A 47\ μF capacitor is charged to 20\ V. Find the energy stored. [2 marks]

SingaporePhysicsSyllabus dot point

How does a capacitor store charge and energy, and how do capacitors combine in circuits?

Define capacitance, calculate the energy stored on a capacitor, and combine capacitors in series and parallel

A focused answer to the H2 Physics learning outcome on capacitance. The definition Q = CV, the energy stored on a capacitor, and the rules for combining capacitors in series and parallel.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to define capacitance, calculate the charge and energy stored on a capacitor, and combine capacitors in series and parallel. A capacitor stores energy in the electric field between its plates, and its combination rules are the mirror image of those for resistors.

The answer

Capacitance

A capacitor stores charge $+Q$ on one plate and $-Q$ on the other when a potential difference $V$ is applied. The capacitance is the charge stored per unit potential difference:

C = \frac{Q}{V}

measured in farads (F), where $1\ \text{F} = 1\ \text{C V}^{-1}$ . The farad is a very large unit, so practical capacitors are usually in microfarads ( $\mu\text{F}$ ) or smaller. Capacitance depends on the geometry of the plates and the material between them, and is constant for a given capacitor.

Energy stored on a capacitor

As a capacitor charges, work is done against the increasing voltage to add more charge. The energy stored is the area under a charge-voltage graph (a triangle):

E = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C}

The factor of $\tfrac{1}{2}$ arises because the voltage rises from zero to $V$ as the capacitor charges, so the average voltage during charging is $\tfrac{1}{2}V$ . Use whichever form matches the quantities you know.

Capacitors in parallel

Capacitors in parallel share the same voltage, and their charges add, so the capacitances add:

C_{\text{total}} = C_1 + C_2 + \dots

This is the opposite of the resistor rule: parallel capacitors give a larger total capacitance.

Capacitors in series

Capacitors in series carry the same charge, and their voltages add, so the reciprocals add:

\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \dots

Again the opposite of resistors: series capacitors give a smaller total capacitance than the smallest individual capacitor.

Worked example

A $100\ \mu\text{F}$ capacitor is charged to $12\ \text{V}$ , then connected in parallel with an uncharged $50\ \mu\text{F}$ capacitor. Find (a) the initial charge and energy, and (b) the combined capacitance.

Step 1: Find the initial charge

Q = CV = 100 \times 10^{-6} \times 12 = 1.2 \times 10^{-3}\ \text{C}

Step 2: Find the initial stored energy

E = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(100 \times 10^{-6})(12)^2 = \tfrac{1}{2}(100 \times 10^{-6})(144) = 7.2 \times 10^{-3}\ \text{J}

Step 3: Find the combined capacitance in parallel

C_{\text{total}} = C_1 + C_2 = 100 + 50 = 150\ \mu\text{F}

Step 4: Comment

The charge is conserved when the capacitors are connected, but it redistributes across the larger combined capacitance, lowering the shared voltage to $V = Q/C_{\text{total}} = 1.2 \times 10^{-3}/(150 \times 10^{-6}) = 8.0\ \text{V}$ . Some energy is dissipated in this redistribution.

Examples in context

Example 1. A camera flash. A capacitor charges slowly from a small battery, storing energy $\tfrac{1}{2}CV^2$ , then discharges rapidly through the flash tube. The slow charge and fast discharge let a low-power battery deliver a brief, high-power burst of light, the standard role of a capacitor as an energy reservoir.

Example 2. Smoothing a power supply. A large capacitor across the output of a rectifier stores charge on the voltage peaks and releases it in the troughs, smoothing the pulsing d.c. into a steadier voltage. The larger the capacitance, the smaller the ripple, which is why power supplies use sizeable smoothing capacitors.

Try this

Q1. Define capacitance and state its SI unit. [2 marks]

Cue. Charge stored per unit potential difference, $C = Q/V$ ; unit farad (F).

Q2. A $47\ \mu\text{F}$ capacitor is charged to $20\ \text{V}$ . Find the energy stored. [2 marks]

Cue. $E = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(47 \times 10^{-6})(20)^2 = 9.4 \times 10^{-3}\ \text{J}$ .

Q3. Two $10\ \mu\text{F}$ capacitors are connected in series. Find the combined capacitance. [2 marks]

Cue. $\dfrac{1}{C} = \dfrac{1}{10} + \dfrac{1}{10} = \dfrac{2}{10}$ , so $C = 5.0\ \mu\text{F}$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA

220\ \mu\text{F}

capacitor is charged to

9.0\ \text{V}

. (a) Find the charge stored. (b) Find the energy stored. (c) The capacitor is then discharged through a resistor. State what happens to the stored energy.

Show worked answer →

(a) Charge: $Q = CV = 220 \times 10^{-6} \times 9.0 = 1.98 \times 10^{-3}\ \text{C}$ .

(b) Energy: $E = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(220 \times 10^{-6})(9.0)^2 = \tfrac{1}{2}(220 \times 10^{-6})(81) = 8.9 \times 10^{-3}\ \text{J}$ .

(c) During discharge the stored electrical energy is transferred to the resistor and dissipated as heat (some as a small amount of electromagnetic radiation), conserving total energy.

Markers reward $Q = CV$ , the energy as $\tfrac{1}{2}CV^2$ , and the statement that the stored energy is dissipated as heat in the resistor during discharge.

Original4 marksTwo capacitors of

4.0\ \mu\text{F}

and

12\ \mu\text{F}

are available. Find their combined capacitance when connected (a) in parallel and (b) in series.