9478 / 2027
Electric Fields overview

Lesson 7 of 7

Energy stored in a capacitor

  • I can use the area under a potential-difference-versus-charge graph and all three equivalent linear-capacitor energy expressions.Syllabus

    Syllabus 9478, 14(k). Use the area under a potential-difference-versus-charge graph and all three equivalent linear-capacitor energy expressions.

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A capacitor is charged to charge Q and p.d. V. Why is the energy stored 1/2 QV and not QV?

  1. One plate stores the charge, so just half of QV is counted.
  2. Half the energy is lost as heat in the capacitor plates during charging.
  3. Each added charge is pushed against a p.d. that rises from zero to V.
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Each added charge is pushed against a p.d. that rises from zero to V.

At the start the p.d. is zero, so the first charge needs almost no work. The p.d. rises steadily to V, so the average p.d. is V/2 and the work done is Q x V/2.

Separating charge requires work as the capacitor's potential difference grows. Its stored energy is the area under a potential-difference-against-charge graph.

Add the work of successive charge transfers

For a capacitor with plate-charge magnitude Q and potential difference V, transferring a small additional charge ΔQ requires work approximately VΔQ at that stage. Adding these small strips gives the energy stored in the ideal capacitor, with zero energy for the uncharged state.

The required graph has V vertically and Q horizontally. For a linear capacitor of constant C, V = Q/C is a straight line through the origin with gradient 1/C. The total area is a triangle:

U = (1/2)QV
Using V = Q/C: U = Q2/(2C)
Using Q = CV: U = (1/2)CV2

These are three equivalent expressions for the same stored energy under the constant-capacitance model. Choose the form suited to the known or fixed quantities.

For charge of order 100 microcoulomb and voltage of order 10 V, QV/2 is of order 10-3 J. One millijoule, mJ, is 10-3 J.

Worked energy area

Read the axes and convert the charge unit

For C = 10.0 microfarad, the V-against-Q model has points (0,0), (40 microcoulomb,4 V), (80 microcoulomb,8 V) and (120 microcoulomb,12 V).

V against Q makes the charging work an area

The axes are reversed from Q(V): V is now vertical and Q horizontal. For this constant-C capacitor, each small addition of plate charge requires work approximately VΔQ at the current p.d.; summing those contributions gives the complete area.

The voltage-charge triangle represents 0.720 millijoules of stored energyPlate-charge magnitude is horizontal in microcoulombs and potential difference vertical in volts. The exact model points are zero and zero, forty microcoulombs and four volts, eighty and eight, and 120 and twelve. A straight line joins the origin to the final point. The triangle between the line and the charge axis is shaded, with horizontal base 120 microcoulombs and height twelve volts. Its energy is one half times 120 times ten to the minus six coulombs times twelve volts, or 720 microjoules, equal to 0.720 millijoules. The slope is one over capacitance, not capacitance. The shading is the capacitor's stored-energy account, not an automatic claim that a connected source transfers only this energy.04080120048120.720 mJV / VQ / µC

The base is 120 × 10-6 C, so the area is 720 µJ = 0.720 mJ. The same stored energy is QV/2, Q2/(2C) or CV2/2 under the stated constant-capacitance condition.

Voltage is vertical and plate-charge magnitude is horizontal. The triangular area gives stored energy. The gradient is 1/C, unlike the Q-against-V graph.
U = (1/2)(120 × 10-6)(12.0)
= 7.20 × 10-4 J = 0.720 mJ

The unconverted numerical area 720 has unit microcoulomb volt, or microjoule. It is not 720 J. The same result follows from (1/2)(10.0 × 10-6)(12.0)2.

State what stays fixed

For the same capacitor, doubling V doubles Q and quadruples U. The 10.0 microfarad device at 24.0 V therefore has Q = 240 microcoulombs and U = 2.88 mJ.

Comparing different capacitances requires a different decision. At fixed voltage U is proportional to C, whereas at fixed charge U is inversely proportional to C.

Alternative capacitor states with the fixed quantity named
CaseCharge and voltageU / mJ
Reference: C = 10.0 microfaradQ = 120 microcoulomb
V = 12.0 V		0.720
C = 20.0 microfarad, same QQ = 120 microcoulomb
V = 6.0 V		0.360
C = 20.0 microfarad, same VQ = 240 microcoulomb
V = 12.0 V		1.44

A disconnected isolated capacitor retains Q only under the ideal no-leakage assumption. A connected ideal source can maintain V by exchanging charge. The fixed-Q and fixed-V rows are separate comparisons; they do not hold both quantities fixed while changing C.

Check your understanding

A 10.0 microfarad capacitor at 12.0 V stores 0.720 mJ and has plate-charge magnitude 120 microcoulomb. Compare a 20.0 microfarad capacitor, first at the same charge and separately at the same voltage.

  1. Its energy is 0.360 mJ at fixed charge and 1.44 mJ at fixed voltage.
  2. Its energy is 0.360 mJ at fixed voltage and 1.44 mJ at fixed charge.
  3. Its energy is 1.44 mJ in both comparisons.
Show the answer

Its energy is 0.360 mJ at fixed charge and 1.44 mJ at fixed voltage.

At Q = 120 microcoulomb, U = Q2/(2C) halves. At V = 12.0 V, U = CV2/2 doubles. The two comparisons do not hold both Q and V fixed.

Distinguish source transfer from stored energy

For an initially uncharged linear capacitor charged through resistance from a constant-voltage source to its final voltage, the source transfers charge Q at source voltage Vs. Its energy transfer is VsQ.

For the 12.0 V, 120 microcoulomb final state, the source transfers 1.44 mJ, while the capacitor stores 0.720 mJ. In this stated resistive charging model, the remaining 0.720 mJ is transferred in the resistance. The capacitor's changing voltage explains why its stored-energy area is triangular even though the source voltage stays constant.

This account depends on the initially uncharged capacitor, constant source voltage and stated charging process. Do not identify every source energy transfer with the capacitor's final stored energy.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

A graph of potential difference V against charge Q for a capacitor is a straight line from the origin to (6.0 mC, 12 V). Use the graph to find the capacitance and the energy stored when the p.d. is 12 V. [3]

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