Chapter 26
capacitance
26-1 Definition of Capacitance
26-2 Calculating Capacitance
26.3
Combinations of Capacitors
26.4
Energy Stored in a Charged Capacitor
26.5
Capacitors with Dielectrics
Norah Ali Al moneef
1
26.1 definition of capacitance
Consider two conductors carrying charges of equal magnitude and opposite sign,
as shown in the Figure. Such a combination of two conductors is called a capacitor.
The conductors are called plates. A potential difference Δ V exists between the conductors
due to the presence of the charges
Capacitor:Two isolated conductors Equal and opposite charges ±Q
Potential difference DV between them.
Experiments show the quantity of electric charge Q on
a capacitor is linearly proportional to the potential
difference between the conductors, that is Q ~ DV. Or
we write Q = C DV
The capacitance C of a capacitor is the ratio of the
magnitude of the charge on either conductor to the
magnitude of the potential difference between them:
C 
C is Always Positive
Q
DV
Units: Coulombs/Volt or Farads
Norah Ali Al moneef
2
2
Example
• A storage capacitor on a random access memory
(RAM) chip has a capacitance of 55 nF. If the
capacitor is charged to 5.3 V, how many excess
electrons are on the negative plate?
We can find the number of excess electrons on the
negative plate if we know the total charge q on the
plate. Then, the number of electrons n=q/e, where e
is the electron charge in coulomb.
:The charge q of the plate is related to the voltage V to
which the capacitor is charged: q=CV.
Norah Ali Al moneef
3
Example
A 0.75 mF capacitor is charged to a voltage
of 16 volts. What is the magnitude of the
charge on each plate of the capacitor?
V = 16 V, C = 0.75 mF = 0.75 x 10-6 F
C = Q/V
Q = CV
Q = (0.75 x 10-6)(16) = 1.2 x 10-5 C
Norah Ali Al moneef
4
26.2 Calculating Capacitance
Capacitance of an isolated sphere
Calculate the capacitance of an isolated spherical conductor of
radius R and charge Q by assuming that the second conductor
making up the capacitor is a concentric hollow sphere of
infinite radius.
Electric potential of the sphere of radius R is
kQ/R and V= 0 at infinity, we have
C=
Q
DV
=
Q
kQ/R
=
R
k
Q
= 4peo R
C is proportional to its radius and independent of both the
charge on the sphere and the potential difference.
The capacitance of a pair of conductors depends on the geometry of the
conductors.
Norah Ali Al moneef
5
Example
(a) If a drop of liquid has capacitance 1.00 pF, what is its
radius ? (b) If another drop has radius 2.00 mm, what is its
capacitance ? (c) What is the charge on the smaller drop if
its potential is 100V ?
C = 4πeo R
R = (9 x 109 N · m2/C2)(1.00 x 10–12 F) = 9 mm
C = 4 π (8.85 x 10-12) x 2.0x10-3 = 0.222 pF
Q =CV = 0.222 pF x 100 V = 2.22 x 10-11 C
Norah Ali Al moneef
6
Example
What is the capacitance of the Earth ?
Think of Earth spherical conductor and the outer
conductor of the “spherical capacitor” may be
considered as a conducting sphere at infinity where V
approaches zero.
C  4p e 0 R


 4p 8.85  10 12 C N  m 2 6.37  10 6 m

C= 7.08 x 10-4 F
A large capacitor !
Norah Ali Al moneef
7
Example
: A 0.75 mF capacitor is charged to a voltage of 16 volts.
What is the magnitude of the charge on each plate
of the capacitor?
V = 16 V, C = 0.75 mF = 0.75 x 10-6 F
C = Q/V or Q = CV
Q = (0.75 x 10-6)(16) = 1.2 x 10-5 C
Norah Ali Al moneef
8
Parallel - Plate Capacitors
A parallel-plate capacitor consists of two
parallel conducting plates, each of area A,
separated by a distance d. When the
capacitor is charged, the plates carry equal
amounts of charge. One plate carries
positive charge, and the other carries
negative charge. And surface charge
density of each plate is s = Q/A
The plates are charged by connection to a battery..
Norah Ali Al moneef
9
Top Sheet:
E
Bottom Sheet:
s
2e 0
E
++++++++++++++
s
2e 0
- - - - - - - -- - - - - -
E 0
E ?
s
E
2e 0
s
E
2e 0
d
E 0
Q  s A
s
s
s
Q
E



2e 0
2e 0
e0
Ae 0
Norah Ali Al moneef
10
10
Variation with A If plates are large, then charges can distribute
themselves over a substantial area, and the amount of charge that can
be stored on a plate for a given potential diff increases as A is
increased. Thus we expect C to be proportional to A
C~A
Variation with d Potential difference DV constant across, E field
increases as d decreases. Imagine d decreases and consider situation
before any charges have had a chance to move in response to this
change. Because no charge move  E the same but over a shorter
distance. DV = Ed means that DV decreases
The difference between this new capacitor voltage and the terminal
voltage of the battery now exists as a potential difference across the
wires connecting the battery to the capacitor.
Norah Ali Al moneef
11
A E field result in the wires that drives more charge onto
the plates, increasing the potential diff. DV until it
matches that of the battery.  potential diff. Across
wire = 0  flow of charges stop.
More charges has accumulated at the capacitor as a
result.
We have d decrease, Q increases. Similarly d increases
 Q decreases.
Capacitance inversely proportional to d.
 C ~ 1/d
Norah Ali Al moneef
12
Assume electric field uniform
between the plates, we have
(see lecture on Gauss’s Law)
Q
• Charge density: s  A
s Q
• Electric field: E  
e0 e0 A
1 Qd
• Potential diff.: Vab  Ed 
e0 A
  b 
Vab  Va  Vb   E  d    E  d   Ed
a
b
The general properties of a
parallel-plate capacitor – that
the capacitance increases as the
plates become larger and
decreases as the separation
increases – are common to all
capacitors.
a
Q
A Only the geometry of the
 e0
• Capacitance: C 
Vab
d plates (A and d) affect the
capacitance.
Norah Ali Al moneef
13
(a) The electric field between the plates of a parallel-plate
capacitor is uniform near the center but nonuniform
near the edges.
(b) Electric field pattern of two oppositely charged
conducting parallel plates.
Norah Ali Al moneef
14
Example
• A huge parallel plate capacitor consists of two
square metal plates of side 50 cm, separated by an
air gap of 1 mm
• What is the capacitance?
e
C = 0A/d
= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)
= 2.21 x 10–9 F
Norah Ali Al moneef
15
Example
• We have a parallel plate capacitor constructed of two parallel plates,
each with area 625 cm2 separated by a distance of 1.00 mm.
• What is the capacitance of this parallel plate capacitor?
e0A
8.85  10-12 F/m  0.0625 m 2
C

d
0.001 m
 5.53  10
10
F
C = 0.553 nF
Result: A parallel plate capacitor constructed out of square
conducting plates 25 cm x 25 cm separated by 1 mm produces a
capacitor with a capacitance of about 0.5 nF.
Norah Ali Al moneef
16
Example
• We have a parallel plate capacitor constructed of two
parallel plates separated by a distance of 1.00 mm.
• What area is required to produce a capacitance of 1.00 F?
1 F  0.001 m
A

e 0 8.85  10-12 F/m
Cd
 1.13  108 m 2
Result: A parallel plate capacitor constructed out of square
conducting plates 10.6 km x 10.6 km (6 miles x 6 miles) separated
by 1 mm produces a capacitor with a capacitance of 1 F.
Norah Ali Al moneef
17
Example.
• What is the AREA of a 1 F capacitor that has a plate
separation of 1 mm?
d = 1 mm = 0.001 m,
eo = 8.85 x 10-12 C2/(Nm2)
A
C  eo
D
A
12
1  8.85 x10
0.001
A  1.13 x 108 m2
Sides 
10629 m
Norah Ali Al moneef
18
example
: Changing Dimensions
A parallel-plate capacitor has plates with equal and
opposite charges ±Q, separated by a distance d, and is
connected to a battery. The plates are pulled apart to a
distance D > d. What happens?
1.
2.
3.
4.
5.
6.
7.
8.
V increases,
V decreases,
V is the same,
V increases,
V decreases,
V is the same,
V increases,
V decreases,
Q increases
Q increases
Q increases
Q is the same
Q is the same
Q is the same
Q decreases
Q decreases
9.
V is the same,
Q decreases
Norah Ali Al moneef
19
Answer: 9. V is the same, Q decreases
With a battery connected to the plates the
potential V between them is held constant
In this situation, since
V=Ed
As d increases, E must decrease.
Since the electric field is proportional to the
charge on the plates, Q must decrease as well.
Norah Ali Al moneef
20
20
Example.
A parallel plate capacitor is constructed with plate of an area
of 0.028 m2 and a separation of 0.55 mm. Find the
magnitude of the charge of this capacitor when the
potential difference between the plate is 20.1 V.
A = 0.028 m2 d
= 0.55 mm = 0.00055 m,
V = 20.1 V
Q = CV
C = eo A/d
C = (8.85 x 10 -12)(0.028) /(0.00055)= 4.51 x 10-10 F
Q = (4.51 x 10-10 )(20.1) = 9.06 x 10-9 C
Norah Ali Al moneef
21
Example
Regarding the Earth and a cloud layer 800 m above the Earth as
the “plates” of a capacitor, calculate the capacitance if the cloud
layer has an area of 1.00 x 1.00 km2. Assume that the air
between the cloud and the ground is pure and dry.
Assume that charge builds up on the cloud and on the ground
until a uniform electric field of 3.00 x 106 N/C throughout the
space between them makes the air break down and conduct
electricity as a lightning bolt. What is the maximum charge the
cloud can hold?
C= eoA/d = 8.85x10-12 x (1000)2/800 = 11.1 nF
Potential between ground and cloud is
DV = Ed = 3.0 x106 x 800 = 2.4 x 109 V
Q = C(DV) = 26.6 C
Norah Ali Al moneef
22
26.3 Combinations of Capacitors
Norah Ali Al moneef
23
Parallel Combination
• When a potential difference V is applied across several
capacitors connected in parallel, that potential
difference V is applied across each capacitor.
• The total charge q stored on the capacitors is the sum of
the charges stored on all the capacitors.
• Capacitors connected in parallel can be replaced with an
equivalent capacitor that has the same total charge q and
the same potential difference V as the actual capacitors.
Q  Q1  Q2
Q1  C1DV1  C1DV
DV  DV1  DV2
Q2  C2 DV2  C2 DV
Q  Ceq DV
Ceq  C1  C2
For parallel capacitors
n
Ceq   C j
j 1
Norah Ali Al moneef
24
Capacitors in Series
•When a potential difference V is applied
across several capacitors connected in series, the
capacitors have identical charge q.
•The sum of the potential differences across all the
capacitors is equal to the applied potential difference
V.
•Capacitors that are connected in series can be
replaced with an equivalent capacitor that has the
same charge q and the same total potential difference
V as the actual series capacitors.
DV  DV1  DV2
Q  Q1  Q2
DV 
Q
Ceq
SERIES:
• Q is same for all capacitors
• Total potential difference = sum of V
For series capacitors
DV1 
Q1
Q

C1
C1
DV2 
Q2
Q

C2
C2
1
1
1
 
Ceq C1 C2
1

Ceq
Norah Ali Al moneef
n
1

j 1 C j
25
Example: Equivalent Capacitance
In series use 1/C=1/C1+1/C2
2.50 mF
20.00 mF
6.00 mF
In series use 1/C=1/C1+1/C2
8.50 mF
In parallel use C=C1+C2
5.965 mF
20.00 mF
Norah Ali Al moneef
26
Example: Equivalent Capacitance
In parallel use C=C1+C2
In parallel use C=C1+C2
In series use 1/C=1/C1+1/C2
Norah Ali Al moneef
27
Example: Equivalent Capacitance
In parallel use Ceq=C+C/2+C/3
C
In series use 1/CA=1/C+1/C
C/2
C/3
In series use 1/CB=1/C+1/C+1/C
Norah Ali Al moneef
28
Example:
Step 1
Step 2
Step 1:
Cp=C1+C2
Cp=0.10 mF+0.20 mF
Cp =0.30 mF
Step 2:
1/Cs=1/C3+1/Cp
Cs 
C3 C p
C3  C p
0.60mF  mF

 0.20mF
0.60mF  mF
Example
Find the Charge stored in the capacitor.
Q
C
V
Q  CV
12V
20μF
C
Q  (20  10
)(12V )
V
4
Q  2.4  10 C
Norah Ali Al moneef
6
30
Example Find the total charge stored in each the
capacitor.
First calculate the total capacitance.
20μF
12V
10μF
Then find the total charge
Q
C
V
Q  CV  (5.45 106 F )(12V )  6.54 105 C
1
1

Cs
i Ci
30μF
1
1
1
1



 C s  5.45mF
Cs 20 10 30
In a series circuit the charge stored on each capacitor is the
same and the voltage is split.
Norah Ali Al moneef
31
Example Find the total charge stored in all the
capacitors.
First find the total capacitance.
12V
20μF
10μF
calculate the total charge.
30μF
Q
C
V
Q  CV  (60 106 F )(12V )  7.2 104 C
C p   Ci
i
C p  20  10  30  60 mF
In a parallel circuit the charge stored on each capacitor can be
different and the voltage must
be the same.
Norah Ali Al moneef
32
Example Find the total capacitance of all the
capacitors.
First find the total capacitance of the
parallel capacitors
C p  8  4  12 mF
12μF
combine this value in series with
12μF capacitor.
12V
8μF
4μF
1
Ctotal
1 1
 
12 12
Ctotal  6mF

Energy Stored in a Capacitor
1
Q2 1 2 1
U C  QV 
 CV   6 106  12
2
2C 2
2
Units: Joules (J)
Norah Ali Al moneef

2
33
Example.
½C
2C
1/3 C
3C
C
2/5 C
Norah Ali Al moneef
34
Example
Step 1
C1
C2
parallel
V
C12
V
C3
C12  C1  C2
C1 = 12.0 mF, C2 = 5.3 mF, C3 = 4.5 mF
Step 2
series
V
C123
C123 
C12C3
C12  C3
C3
1
1
1


C123 C12 C3
C123 = (12 + 5.3)4.5/(12+5.3+4.5) mF = 3.57 mF
Example
C4
C45
C5
C1
V
C2
C3
C3
C3
C6
C6
parallel
C1456  C1  C45  C6
series
CC
C45  4 5
C4  C5
parallel
C23  C2  C3
Example
C45 
C1456
C1456  C1  C45  C6
C23  C2  C3
series
V
C23
C4C5
C4  C5
C123456 
Complete solution


CC
 C1  4 5  C6 (C2  C3 )
C 4  C5

C123456  
CC
C1  4 5  C6  C2  C3
C 4  C5
C1456C23
23
C1456  C23
23
26.4 Energy Stored in Capacitor
Suppose that, at a given instant, a charge q′ has
been transferred from one plate of a capacitor to
the other. The potential difference DV between the
plates at that instant will be q / C.
If an extra increment of charge dq is then
transferred, the increment of work required will
be,
- +
dW = DV dq
Norah Ali Al moneef
38
38
Energy To Charge Capacitor
1. Capacitor starts uncharged.
2. Carry +dq from bottom to top.
Now top has charge q = +dq,
bottom -dq
3. Repeat
4. Finish when top has charge q =
+Q,
bottom
-Q
At some point top plate has +q,
bottom has –q
Potential difference is DV = q /
C
Work done lifting another dq is
dW = dq DV
Norah Ali Al moneef
+q
-q
39
39
So work done to move dq is:
q
1
dW  dq DV  dq

q dq
C
C
Total energy to charge to
Q
q=W
Q:  dW  1 q dq 

1 Q2
C 
C 2
0
A plot of potential difference versus charge for
a capacitor is a straight line having a slope 1/C.
The work required to move charge dq through
the potential difference DV across the capacitor
plates is given by the area of the shaded
rectangle. The total work required to charge
the capacitor to a final charge Q is the
triangular area under the straight line, W =
QDV/2.
1V = 1 J/C hence the unit for the area is joule J.
Norah Ali Al moneef
40
Energy Stored in Capacitor
Work done in charging the capacitor = electric potential
energy U stored in the capacitor.
Q
Since C 
DV
2
Q
1
1
U 
 Q DV  C DV
2C
2
2
2
Where is the energy stored???
Norah Ali Al moneef
41
41
Energy Stored in Capacitor
Energy stored in the E field!
C
Parallel-plate capacitor:
eo A
d
and V  Ed
Ad = Volume occupied by the E field. This lead to a new quantity
known as Energy Density
uE = U/Volume = U/Ad
1 eo A
1
2
U  CV 
2 d
2
 Ed 
2

eo E 2
2
 ( Ad )  uE  (volume)
uE  E field energy density 
eoE2
2
The energy stored in a capacitor can be put to a number of uses: a
camera flash; a cardiac defibrillator; and others. In addition, capacitors
form an essential part of most electrical devices used today.
Norah Ali Al moneef
42
42
Example
• Suppose the capacitor shown here is charge to Q
and then the battery is disconnected.
• Now suppose I pull the plates further apart so that the final
separation is d1.
A
• How do the quantities Q, C, E, V, U change?
++++
• Q: remains the same.. no way for charge to leave. d - - - - -
•
•
•
•
C:
E:
V:
U:
decreases.. since capacitance depends on geometry
remains the same... depends only on charge density
increases.. since C , but Q remains same (or d  but E the same)
increases.. add energy to system by separating
• How much do these quantities change?
Answers:
d
C1  C
d1
d1
V1  V
d
Norah Ali Al moneef
d1
U1  U
d
43
Example.
• Suppose the battery (V) is kept
attached to the capacitor.
V
A
++++
d -----
• Again pull the plates apart from d to d1.
• Now what changes?
•
•
•
•
•
C:
V:
Q:
E:
U:
decreases (capacitance depends only on geometry)
must stay the same - the battery forces it to be V
must decrease, Q=CV charge flows off the plate
s
must decrease ( E  V, E  )
E0
D
2
1
must decrease ( U  CV
)
2
• How much do these quantities change?.. exercise for student!!
Answers:
d
d
E

E
C1  C
1
d1 Norah Ali Al moneef d1
U1 
d
U
d1
44
example
A parallel-plate capacitor, disconnected from a
battery, has plates with equal and opposite charges,
separated by a distance d.
Suppose the plates are pulled apart until separated
by a distance D > d.
How does the final electrostatic energy stored in
the capacitor compare to the initial energy?
1. The final stored energy is smaller
2. The final stored energy is larger
3. Stored energy does not change.
Norah Ali Al moneef
45
45
Answer: 2. The stored energy increases
As you pull apart the capacitor plates you
increase the amount of space in which the E field
is non-zero and hence increase the stored energy.
Where does the extra energy come from? From
the work you do pulling the plates apart.
Norah Ali Al moneef
46
46
Example
A parallel-plate capacitor is charged and then
disconnected from a battery. By what fraction does the
stored energy change (increase or decrease) when the
plate separation is doubled ?
U = Q2/2C
then
C1= C2/2
and C = eoA/d
and d2 = 2 d1
and the energy stored doubles.
Norah Ali Al moneef
47
example
In a typical defibrillator, a 175 mF, is charged until the
potential difference between the plates is 2240 V. A.)
What is the charge on each plate?
V = 2240 V, C = 175 mF = 175 x 10-6 F
Q = CV = (175 x 10-6)(2240) = 0.392 C
B.) Find the energy stored in the charged up defibrillator.
U = ½ CV2 U = Q2/(2C)
U = ½ QV = ½ (0.392)(2240)
U = 439 J
Norah Ali Al moneef
48
Example
Consider the circuit as shown, where C1 = 6.00mF and C2= 3.00
mF and DV =20.0V. Capacitor C1 is first charged by closing of
switch S1. Switch S1 is then opened and the charged capacitor is
connected to the uncharged capacitor by the closing of S2.
Calculate the initial charge acquired by C1 and the final charge
on each.
S1 close, S2 open 
C = Q/V  Q = 120 mC
After S1 open, S2 close 
Q1 + Q2 = 120 mC
Same potential  Q1 /C1 = Q2 / C2 
(120-Q2)/C1= Q2/C2
(120 - Q2)/6 = Q2/ 3  Q2 = 40 mC
 Q 1= 80 mC
Norah Ali Al moneef
49
example
--Find the charge on (Q) and potential
difference (DV) across each capacitor. What
is the total energy stored in the system?
Cequ  C2  C8 (parallel)
= 2mF + 8mF
= 10mF
1
1
1
1



Cequ
C5
C10
C2

(series)
1
1
1


5mF
10mF
2mF
8

10mF

 C equ 

5mF
4

50
Q2 1
UC 
 C(DV ) 2
2C 2
(15mF) 2

 90mJ
2(5 /4mF )
Q  CDV
5
Qtot  ( mF)(12V )  15mC
4

Q5  Q10  Q2  Qtot  15mC (in series)


DV5 
Q5 15mC

 3V
C5 5mF
DV10 
Q10 15mC


 1.5V
C10 10mF
DV2 
Q2 15mC

 7.5V
C2 2mF


DVC 
DV2 = DV8 = DV10 =1.5V (in parallel)

QC
CC
(always)
QC  CC DVC
(always)
Q2  C2DV2  (2mF)(1.5V)  3mC
Q8  C8DV8  (8mF)(1.5V)
 12mC



51
26.5 Capacitors with Dielectrics
A dielectric is a nonconducting
material, such as rubber, glass, or
waxed paper. Dielectric constant is a
property of a material and varies
from one material to another.
A charged capacitor (a) before and (b) after insertion of a
dielectric between the plates. The charge on the plates
remains unchanged, but the potential difference
decreases from DVo to DV = DVo/k.
Thus the capacitance increases from Co to kCo.
Note no battery is involved in this example.
Norah Ali Al moneef
52
When a dielectric is inserted between the plates of a capacitor, the
capacitance increases.
If the dielectric completely fills the space between the plates, the
capacitance increases by a dimensionless factor k , which is called
the dielectric constant.
Capacitance increases by the factor k when dielectric completely
fills the region between the plates.
If the dielectric is introduced while the potential difference is being
maintained constant by a battery, the charge increases to a value
Q = k Qo . The additional charge is supplied by the battery and the
capacitance again increases by the factor k.
C = k
For parallel plate capacitor:
Norah Ali Al moneef
eoA
d
53
Dielectrics
• Definition:
V0
V
k
E0
E
k
The dielectric constant of a material is the ratio of the capacitance
when filled with the dielectric to that without it:
k 
C
C0
k values are always > 1 (e.g., glass = 5.6; water = 78)
They INCREASE the capacitance of a capacitor (generally good, since
it is hard to make “big” capacitors)
They permit more energy to be stored on a given capacitor than
otherwise with vacuum (i.e., air):
CV 2 kC0V 2
U 
Norah Ali Al moneef
2

2
 kU
54
Dielectric Strength
If the electric field in a dielectric becomes too large, it can
tear the electrons off the atoms, thereby enabling the
material to conduct. This is called dielectric breakdown;
the field at which this happens is called the dielectric
strength. For any given separation d, the maximum voltage
that can be applied to a capacitor without causing a
discharge depends on the dielectric strength (maximum
electric field) of the dielectric.
. If magnitude of the electric field in the dielectric
exceeds the dielectric strength, then the insulating
properties break down and the dielectric begins to
conduct.
Norah Ali Al moneef
55
the advantages of dielectric material in a capacitor?
•Increase the capacitance
• Increase the maximum operating voltage
• Possible mechanical support between the plates, which allows the
plates to be close together without touching, thereby decreasing d
and increasing C.
Norah Ali Al moneef
56
Example.
Two identical parallel plate capacitors are given
the same charge Q, after which they are
disconnected from the battery. After C2 has
been charged and disconnected it is filled with
a dielectric.
Compare the voltages of the two capacitors.
a) V1 > V2
b) V1 = V2
c) V1 < V2
Compare the electric fields between the plates of both capacitors.
a) E1 > E2
b) E1 = E2
c) E1 < E2
Norah Ali Al moneef
57
Example.
When we insert the dielectric into the
capacitor C2 we do:
a) positive work
b) negative work
c) no work
Recall the meaning of negative work is the energy of the system is
reduced. The dielectric is sucked into the capacitor. When the charge
is constant, the total energy of the capacitor decreases because the
presence of the dielectric increases the capacitance . It turns out
that the dielectric is pulled in even if the voltage is held constant,
e.g., via a battery. On a microscopic scale the force on the dielectric
arises due to “fringing fields” at the edges of the capacitor.
Norah Ali Al moneef
58
Examples:
(a) A parallel-plate capacitor is connected to a power supply that
maintains a constant potential difference V0 between the two
plates. If the gap between the plates is first filled with air
(≈ vacuum) and then a dielectric with dielectric constant κ=3 is
inserted, how do the following quantities change:
•The capacitance of the capacitor?
•The charge on the positive plate of the capacitor?
•The electric field between the capacitor plates?
•The energy stored in the capacitor?
C
ke 0 A
d
 If A and d don’t change then: if κ ↑3  C↑3
Q  CV  If V doesn’t change then: if C↑3
Norah Ali Al moneef
Q↑3
59
Example.
Two identical parallel plate capacitors are connected in series to a
battery as shown below. If a dielectric is inserted in the
lower capacitor, which of the following increase for that
capacitor?
A.
B.
C.
D.
E.
I and III.
I, II and IV.
I, II and III.
All except II.
All increase.
q  CV
C
I.
Capacitance of capacitor
II.
Voltage across capacitor
III.
Charge on capacitor
IV.
Energy stored on capacitor
C
V
ke 0 A
d
q2 1
U
 2 CV 2
2C
k
C
A Closer Look
•
Insert dielectric

Capacitance goes up by k

Charge increases

Charge on upper plate comes from upper
capacitor, so its charge also increases.

Since q’ = CV1 increases on upper
capacitor, V1 must increase on upper
capacitor.

Since total V = V1 + V2 = constant, V2 must
decrease.
q
q’
C
V
V
q’
k
CkC V
Example.
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the charges on
the plates remain constant.
What effect does adding the dielectric have on the
energy stored in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide
Example.
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the potential
difference between the plates remains constant.
What effect does adding the dielectric have on the
amount of charge on each of the capacitor plates?
A. The amount of charge increases.
B. The amount of charge remains the same.
C. The amount of charge decreases.
D. not enough information given to decide
Example.
You slide a slab of dielectric between the plates of a parallel-plate
capacitor. As you do this, the potential difference between the
plates remains constant.
What effect does adding the dielectric have on the energy stored
in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide
Norah Ali Al moneef
65
Example.
A parallel-plate capacitor is attached to a battery
that maintains a constant potential difference V
between the plates. While the battery is still
connected, a glass slab is inserted so as to just fill
the space between the plates. The stored energy
1. increases.
2. decreases.
3. remains the same.
Norah Ali Al moneef
66
Example.
A parallel plate capacitor is charged to a total charge Q and the battery removed. A slab of
material with dielectric constant k in inserted between the plates. The charge stored in the
capacitor
1. Increases
2. Decreases
3. Stays the Same
+ + + + + + +
k
- - - - - - -
Answer: 3. Charge stays the same
Since the capacitor is disconnected from a battery there
is no way for the amount of charge on it to change.
Norah Ali Al moneef
67
67
Example.
A parallel plate capacitor is charged to a total charge Q and the
battery removed. A slab of material with dielectric constant k in
inserted between the plates. The energy stored in the capacitor
+ + + + + + + +
1. Increases
2. Decreases
3. Stays the Same
k
- - - - - - - -
Answer: 2. Energy stored decreases
The dielectric reduces the electric field and hence reduces the
amount of energy stored in the field.
The easiest way to think about this is that the capacitance is
increased while the charge remains the same so U = Q2/2C 2
1
1
E
2
uE ,0  e 0 E  ke 0     u E ,0
Also from energy density:
2
2
k 
Norah Ali Al moneef
68
68
Calculating Capacitance
 Units
1 F = 1 C2/N m (Note [e]C2/N m2)
1 mF = 10-6 F, 1 pF = 10-12 F
e0 = 8.85 x 10-12 F/m
 Example
: Size of a 1-F capacitor ,calculate its area
d  1 mm , C  1.0 F
(1.0 F)(1.0 10 3 m)
8
2
A


1
.
1

10
m
e0
8.85 10 12 F/m
Cd
 Example
A parallel - palte capacitor in vacuum
d  5.00 mm , A  2.00 m 2 , V  10,000 V  10.0 kV
Calculate the Capacitance ,the charge on the plates
and the electric field
A (8.85 10 12 F/m)(2.00 m 2 )
C  e0 
d
5.00 10 3 m
 3.54 10 5 F  0.00354 mF
Q  CVab  (3.54 109 C/V)(1.00 104 V)
 3.54 105 C  35.4 mC
s
Q
3.54 105 C
E 

e 0 e 0 A (8.85 1012 C2 / N  m 2 )( 2.00 m 2 )
 2.00 106 N/C
Example.
Find the capacitance of a 4.0 cm diameter sensor
immersed in oil if the plates are separated by 0.25 mm.
e r  4.0 for oil
C  8.85 10
The plate area is
12
 er A 
F/m 

d


A  πr 2  p 0.02 m 2  1.26 103 m 2
The distance between the plates is


0.25 103 m
  4.0  1.26 103 m 2 
C  8.85 1012 F/m 

0.25 103 m





178 pF
Capacitors in Series and Parallel
example
In this problem you try to measure dielectric constant of a material. First
a parallel-plate capacitor with only air between the plates is charged by
connecting it to a battery. The capacitor is then disconnected from the
battery without any of the charge leaving the plates.
(a) Express the capacitance C0 in terms of the potential difference V0
between the plates and the charge Q if air is between the plates.
(b) Express the dielectric constant k in terms of the capacitance C0 (air gap)
and the capacitance C with material of the dielectric constant k).
(c) Using the results of (a) and (b), express the ratio of the potential
difference V/V0 if Q is the same, where V is the potential difference
between the plates and a dielectric material dielectric constant is k
fills the space between them.
(d) A voltmeter reads 45.0 V when placed across the capacitor. When
dielectric material is inserted completely filling the space, the voltmeter
reads 11.5 V. Find the dielectric constant of this material.
(a)
(b)
(c)
(d)
C0  Q /V0
k  C / C0
V0 / V  C / C0  k  V / V0  1 / k
From (c)
k  V0 / V  45.0 / 11.5  3.91
Example.
If a 22 mF capacitor is connected to a 10 V source,
the charge is
220 mC
Example.
If a 0.001 mF capacitor is connected in series
with an 800 pF capacitor, the total capacitance
is
444 pF
C1
0.001 µF
C2
800 pF
Example.
If a 0.001 mF capacitor
is connected in parallel
with an 800 pF
capacitor, the total
capacitance is
1800 pF
C1
C2
0.001 µF
800 pF
Example.
If a 0.015 mF capacitor is in series with a 6800 pF
capacitor, the total capacitance is
a. 1568 pF
b. 4678 pF
c. 6815 pF
d. 0.022 mF
Example.
Two capacitors that are initially uncharged are
connected in series with a dc source. Compared
to the larger capacitor, the smaller capacitor will
have
a. the same charge
b. more charge
c. less voltage
d. the same voltage
Example.
(a) Calculate the capacitance of a parallel-plate
capacitor whose plates are 20 cm × 3.0 cm and are
separated by a 1.0-mm air gap. (b) What is the charge
on each plate if a 12-V battery is connected across the
two plates? (c) What is the electric field between the
plates? (d) Estimate the area of the plates needed to
achieve a capacitance of 1 F, given the same air gap d.
Solution: a. C = 53 pF.
b. Q = CV = 6.4 x 10-10 C.
c. E = V/d = 1.2 x 104 V/m.
d. A = C d /ε0 = 108 m2.
Example.
Determine the capacitance of a single capacitor that
will have the same effect as the combination shown.
Solution: First, find the equivalent capacitance of the
two capacitors in parallel (2C);
then the equivalent of that capacitor in series with
the third (2/3 C).
Example
A camera flash unit stores energy in a 150-μF capacitor
at 200 V. (a) How much electric energy can be stored?
Solution:
a. U = ½ CV2 = 3.0 J.
Example.
The plates of a parallel-plate capacitor have area A, separation x,
and are connected to a battery with voltage V. While connected
to the battery, the plates are pulled apart until they are separated
by 3x. (a) What are the initial and final energies stored in the
capacitor? (b) How much work is required to pull the plates apart
(assume constant speed)? (c) How much energy is exchanged
with the battery?
a. U = ½ CV2, and increasing the plate separation
decreases the capacitance, so the initial and final
energies can be calculated from this.
b. The work is equal to the negative of change in energy.
c. The energy of the battery is increased by the work
done minus the change in potential energy of the
capacitor (which is negative).
Example
A parallel-plate capacitor, filled with a dielectric with K = 3.4, is
connected to a 100-V battery. After the capacitor is fully charged,
the battery is disconnected. The plates have area A = 4.0 m2 and
are separated by d = 4.0 mm
(a) Find the capacitance, the charge on the capacitor, the
electric field strength, and the energy stored in the capacitor.
. (b) The dielectric is carefully removed, without
changing the plate separation nor does any charge
leave the capacitor. Find the new values of capacitance,
electric field strength, voltage between the plates, and
the energy stored in the capacitor.
Solution: a. C = Kε0A/d = 3.0 x 10-8 F. Q = CV = 3.0 x 10-6 C.
E = V/d = 25 kV/m. U = ½ CV2 = 1.5 x 10-4 J.
B. Now C = 8.8 x 10-9 F, Q = 3.0 x 10-6 C (no change), V = 340 V,
E = 85 kV/m, U = 5.1 x 10-4 J. The increase in energy comes from the
work it takes to remove the dielectric.
Example
Capacitor C1 is connected across a
battery of 5 V. An identical
capacitor C2 is connected across a
battery of 10 V. Which one has
more charge?
1) C1
2) C2
3) both have the same charge
4) it depends on other factors
Since Q = CV and the two capacitors are
identical, the one that is connected to
the greater voltage has more charge,
which is C2 in this case.
Example
What must be done to a
capacitor in order to
increase the amount of
charge it can hold (for a
constant voltage)?
1) increase the area of the plates
2) decrease separation between the plates
3) decrease the area of the plates
4) either (1) or (2)
5) either (2) or (3)
+Q –Q
Since Q = CV, in order to increase the charge that a
capacitor can hold at constant voltage, one has to
increase its capacitance. Since the capacitance is
given by
, that can be
C done
e 0 Aby either
increasing A or decreasing d.
d
Example
A parallel-plate capacitor
initially has a voltage of 400 V
and stays connected to the
battery. If the plate spacing is
now doubled, what happens?
1) the voltage decreases
2) the voltage increases
3) the charge decreases
4) the charge increases
5) both voltage and charge change
Since the battery stays connected, the voltage
must remain constant! Since
,
A spacing d is doubled, the capacitance
when
C
 e 0the
d
C is halved. And since Q = CV, that means the
charge must decrease.
+Q
–Q
Example
A parallel-plate capacitor initially has a
potential difference of 400 V and is
then disconnected from the charging
battery. If the plate spacing is now
doubled (without changing Q), what is
the new value of the voltage?
1) 100 V
2) 200 V
3) 400 V
4) 800 V
5) 1600 V
+Q –Q
Once the battery is disconnected, Q has to remain
constant, since no charge can flow either to or from
the battery. Since
, when the
Cspacing
 e 0 Ad is doubled, the capacitance C is halved.
d
And since Q = CV, that means the voltage must
double.
Example
1) Ceq = 3/2C
What is the equivalent capacitance, Ceq ,
2) Ceq = 2/3C
of the combination below?
3) Ceq = 3C
4) Ceq = 1/3C
5) Ceq = 1/2C
The 2 equal capacitors in series add up as
inverses, giving 1/2C. These are parallel to
the first one, which add up directly. Thus,
the total equivalent capacitance is 3/2C.
o
Ceq
C
o
C
C
Example
How does the voltage V1 across
the first capacitor (C1) compare to
the voltage V2 across the second
capacitor (C2)?
The voltage across C1 is 10 V. The
combined capacitors C2 + C3 are
parallel to C1. The voltage across C2
+ C3 is also 10 V. Since C2 and C3
are in series, their voltages add.
Thus the voltage across C2 and C3
each has to be 5 V, which is less
than V1.
1) V1 = V2
2) V1 > V2
3) V1 < V2
4) all voltages are zero
C2 = 1.0 mF
C1 = 1.0 mF
C3 = 1.0 mF
10 V
Follow-up: What is the current in
this circuit?
Example
1) Q1 = Q2
How does the charge Q1 on the
first capacitor (C1) compare to
the charge Q2 on the second
capacitor (C2)?
2) Q1 > Q2
3) Q1 < Q2
4) all charges are zero
We already know that the voltage
C2 = 1.0 mF
across C1 is 10 V and the voltage
across both C2 and C3 is 5 V each.
Since Q = CV and C is the same for
all the capacitors, we have V1 > V2
and therefore Q1 > Q2.
10 V
C1 = 1.0 mF
C3 = 1.0 mF
How much energy is stored in a 2.0 mF capacitor that has
been charged to 5000 V? What is the average power
dissipation if the capacitor is discharged in 10 ms?
The spacing between the plates of a 1.0 mF capacitor is
1 mm.
(a)What is the surface area A of the plates?
(b)How much charge is on the plates if this capacitor is
attached to a 1.5 V battery?
Two flat parallel plates are d = 0.40 cm apart. The
potential difference between the plates is 360 V. The
electric field at the point P at the center is
approximately
A.90 kN/C.
B.180 N/C.
C.0.9 kN/C.
D.Zero.
E. 3.6  105 N/C
Two large metallic plates are parallel to each other and
charged. The distance between the plates is d. The
potential difference between the plates is V. The
magnitude of the electric field E in the region between the
plates and away from the edges is given by
A. d/V.
B. V2/d.
C. d ∙V.
D. V/d2.
E. V/d .
A capacitor of capacitance C holds a charge Q
when the potential difference across the plates
is V. If the charge Q on the plates is doubled to
2Q,
A. the capacitance becomes (1/2)V.
B. the capacitance becomes 2C.
C. the potential changes to (1/2)V.
D.the potential changes to 2V.
E. the potential does not change.
If a capacitor of capacitance 2.0 µF is given a
charge of 1.0 mC, the potential difference across
the capacitor is
A.0.50 kV.
B.2.0 V.
C.2.0 µV.
D.0.50 V.
E. None of these is correct.
If the area of the plates of a parallel-plate
capacitor is doubled, the capacitance is
A.not changed.
B.doubled.
C.halved.
D.increased by a factor of 4.
E. decreased by a factor of 1/4.
An 80-nF capacitor is charged to a potential of
500 V. How much charge accumulates on each
plate of the capacitor?
A.4.0  10–4 C
B.4.0  10–5 C
C.4.0  10–10 C
D.1.6  10–10 C
E. 1.6  10–7 C
As the voltage in the circuit is increased (but not to the
breakdown voltage), the capacitance
A.increases.
B.decreases.
C.does not change.
Doubling the potential difference
across a capacitor
A.doubles its capacitance.
B.halves its capacitance.
C.quadruples the charge stored on the capacitor.
D.halves the charge stored on the capacitor.
E.does not change the capacitance of the
capacitor.
If the area of the plates of a parallel plate
capacitor is halved and the separation between
the plates tripled, then by what factor does the
capacitance change?
A.It increases by a factor of 6.
B.It decreases by a factor of 2/3.
C.It decreases by a factor of 1/6.
D.It increases by a factor of 3/2.
E. It decreases by a factor of ½.
Which of the following statements is
false?
A. In the process of charging a capacitor, an electric field is
produced between its plates.
B. The work required to charge a capacitor can be thought of as
the work required to create the electric field between its
plates.
C. The energy density in the space between the plates of a
capacitor is directly proportional to the first power of the
electric field.
D. The potential difference between the plates of a capacitor is
directly proportional to the electric field.
E. None of these is false.
Which of the following statements about a
parallel plate capacitor is false?
A.The two plates have equal charges of the same sign.
B. The capacitor stores charges on the plates.
C. The capacitance is proportional to the area of the
plates.
D.The capacitance is inversely proportional to the
separation between the plates.
E. A charged capacitor stores energy.
If you increase the charge on a parallel-plate capacitor
from 3 µC to 9 µC and increase the plate separation
from 1 mm to 3 mm, but keep all other properties the
same, the energy stored in the capacitor changes by a
factor of
A.27.
B.9.
C.3.
D.8.
E. 1/3.
The energy stored in a capacitor is directly
proportional to
A. the voltage across the capacitor.
B. the charge on the capacitor.
C. the reciprocal of the charge on the capacitor.
D.the square of the voltage across the capacitor.
E. None of these is correct.
A parallel plate capacitor is constructed using two
square metal sheets, each of side L = 10 cm. The plates
are separated by a distance d = 2 mm and a voltage
applied between the plates. The electric field strength
within the plates is E = 4000 V/m. The energy stored in
the capacitor is
A.0.71 nJ.
B.1.42 nJ.
C.2.83 nJ.
D.3.67 nJ.
E. Zero.
A circuit consists of a capacitor, a battery, and a switch,
all connected in series. Initially, the switch is open and
the capacitor is uncharged. The switch is then closed
and the capacitor charges. While the capacitor is
charging, how does the net charge within the battery
change?
A.It increases.
B.It decreases.
C.It stays the same
Several different capacitors are hooked
across a DC battery in parallel. The
A.directly proportional to its capacitance.
charge on each capacitor is
B.inversely proportional to its capacitance.
C.independent of its capacitance.
Several different capacitors are hooked
across a DC battery in parallel. The
voltage across each capacitor is
A.directly proportional to its capacitance.
B.inversely proportional to its capacitance.
C.independent of its capacitance.
Several different capacitors are hooked
across a DC battery in series. The charge on
each capacitor is
A.directly proportional to its capacitance.
B.inversely proportional to its
capacitance.
C.independent of its capacitance.
Several different capacitors are hooked
across a DC battery in series. The
voltage across each capacitor is
A.directly proportional to its capacitance.
B.inversely proportional to its
capacitance.
C.independent of its capacitance.
If C1 < C2 < C3 < C4 for the combination
of capacitors shown, the equivalent
capacitance is
A.less than C1.
B.more than C4.
C.between C1 and C4.
If C1 < C2 < C3 < C4 for the combination
of capacitors shown, the equivalent
capacitance is
A.less than C1.
B.more than C4.
C.between C1 and C4.
The equivalent capacitance of two
capacitors in series is
A.the sum of their capacitances.
B.the sum of the reciprocals of their
capacitances.
C.always greater than the larger of their
capacitances.
D.always less than the smaller of the
capacitances.
E. described by none of the above.
The equivalent capacitance of
three capacitors in series is
A.the sum of their capacitances.
B.the sum of the reciprocals of their
capacitances.
C.always greater than the larger of their
capacitances.
D.always less than the smaller of the
capacitances.
E. described by none of the above.
The equivalent capacitance of two
capacitors in parallel is
A.the sum of the reciprocals of their
capacitances.
B.the reciprocal of the sum of the reciprocals of
their capacitances.
C.always greater than the larger of their
capacitances.
D.always less than the smaller of the two
capacitances.
E. described by none of the above.
The capacitance of a parallel-plate
capacitor
A.is defined as the amount of work required to
move a charge from one plate to the other.
B.decreases if a dielectric is placed between its
plates.
C.is independent of the distance between the
plates.
D.has units of J/C.
E.is independent of the charge on the capacitor.
A capacitor is connected
to a battery as shown.
When a dielectric is
inserted between the
plates of the capacitor,
A.only the capacitance changes.
B.only the voltage across the capacitor
changes.
C.only the charge on the capacitor changes.
D.both the capacitance and the voltage change.
E.both the capacitance and the charge change.
Two identical
capacitors A and B
are connected
across a battery, as
shown. If mica (k =
5.4) is inserted in B,
A. both capacitors will retain the same charge.
B. B will have the larger charge.
C. A will have the larger charge.
D. the potential difference across B will increase.
E. the potential difference across A will increase.
If a dielectric is inserted between the plates of a
parallel-plate capacitor that is connected to a
100-V battery, the
A.voltage across the capacitor decreases.
B.electric field between the plates decreases.
C.electric field between the plates increases.
D.charge on the capacitor plates decreases.
E.charge on the capacitor plates increases.
A charged capacitor has an initial electric field E0 and potential
difference V0 across its plates. Without connecting any source of emf,
you insert a dielectric (k > 1) slab between the plates to produce an
electric field Ed and a potential difference Vd across the capacitor. The
pair of statements that best represents the relationships between the
magnitude of the electric fields and potential differences is
A. Ed > E0; Vd > V0.
B. Ed = E0; Vd > V0.
C. Ed > E0; Vd = V0.
D. Ed < E0; Vd > V0.
E. Ed < E0; Vd < V0.
Does the capacitance always increase
when a dielectric is inserted into the gap of
a capacitor?
A.Yes, it always increases.
B.No, it always decreases.
C.No, it may increase or decrease depending on
the dielectric constant of the material.
An external electric field, E, is applied to a region which
contains a dielectric. Which of the following statements
is true?
A. The electric field within the dielectric is less than E.
B. The dielectric produces an electric field in the opposite
direction to E.
C. The molecules in the dielectric become polarized.
D.The electric field will produce a torque on molecules in
the dielectric that have permanent dipoles.
E. All the above statements are true.
In real life we want to store more charge at lower voltage,
hence large capacitances are needed
Increased area, decreased separations, “stronger”
insulators
Electronic circuits – like a shock absorber in the car, capacitor smoothes power
fluctuations
Response on a particular frequency – radio and TV broadcast and receiving
Undesirable properties – they limit high-frequency operation
Summary
• Capacitance says how much charge is on an
arrangement of conductors for a given potential. q  CV
• Capacitance depends only on geometry
– Parallel Plate Capacitor
– Isolated Sphere
C 
e0 A
d
C  4pe0 R
• Units, F (farad) = C2/Nm or C/V (note e0 = 8.85 pF/m)
• Energy and energy density stored by capacitor
.
U 
1
2
CV 2
u
1
2
e0E2
Capacitors: series equivalent
DV1  DV2  DVequivalent
Q1  Q2  Qequivalent


Q1 Q2 Qequivalent


C1 C2 Cequivalent
1
1
1


C1 C2 Cequivalent
Phys 133 -- Chapter 30
129
Capacitors: parallel equivalent
DV1  DV2  DVequivalent
Q1  Q2  Qequivalent
C1DV  C2 DV2  CequivalentDVequivalent

C1  C2  Cequivalent

130
Capacitors
series
DV
DV1+DV2= DVeq
DV1=DV2
Q
Q1=Q2
Q1+Q2= Qeq
1
1
1



Ceq
C1 C2
n
1
1

Ceq j 1 C j

parallel

Ceq  C1  C2 
n
Ceq   C j
j 1
131
Dielectrics change the potential difference
• The potential between to
parallel plates of a capacitor
changes when the material
between the plates changes.
k is a unitless number
C
K
C0

V0
V
K
E0
E
K

Field lines as dielectrics change
• Moving from part (a) to
part (b) of Figure 24.15
shows the change induced
by the dielectric.
In vacuum, energy density is
1
u  e0 E 2
2
In dielectric
e  Ke 0
C  Ke 0
A
A
e
d
d
1
u  eE 2
2
•
With battery attached, V=const, so
more charge flows to the capacitor

q  CV
With battery disconnected, q=const,
so voltage (for given q) drops.
V
q  kCV
V 
Norah Ali Al moneef
q
C
q
kC
134
, capacitors with and without dielectrics
• If capacitor is disconnected from circuit, inserting a
dielectric changes decreases electric field, potential
and increases capacitance, but the amount of charge
on the capacitor is unchanged.
• If the capacitor is hooked up to a power supply with
constant voltage, the voltage must remain the same,
but capacitance and charge increase
Norah Ali Al moneef
135
Applications of Capacitors – Camera Flash
• The flash attachment on a camera uses a capacitor
– A battery is used to charge the capacitor
– The energy stored in the capacitor is released when the button
is pushed to take a picture
– The charge is delivered very quickly, illuminating the subject
when more light is needed
• Defibrillators
– When fibrillation occurs, the heart produces a rapid, irregular
pattern of beats
– A fast discharge of electrical energy through the heart can
return the organ to its normal beat pattern
• In general, capacitors act as energy reservoirs that can
slowly charged and then discharged quickly to provide
large amounts of energy in a short pulse