Capacitors: Combinations, Energy, Charging and Discharging – Electric Force and Electric Field (Senior 4 Physics) | Ganzaa

Electric Force and Electric Field

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Senior 4 · Physics

Electric Force and Electric Field

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Lesson 5 of 5

Capacitors: Combinations, Energy, Charging and Discharging

25 min

A flash from a small battery

A camera flash needs a burst of several thousand watts for about a thousandth of a second, far more power than its small battery can give. So the battery spends a few seconds charging a capacitor, a store of separated charge, to a high voltage, and the capacitor then empties through the flash tube almost at once. Capacitors are in nearly every electronic circuit: they smooth the output of phone chargers, set the timing of blinking lights, sense your finger on a touchscreen and store the bits in computer memory. In this lesson you define capacitance, combine capacitors, find the energy they store and describe how they charge and discharge through a resistor.

Capacitance

A capacitor is two conductors separated by an insulator. Connected to a supply, it takes charge $+ Q$ on one plate and $- Q$ on the other; no charge crosses the gap. The charge stored is proportional to the voltage, and the ratio is the capacitance:

C = \frac{Q}{V}

The unit is the farad (F), one coulomb per volt. A farad is very large: practical capacitors range from picofarads ( $10^{-12}\,\mathrm{F}$ ) to thousands of microfarads.

The parallel-plate capacitor. For plates of area $A$ a distance $d$ apart (Figure 5.1), the uniform field between them is $E = \frac{Q}{\varepsilon_0 A}$ , so the voltage is $V = Ed = \frac{Qd}{\varepsilon_0 A}$ , and

C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d} \qquad\text{or, with an insulator (dielectric) between the plates,}\qquad C = \frac{\varepsilon_0\varepsilon_r A}{d}

Large plates, close together, with a dielectric of high relative permittivity, give a large capacitance. The dielectric's molecules are polarised by the field and partly cancel it, so the same charge needs a smaller voltage.

Figure 5.1: A parallel-plate capacitor and its circuit symbol.

Definition

The capacitance of a capacitor is the charge stored on either plate per unit potential difference between the plates: $C = \frac{Q}{V}$ , in farads ( $1\,\mathrm{F} = 1\,\mathrm{C\,V^{-1}}$ ). A dielectric is an insulator placed between the plates; it multiplies the capacitance by its relative permittivity $\varepsilon_r$ . Example: a capacitor storing $60\,\mu\mathrm{C}$ at 12 V has $C = \frac{60 \times 10^{-6}}{12} = 5.0\,\mu\mathrm{F}$ .

Worked example 1: a parallel-plate capacitor

Two plates of area $0.020\,\mathrm{m^{2}}$ are 1.0 mm apart in air. Find the capacitance and the charge at 100 V. What is the capacitance if the gap is filled with a dielectric of $\varepsilon_r = 5.0$ ?

C = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 0.020}{1.0 \times 10^{-3}} = 1.8 \times 10^{-10}\,\mathrm{F} = 177\,\mathrm{pF}, \qquad Q = CV = 1.8 \times 10^{-8}\,\mathrm{C}

With the dielectric, $C = 5.0 \times 177\,\mathrm{pF} = 885\,\mathrm{pF}$ .

Capacitors in series and in parallel

Parallel. Capacitors connected side by side (Figure 5.2) all have the same voltage $V$ , and the total charge is the sum: $Q = Q_{1} + Q_{2} + Q_{3} = (C_{1} + C_{2} + C_{3}) V$ . So

C = C_{1} + C_{2} + C_{3}

Series. Capacitors in a line all carry the same charge $Q$ : the charge on the inner plates is only moved, not added. The voltages add: $V = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3}$ . So

\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}

These are the opposite of the rules for resistors: parallel capacitors add like resistors in series. A series combination always has less capacitance than its smallest member.

Figure 5.2: Capacitors in series and in parallel.

Worked example 2: a network across 12 V

Capacitors of $2.0\,\mu\mathrm{F}$ and $3.0\,\mu\mathrm{F}$ are connected in parallel, and this pair is connected in series with a $5.0\,\mu\mathrm{F}$ capacitor across a 12 V supply. Find the total capacitance and the charge on each capacitor.

Parallel pair: $2.0 + 3.0 = 5.0\,\mu\mathrm{F}$ . In series with $5.0\,\mu\mathrm{F}$ : $\frac{1}{C} = \frac{1}{5.0} + \frac{1}{5.0} \Rightarrow C = 2.5\,\mu\mathrm{F}$ .

Total charge $Q = CV = 2.5 \times 12 = 30\,\mu\mathrm{C}$ , which is the charge on the $5.0\,\mu\mathrm{F}$ capacitor and on the pair as a whole. Voltage across the $5.0\,\mu\mathrm{F}$ : $\frac{30}{5.0} = 6.0\,\mathrm{V}$ ; across the pair: $12 - 6.0 = 6.0\,\mathrm{V}$ . So $Q_{2.0} = 2.0 \times 6.0 = 12\,\mu\mathrm{C}$ and $Q_{3.0} = 3.0 \times 6.0 = 18\,\mu\mathrm{C}$ (check: $12 + 18 = 30$ ).

Energy stored in a capacitor

Charging a capacitor means pushing each extra bit of charge $\Delta q$ onto the plates against the voltage $v = \frac{q}{C}$ already there, which takes work $v\,\Delta q$ . On a graph of $V$ against $Q$ , a straight line through the origin, this work is the area under the line: a triangle of area $\tfrac{1}{2}QV$ . So the energy stored is

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

The energy is stored in the electric field between the plates. Only half the energy taken from the battery is stored; the other half is dissipated as heat in the circuit while charging.

Worked example 3: a camera flash

A flash capacitor of $100\,\mu\mathrm{F}$ is charged to 300 V (typical values) and discharged through the flash tube in 1.0 ms. Find the energy stored and the average power of the flash.

W = \tfrac{1}{2}CV^{2} = \tfrac{1}{2} \times 100 \times 10^{-6} \times 300^{2} = 4.5\,\mathrm{J}, \qquad P = \frac{W}{t} = \frac{4.5}{1.0 \times 10^{-3}} = 4.5\,\mathrm{kW}

The energy is small, but delivered so quickly that the power is thousands of watts. That is what capacitors are good at; batteries store far more energy but release it slowly.

Charging and discharging through a resistor

Connect a charged capacitor across a resistor $R$ . The current $I = \frac{V}{R}$ is largest at the start; as charge flows off, $V$ falls, so the current falls, so the charge flows off more slowly. The rate of loss is proportional to what is left, $\frac{\mathrm{d}Q}{\mathrm{d}t} = -\frac{Q}{RC}$ , which gives an exponential decay:

Q = Q_0\,e^{-\frac{t}{RC}}, \qquad V = V_0\,e^{-\frac{t}{RC}}, \qquad I = I_0\,e^{-\frac{t}{RC}}

The product $\tau = RC$ is the time constant (in seconds, since $\Omega \times \mathrm{F} = \frac{\mathrm{V}}{\mathrm{A}} \times \frac{\mathrm{C}}{\mathrm{V}} = \mathrm{s}$ ). After one time constant the charge, voltage and current have fallen to $e^{-1} = 0.37$ of their starting values; after $5 R C$ the capacitor is practically empty. The time to halve is $t_{\frac{1}{2}} = RC\ln 2 = 0.69RC$ .

When an uncharged capacitor is charged from a supply $V_{0}$ through $R$ , the voltage rises as $V = V_0\left(1 - e^{-\frac{t}{RC}}\right)$ , reaching 63% of $V_{0}$ after one time constant, while the current falls exponentially from $\frac{V_0}{R}$ (Figure 5.3).

Figure 5.3: Discharging and charging a capacitor through a resistor; time is measured in time constants RC.

Worked example 4: how long does it take to discharge?

A $470\,\mu\mathrm{F}$ capacitor charged to 9.0 V is discharged through a $10\,\mathrm{k\Omega}$ resistor. Find the time constant, the voltage after one time constant, and the time for the voltage to fall to 1.0 V.

\tau = RC = 10 \times 10^{3} \times 470 \times 10^{-6} = 4.7\,\mathrm{s}, \qquad V(4.7\,\mathrm{s}) = 9.0\,e^{-1} = 3.3\,\mathrm{V}

1.0 = 9.0\,e^{-t/4.7} \;\Rightarrow\; t = 4.7\ln 9.0 = 10.3\,\mathrm{s}

Applications of capacitors

Storing and releasing energy quickly: camera flashes, defibrillators that shock a stopped heart back into rhythm, and backup supplies for memory.
Smoothing: in a phone charger, a large capacitor charges at each peak of the rectified supply and discharges between peaks, filling in the dips to give a steady d.c. voltage.
Timing: the time constant $R C$ sets how long a light stays on or how fast an LED blinks.
Sensing: a touchscreen measures the tiny change in capacitance when a finger, a conductor, comes near; some microphones and keyboards work the same way.
Tuning and filtering: with inductors and resistors, capacitors select radio frequencies and separate a.c. from d.c. signals (S6).

Capacitors of many kinds and sizes, from tiny ceramic discs to large metal cans. Every one has the same basic plan: two conductors separated by a thin insulator, with a lead connected to each. Photo: public domain, via Wikimedia Commons.

Warning

Safety and common mistakes. (1) Large capacitors in televisions, microwave ovens and power supplies can hold a dangerous charge long after the appliance is unplugged: never open such equipment. (2) Using the resistor rules for capacitors: parallel capacitors add; series capacitors combine as reciprocals. (3) Forgetting that capacitors in series share the same charge, not the same voltage. (4) Writing the stored energy as $Q V$ : it is $\tfrac{1}{2}QV$ . (5) Mixing μF and F, or kΩ and Ω, in $R C$ .

Tip

The 37–63 rule. In one time constant a discharging capacitor loses 63% of what it had (37% is left); a charging one gains 63% of the way to full. Five time constants is "practically finished".

RC Circuits: Time Constant Explained, Capacitor Charging and Discharging (The Organic Chemistry Tutor). Watch how the time constant RC is used to find the voltage and current at any time, as in Worked example 4.

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Practice

Find the capacitance of two plates, each of area $1.0\,\mathrm{m^{2}}$ , 1.0 mm apart in air. What plate area would give a capacitance of 1.0 F at the same spacing? What does this tell you about the farad?
Find the combined capacitance of $3.0\,\mu\mathrm{F}$ and $6.0\,\mu\mathrm{F}$ (a) in series, (b) in parallel.
Capacitors of $4.0\,\mu\mathrm{F}$ and $12\,\mu\mathrm{F}$ are connected in series across 24 V. Find the charge on each and the voltage across each.
Find the energy stored in a $2200\,\mu\mathrm{F}$ capacitor charged to 12 V.
A $100\,\mu\mathrm{F}$ capacitor charged to 12 V discharges through a $220\,\mathrm{k\Omega}$ resistor. Find the time constant, the voltage after 10 s and the time for the voltage to halve.
Show that the unit of $R C$ is the second.
Explain why a capacitor charging through a resistor gains charge quickly at first and then more and more slowly.
A student says: "A capacitor is just a small battery." Evaluate this statement.

Show answersHide answers

$C = \frac{8.85 \times 10^{-12} \times 1.0}{1.0 \times 10^{-3}} = 8.9\,\mathrm{nF}$ ; for 1.0 F, $A = \frac{Cd}{\varepsilon_0} = \frac{1.0 \times 1.0 \times 10^{-3}}{8.85 \times 10^{-12}} = 1.1 \times 10^{8}\,\mathrm{m^{2}}$ (about 11 km by 10 km). The farad is a huge unit.
(a) $\frac{1}{C} = \frac{1}{3.0} + \frac{1}{6.0} = \frac{1}{2.0}$ , $C = 2.0\,\mu\mathrm{F}$ ; (b) $9.0\,\mu\mathrm{F}$ .
$C = \frac{4.0 \times 12}{4.0 + 12} = 3.0\,\mu\mathrm{F}$ ; $Q = 3.0 \times 24 = 72\,\mu\mathrm{C}$ on each; $V_1 = \frac{72}{4.0} = 18\,\mathrm{V}$ , $V_2 = \frac{72}{12} = 6.0\,\mathrm{V}$ .
$W = \tfrac{1}{2} \times 2200 \times 10^{-6} \times 12^{2} = 0.16\,\mathrm{J}$ .
$\tau = 220 \times 10^{3} \times 100 \times 10^{-6} = 22\,\mathrm{s}$ ; $V = 12\,e^{-\frac{10}{22}} = 7.6\,\mathrm{V}$ ; $t_{\frac{1}{2}} = 0.693 \times 22 = 15\,\mathrm{s}$ .
$\Omega \times \mathrm{F} = \frac{\mathrm{V}}{\mathrm{A}} \times \frac{\mathrm{C}}{\mathrm{V}} = \frac{\mathrm{C}}{\mathrm{A}} = \frac{\mathrm{A\,s}}{\mathrm{A}} = \mathrm{s}$ .
The current is $\frac{V_0 - V_C}{R}$ . At first the capacitor's voltage $V_{C}$ is zero, so the current is largest. As charge builds up, $V_{C}$ rises and opposes the supply, the difference $V_{0} - V_{C}$ shrinks, and the current, and so the rate of charging, falls towards zero.
Partly true: both store energy and can drive a current. But a capacitor stores energy in the electric field of separated charge, can release it in a fraction of a second, and its voltage falls steadily as it discharges; a battery stores energy chemically, keeps a nearly steady voltage, and stores far more energy for its size. So a capacitor suits short bursts and smoothing; a battery suits steady supply over hours.

Quick check

Key points

$C = \frac{Q}{V}$ (farad); parallel plates $C = \frac{\varepsilon_0\varepsilon_r A}{d}$ .
Parallel: $C = C_1 + C_2 + \dots$ (same $V$ ); series: $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$ (same $Q$ ).
Stored energy $W = \tfrac{1}{2}QV = \tfrac{1}{2}CV^{2} = \frac{Q^{2}}{2C}$ .
Through a resistor: discharge $V = V_0 e^{-\frac{t}{RC}}$ , charge $V = V_0(1 - e^{-\frac{t}{RC}})$ ; time constant $R C$ , half-time $0.69 R C$ . Used in flashes, smoothing, timing and touch sensing.

1A flash from a small battery

2Capacitance

3Capacitors in series and in parallel

4Energy stored in a capacitor

5Charging and discharging through a resistor