The document provides information about capacitors and capacitance. It defines capacitance as the ratio of the charge stored (Q) to the potential difference (V) between capacitor plates (C = Q/V). It describes how capacitance is measured in Farads and depends on the area of plates, distance between plates, and dielectric material. Capacitors can be arranged in series or parallel circuits, and formulas are provided to calculate the equivalent capacitance in these arrangements. Energy is stored in capacitors and the amount depends on the capacitance and potential difference according to the formula E = 1/2 CV^2.

1. 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. 2
Capacitors
symbol

3. 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 circuits
because they can transmit alternating currents.

4. 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. 5
Capacitance
 When an isolated spherical conductor is connected to a high
voltage supply, it is found that as the potential is increased, the
charge 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 of
the conductor and known as the capacitance
 The capacitance C of a capacitor is defined as the ratio of the
charge Q stored on either plate to the potential difference V
between the plates.
Thus: C = Q/V
 It can also be defined as the charge stored per unit p.d. applied to
the 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 a
conductor or the extent to which a capacitor can store charge. The
value 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. 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. 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. 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. 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. 10
Charging and discharging circuit
A
C
V
switch
R

11. 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 I0
is 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 I0
to 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 I0
flows 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. 12
I
+ charging C
O
time
- discharging C

13. 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.f
value 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. 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. 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 the
different paths.
• It can be proved that the effective capacitance C of such
connections is given by: C = C1
+ C2
+ C3
+ ... + Cn
• 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 three
capacitors is: Q = Q1
+ Q2
+ Q3
= C1
V0
+ C2
V0
+ C3
V0
since Q =
CV
• If the effective capacitance is C, then
Q = CV0
= C1
V0
+ C2
V0
+ C3
V0
giving
C = C1
+ C2
+ C3
• i.e in a parallel circuit the effective capacitance is the sum of
the individual capacitances (compare this with resistors)

16. 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. 17
Capacitors in Series - derivation
• When charged in series, each capacitor stores the same
amount of charge as the current through all of them is the
same.
• The effective/combined capacitance C for series connections
of capacitors is: 1/C = 1/C1
+ 1/C2
+ 1/C3
+ ...+ 1/Cn
• 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. 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. 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. 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. 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 graph
V
V0
0 Q
0 Q0

22. 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. 23
Capacitors and resistors on printed circuit
boards

24. 24
Resuscitator

25. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 35
Exercises
1) A capacitor, C1, of 4 F is charged to a potential difference of 50 V. Another
capacitor, C2, of 6F 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 2F 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 ]