C
CraftExam
medium difficulty9 min

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.

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

Concept Practice Coming Soon

Understand it

  • 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 CV2, U = 1/2 QV, and U = Q2/(2C).
  • The correct form depends on which quantities are known or held constant.

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.

Formula and rule

U = 1/2 CV2 = 1/2 QV = Q2/(2C), with U in joule, C in farad, Q in coulomb, and V in volt.

Energy density between parallel plates is u = 1/2 ε0 E2 in joule per cubic metre for vacuum or air approximation.

Use the form matching the constant quantity in the problem.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A 4 microfarad capacitor is charged to 100 V.

  1. 1U = 1/2 CV2 = 1/2(4 × 10−6)(100)2 = 2.0 × 10−2 J.
  2. 2The capacitor stores 0.020 J of electrostatic energy in its electric field.

Method you can reuse

List given Q, V, and C; convert microfarad to farad; choose U = 1/2 CV2, 1/2 QV, or Q2/(2C); substitute with SI units; write energy in joule; interpret changes using constant V or constant Q when conditions change.

Visual tip

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.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

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

Think first, then show answer
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.

Common wrong answer

Using Q2/(2C) when voltage is fixed after capacitance changes can lead to the wrong comparison unless Q is recalculated.

Don't mix this up

Students often use U = CV2 instead of U = 1/2 CV2, missing the factor one-half that comes from gradual charging.

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.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

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.
  1. 1

    Enough for 1 mark

    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

    Add for 2 marks

    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.

  3. 3

    Add for 3 marks

    As charge is added to a capacitor, its potential difference increases, so later charges require more work.

Practice

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