1

Capacitance

18. Capacitance

Content

18.1 Capacitors and capacitance

18.2 Energy stored in a capacitor

Learning Outcomes

Candidates should be able to:

(a) show an understanding of the function of capacitors in simple circuits.

(b) define capacitance and the farad.

(c) recall and solve problems using C = Q/V.

(d) derive, using the formula C = Q/V, conservation of charge and the addition of p.ds,

formulae for capacitors in series and in parallel.

(e) solve problems using formulae for capacitors in series and in parallel.

* (f) deduce from the area under a potential-charge graph, the equation W = ½QV and

hence W = ½CV2 .

2

Capacitors

symbol

3

Capacitors - another electrical component A capacitor is any device that is capable of storing charge.

A practical capacitor is normally made up of two parallel metal plates(called a parallel plate capacitor) separated by an insulator or dielectric e.g. air, mica, waxed paper, polythene, oil etc.

An isolated conductor carrying net charge is considered as storing charges

and hence functioning as a capacitor

A dielectric increases the amount of charge that can be stored

Capacitors are used in electrical and electronic devices same as resistors

Capacitors are generally non-polarity sensitive. However polarity sensitive capacitors are known as electrolytic capacitors and must be connected with the correct polarity or they will be damaged

Some uses are to store charge, store energy, coupled with an inductor to

tune a radio circuit, to power electromagnets in supercolliders, filter out high

frequency radio waves, camera flashes, in computers to function when there is a power failure, to save valuable data

• Capacitors are widely used in alternating current and radio circuitsbecause they can transmit alternating currents.

4

Breakdown potential in capacitor dielectrics

• When an external voltage applied across a capacitor containing a gaseous dielectric is increased gradually, there will reach a potential whereby the gas molecules are ionized resulting in electrical conduction and hence a drop in the p.d. across the capacitor.

• The potential applied that causes this effect is called the breakdown potential.

5

Capacitance

When an isolated spherical conductor is connected to a highvoltage supply, it is found that as the potential is increased, thecharge stored on the sphere also increases i.e. Q α V and hence Q= CV where the gradient C is a constant depending on the size ofthe conductor and known as the capacitance

The capacitance C of a capacitor is defined as the ratio of thecharge Q stored on either plate to the potential difference Vbetween the plates.

Thus: C = Q/V

It can also be defined as the charge stored per unit p.d. applied tothe capacitor. The unit of C is farads (F).

1 farad is 1 coulomb per volt. i.e F = C V-1

Capacitance is a measure of the charge storing ability of aconductor or the extent to which a capacitor can store charge. Thevalue of C in radio and hi-fi systems are usually from micro-farads(F = 10-6 F) to pico-farads (pF = 10-12 F).

6

Factors affecting capacitance

• The material used as a dielectric affects the capacitance of a capacitor

• Experiment shows that capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them

i.e. C α A/d

• For a capacitor with air or a vacuum between the plates, the constant of proportionality is the permittivity of free space ε0 whose value is 8.85 x 10-12 C2 N-1 m-2

thus C = ε0A/d and the unit of permittivity is F m-1

• A quantity called relative permittivity εr is introduced to account for the fact that the use of a dielectric increases the capacitance

• The relative permittivity is defined as the capacitance of a parallel plate capacitor with the dielectric between the plates divided by the capacitance of the same capacitor with a vacuum between the plates

• Including the relative permittivity factor, the full expression for the capacitance of a parallel plate capacitor is

C = ε0εr A/d

7

Relative permittivity of some materials

Materials Relative permittivity

Air 1.0005

Polythene 2.3

Sulphur 4

Paraffin oil 4.7

Mica 6

Barium titanate 1200

8

Example

• A parallel plate air-filled capacitor has

square plates of side 30 cm that are a

distance 1.0 mm apart. Given that ε0 = 8.85

x 10-12 C2 N-1 m-2 calculate the capacitance

of the capacitor.

Solution

Using C = ε0εr A/d = 8.0 x 10-10 F

9

Q-V graph - relationship between charge and

potential• If the charge and the p.d. is measured at various times of the charging

process, the following Q - V graph is obtained.

• The gradient is equal to a quantity known as capacitance.

Capacitance C = Q/V

Q

V

10

Charging and discharging circuit

A

CV

switch

R

11

Charging/Discharging of a parallel plate

capacitor In a circuit where a battery or energy source is connected

through a switch to a parallel plate capacitor and an ammeter in series, a maximum current I

0is observed initially which then

decreases gradually to zero as time increases.

No current flows through the capacitor. The time period during which the current drops from a maximum value I

0to zero is the

charging stage.

When charging a capacitor work is done by the battery to move charge on to the capacitor hence energy is transferred from the power supply and stored as electric potential energy in the capacitor

If this battery is then disconnected and the fully charged capacitor is now connected to an external resistor, a maximum current of initial value I

0flows through the circuit in the opposite

direction to that of the charging stage.

The current similarly will drop to zero after a certain time. This time period is said to be the discharging stage.

• This mechanism is known as induction

12

I

+ charging C

O

time

- discharging C

13

Induction• In the same circuit earlier with the switch is on, electrons

from the metal plate of the capacitor and connecting leads will be attracted to the + ve terminal of the battery.

• Electrons in the connecting leads joined to the – ve terminal will be pushed to the other metal plate i.e + ve plate

• As time goes by, more electrons are deposited on one plate than the other plate hence becoming more positive.

• A potential difference is set up across the plates

• This p.d. will become constant when it reaches the e.m.fvalue of the battery

• The induction process will then stop.

• This takes only a fraction of a second if there is negligible resistance in the circuit.

• If a resistor is connected in series, it will slow down the entire process. It will then take a longer time to charge the capacitor

14

Arrangement of Capacitors in Circuit

Capacitors can be arranged in a series

or parallel manner in a circuit similar to

resistors. Such arrangements will give

rise to a resultant or combined

capacitance for each that can be

calculated

15

Capacitors in Parallel - derivation• For capacitors charged in parallel generally, each capacitor stores a

different amount of charge as the current could be different in thedifferent paths.

• It can be proved that the effective capacitance C of suchconnections is given by: C = C

1+ C

2+ C

3+ ... + C

n

• For 3 capacitors connected in parallel,

VAB

= VCD

= VEF

= V0

, upon complete charging

• By the law of conservation of charge, total charge Q for the threecapacitors is: Q = Q

1+ Q

2+ Q

3= C

1V

0+ C

2V

0+ C

3V

0since Q =

CV

• If the effective capacitance is C, then

Q = CV0

= C1V

0+ C

2V

0+ C

3V

0giving

C = C1

+ C2

+ C3

• i.e in a parallel circuit the effective capacitance is the sum ofthe individual capacitances (compare this with resistors)

16

Example

Calculate the effective capacitance for

a) 10 µF and 40 µF in parallel

Ans: 50 µF

b) 10 µF , 40 µF and 8 µF in parallel

Ans: 58 µF

17

Capacitors in Series - derivation

• When charged in series, each capacitor stores the sameamount of charge as the current through all of them is thesame.

• The effective/combined capacitance C for series connections of capacitors is: 1/C = 1/C

1+ 1/C

2+ 1/C

3+ ...+ 1/C

n

• For 3 capacitors in series,

V0

= VAB

+ VCD

+ VEF

V0

= Q/C1

+ Q/C2

+ Q/C3

since V = Q/C

Q/C = Q/C1

+ Q/C2

+ Q/C3

1/C = 1/C1

+ 1/C2

+ 1/C3

• i.e in a series circuit the effective capacitance is the

reciprocal of the sum of the reciprocals of the individual

capacitances (again compare this with resistors)

18

Example

Calculate the effective capacitance for

a) 10 µF and 40 µF in series

Ans: 8 µF

b) 10 µF , 40 µF and 8 µF in series

Ans: 4 µF

19

Summary of capacitors in series and parallel

• For capacitors in parallel, the combined

capacitance is greater than the largest capacitor

• For capacitors in series, the combined capacitance

is less than the smallest capacitance

• For 2 identical capacitance in parallel, the

combined capacitance is equal to double that of a

single capacitance

• For 2 identical capacitors in series, the combined

capacitance is equal to half that of a single

capacitance

20

Example

Calculate the charge stored in a capacitor connected to a battery of e.m.f 10 V if the capacitance is 470 µF.

Solution

Q = CV

= 470 x 10-6 F x 10 V since F = C V-1

= 4.7 x 10-3

= 4.7 mC

21

Energy stored

• In the Q-V graph, the area under the graph represents the energy

stored.

• E = ½ QV = ½ CV2 = ½ Q2/C = Q2/2C

• Theory: if work is needed to separate the charges across the gap, then

work done = Fd

W = Fd = Eqd = qd x V/d = qV

Hence, total work W = integration of q.dV

= area under graphV

V0

0 Q

0 Q0

22

Examples

• Calculate the energy stored for a 470 uF capacitor at 10 V.

Solution

E = ½ CV2 = ½ x 470 x 10-6 F x (10 V)2

= 0.0235 J

Try this for the units:

(F x V2 = C V-1 x V2 = C V = J since, energy or work done = Fd = C V)

• A camera flash lamp uses a 5000 uF capacitor which is

charged by a 9 V battery. Calculate the energy transferred

when the capacitor is fully discharged through the lamp

Solution

E = ½ CV2 = ½ x 5000 x 10-6 x (9)2

= 0.203 J

23

Capacitors and resistors on printed circuit

boards

24

Resuscitator

25

Rating

• A capacitor is usually rated as value µF,

value V

e.g. 470 µF, 16 V

• What does the rating mean?

For every 1 V increase in pd, there will be

an increase of 470 µC of charge

• 16 V is the max safe operating p.d. beyond

which the capacitor will break down.

26

Example

A capacitor of capacitance 220 µF is connected to a battery of e.m.f. 20 V. After being fully charged, it is disconnected from the battery and is connected to an empty capacitor of capacitance 470 µF .

What is:

(a) the final voltage across the capacitors.

Ans: 6.38 V

(b) the charge on each capacitor ?

Ans : 1.41 mC; 3 mC

27

Example

You are provided with capacitors of ratings

48 µF, 25 V.

Show the combinations of capacitors to

produce

(a) a capacitor of rating 96 µF, 25V

(b) a capacitor of rating 24 µF, 50V

(c) a capacitor of rating 72 µF, 50 V

(d) a capacitor of rating 16.0 µF, 75V

28

Capacitors in A.C. Circuits

• Capacitors are widely used in alternating current and radio circuits because they can transmit alternating currents.

• Charge cannot flow through the dielectric of the capacitor due to one direction flow of the d.c.

• When an alternating voltage V of frequency f is applied to a capacitor, an a.c. current flows in the circuit even though the capacitor has an insulator between its plates.

• When f is high, such as 50 or 100 Hz, a steady current I flows through the circuit where :

I = fQ since I = Q/t

29

Graph of discharge current/charge against time

• A graph of discharge current/charge against time is seen to change very

rapidly at first and then move more slowly

• Analysis shows that the graph is exponential which has an equation of

the form x = x0e-kt where x is the quantity that is decaying, x0 the value

of x at t = 0, e = 2.718 (the root of natural log) and k is a constant

characteristic of the decay

• A large value of k means that the decay is rapid and a small value of k

means a slow decay

• The solution for the discharge of a capacitor of capacitance C through a

resistor is of the form Q = Q0e-t/CR where Q0 is the charge on the plates

at t = 0 and Q is the charge at time t

• Since C is proportional to V, the equation for discharge of a capacitor

may be written as V = V0e-t/CR and I = I0e

-t/CR

• As time progresses, the exponential curve gets closer and closer to the

time axis but never actually meets it, thus it is not possible to quote a

time for the capacitor to discharge completely

30

Time constant CR

• The quantity CR in the decay equation may be used to give an

indication of whether the decay is fast or slow and is called the time

constant of the capacitor-resistor circuit

• Since C = Q/V and R = V/I, CR = Q/I = t which is in seconds

• To find the charge Q on the capacitor plates after a time t = CR we

substitute t = CR in the exponential equation Q = Q0e-t/CR giving

Q = Q0e-CR/CR = Q0e

-1 = Q0/e = Q0/2.718 =

• Hence the time constant is the time for the charge to have dcecreased

to 1/e (or 1/2.718) of its initial charge

• This means that in 1 time constant, the charge stored by the capacitor

drops to roughly one-third of its initial value. In the next time constant

it will drop by the same ratio, to about one-ninth of the value at the

beginning of the decay

31

Example

• A 500 uF capacitor is connected to a 10 V supply, and is then

discharged through a 100 kΩ resistor. Calculate:

a) the initial charge stored by the cpacitor

b) the initial discharge current

c) the value of the time constant

d) the charge on the plates after 100 s

e) the time at which the remaining charge is 2.5 x 10-3 C

Solution

a) Q = CV, so Q = 500 x 10-6 x 10 = 5.0 x 10-3 C

b) I = V/R, so I = 10/(100 x 103) = 1.0 x 10-4 A

c) CR = 500 x 10-6 x 100 x 103 = 50 s

d) after 50 s, the charge on the plates is Q0/2.718 = 1.8 x 10-3 C

after another 50 s, the charge is 1.8 x 10-3/2.718 = 6.8 x 10-4 C

e) using Q = Q0e-t/CR , 2.5 x 10-3 = 5.0 x 10-3e-t/50 or 0.50 = e-t/50

Taking natural log on both sides, -0.693 = -t/50 or t = 35 s

32

Worked examples

• Define the capacitance of a capacitor

• The figure below shows a circuit which is used to determine the capacitance of a

capacitor C. The switch S causes the capacitor to be alternatively charged by the

battery and completely discharged through the ammeter. The battery has an e.m.f. of

6.0 V and negligible internal resistance.

(a) Describe, in terms of the flow of charged particles, what happens in the circuit

when the capacitor is being charged.

(b) The switch S is set so that the charge/discharge process occurs 50 times per

second and the steady reading on the ammeter is 0.14 mA. Calculate the capacitance

of C.

(c) Calculate the maximum energy stored by the capacitor.

AC6 V

switch

33

Solution

• C = Q/V where Q is the charge and V is the potential difference.

• (a) Electrons flow away from the -ve battery terminal or towards the +ve terminal.

(b) Q = It = 1/f (I) = (0.14 x 10-3)/50 = 2.8 x 10-6

C

C = Q/V = (2.8 x 10-6)/6 = 4.7 x 10-7 F = 0.47 F

(c) Energy stored = ½ CV2 = 0.5 x 4.7 x 10-7 x 62 = 8.4 x 10-6 J

34

Summary

• Capacitors store charge or energy

• Charged by the process of induction

• Capacitors have a rating of capacitance and maximum safe voltage

• Capacitance = charge/p.d. across plates = C = Q/V

• Energy stored = area under graph = ½ QV

• Capacitance increases in parallel connections but decreases in series

connections

• Q = Q0e-t/CR , I = I0e

-t/CR , V = V0e-t/CR

35

Exercises

1) A capacitor, C1, of 4 F is charged to a potential difference of 50 V. Another capacitor, C2, of 6F is charged to a potential difference of 100V.

(a) Find the energy stored in each capacitor. [ 5x10-3 J; 30x10-3 J ]

(b) If the two capacitors are now joined, with plates of like charge connected together, what is the final common p.d.? [ 80 V ]

(c) What is the energy stored after the connection?[ 32x10-3 J ]

(d) What happens to the difference in energy stored? [ loss energy = 3x10-3 J ]

2) A 2 F capacitor is charged to a potential of 200 V and then isolated. When it is connected in parallel with a second capacitor which is initially uncharged, the common potential becomes 40 V. Find the capacitance of the second capacitor.

[ 8 F ]

3) Given a number of capacitors each with a capacitance of 2F and a maximum safe working potential difference of 10 V, how would you construct capacitors of

(a) 1 F capacitance, suitable for use up to 20 V

(b) 2 F capacitance, suitable for use up to 20 V?

4) A capacitor of capacitance 160 F is charged to a potential difference of 200 V and then connected across a discharge tube, which conducts until the potential difference across it has fallen to 100 V. Calculate the energy dissipated in the tube. [ 2.4 J ]