a battery, voltage remains constant when a dielectric is inserted. If it is disconnected, charge remains constant. This changes how stored energy is calculated.

Rule to remember

U = 1/2 CV^2 = 1/2 QV = Q^2/(2C), with U in joule, C in farad, Q in coulomb, and V in volt. Energy density between parallel plates is u = 1/2 epsilon0 E^2 in joule per cubic metre for vacuum or air approximation. Use the form matching the constant quantity in the problem.

Memory hook

Capacitor energy always has a half; choose the version that uses what stays known or constant.

Examples and method

Worked example

A 4 microfarad capacitor is charged to 100 V. U = 1/2 CV^2 = 1/2(4 x 10^-6)(100)^2 = 2.0 x 10^-2 J. The capacitor stores 0.020 J of electrostatic energy in its electric field.

Method to apply

List given Q, V, and C; convert microfarad to farad; choose U = 1/2 CV^2, 1/2 QV, or Q^2/(2C); substitute with SI units; write energy in joule; interpret changes using constant V or constant Q when conditions change.

Diagram support

A capacitor-energy sketch can show plates, electric field between plates, and the idea that energy is stored in the field region. For energy density, mark field E between plates.

How CBSE asks it

Questions ask direct energy calculation, derivation from charging work, comparison after dielectric insertion, or energy density in a uniform electric field.

Avoid common mistakes

Common confusion

Students often use U = CV^2 instead of U = 1/2 CV^2, missing the factor one-half that comes from gradual charging.

Common wrong answer

Using Q^2/(2C) when voltage is fixed after capacitance changes can lead to the wrong comparison unless Q is recalculated.

Exam tip

Before choosing the energy formula, check whether Q, V, or C is given. In dielectric questions, check whether the battery is connected or disconnected.

Quick check

Why is the energy stored in a capacitor not simply QV?

The potential difference is not V throughout charging; it rises gradually from zero to V. The average potential difference during charging is V/2, so the work stored is Q times V/2, giving U = 1/2 QV.

Answer writing and exam use

1-mark answer

Energy stored in a charged capacitor is the work done in charging it and is stored in the electric field between its plates.

2-mark answer

Energy stored in a charged capacitor is the work done in charging it and is stored in the electric field between its plates. U = 1/2 CV^2 = 1/2 QV = Q^2/(2C), with U in joule, C in farad, Q in coulomb, and V in volt. Energy density between parallel plates is u = 1/2 epsilon0 E^2 in joule per cubic metre for vacuum or air approximation. Use the form matching the constant quantity in the problem. If a capacitor remains connected to a battery, voltage remains constant when a dielectric is inserted. If it is disconnected, charge remains constant. This changes how stored energy is calculated.

3-mark answer

As charge is added to a capacitor, its potential difference increases, so later charges require more work. The total work done becomes electrostatic energy stored in the capacitor. The same energy can be written in three equivalent forms: U = 1/2 CV^2, U = 1/2 QV, and U = Q^2/(2C). The correct form depends on which quantities are known or held constant. U = 1/2 CV^2 = 1/2 QV = Q^2/(2C), with U in joule, C in farad, Q in coulomb, and V in volt. Energy density between parallel plates is u = 1/2 epsilon0 E^2 in joule per cubic metre for vacuum or air approximation. Use the form matching the constant quantity in the problem. A 4 microfarad capacitor is charged to 100 V. U = 1/2 CV^2 = 1/2(4 x 10^-6)(100)^2 = 2.0 x 10^-2 J. The capacitor stores 0.020 J of electrostatic energy in its electric field. Questions ask direct energy calculation, derivation from charging work, comparison after dielectric insertion, or energy density in a uniform electric field. Using Q^2/(2C) when voltage is fixed after capacitance changes can lead to the wrong comparison unless Q is recalculated.
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