19.2 Energy stored in a capacitor

Learning objectives

19.2.1The electric potential energy stored in a capacitor• determine the electric potential energy stored in a capacitor from the area under the potential–charge graph 19.2.2W = 1/2 QV = 1/2 CV 2• recall and use W = 1/2 QV = 1/2 CV 2

$For a small added charge dQ at potential V: dW=V dQ Therefore stored energy W is the area under a graph of V (vertical axis) against Q (horizontal axis).$

For a fixed linear capacitor, V=Q/C, so the V–Q graph is a straight line from (0,0) to (Q,V). The area is a triangle.

$W=area=(1/2)×base×height=(1/2)QV$

If the graph reaches V=8.0 V at Q=1.2×10⁻⁴ C, W=(1/2)(1.2×10⁻⁴)(8.0)=4.8×10⁻⁴ J.

For charging from (Q1,V1) to (Q2,V2) on a straight line, added energy is the trapezium area ΔW=(1/2)(V1+V2)(Q2-Q1). For a curved graph, estimate or integrate the area under the curve.

Check axes before taking area. On V against Q, area has units V C=J. The final product QV is a rectangle and is twice the triangular area when charging a linear capacitor from zero.

For a linear capacitor, stored energy W=½QV=½CV²=Q²/(2C). The half factor comes from the average voltage during charging.

Choose the form matching known Q, V or C and keep units in farads, volts and coulombs.

A 100 μF capacitor at 20 V stores 0.020 J.

Using QV without the half factor doubles the energy for a capacitor charged from zero.

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—19.3 Discharging a capacitorRead

19.2 Energy stored in a capacitor

Learning objectives

Stored capacitor energy is the area under a potential–charge graph

Capacitor energy is W=½QV=½CV²=Q²/(2C)

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