CAPACITANCE:
As we’ve seen, when an object has a charge Q, it will
CHAPTER 24
ELECTROSTATIC ENERGY
and CAPACITANCE
• Capacitance and capacitors
• Storage of electrical energy
• Energy density of an electric field
• Combinations of capacitors
• In parallel
• In series
• Dielectrics
• Effects of dielectrics
• Examples of capacitors
(
)
have a potential V = U Q , where U is the work done to
assemble the charge. Conversely, if an object has a
potential V it will have a charge Q. The capacitance (C)
of the object is the ratio Q V.
+
+
+
+
Example: A charged spherical
+
R
+
conductor with a charge Q. The
electrostatic potential of the sphere is
Q
V= k .
R
Therefore, the capacitance of the charged sphere is
Q
Q
R
C = = Q = = 4πε!R
V k
k
R
UNITS:
Capacitance ⇒ Coulombs/Volts
⇒ Farad (F).
Example: A sphere with R = 6.0 cm (= 0.06m)
C = 6.67 × 10−12 F (⇒ 6.67pF).
Capacitance is a measure of the “capacity” that an
object has for “holding” charge, i.e., given two objects at
the same potential, the one with the greater capacitance
will have more charge. As we have seen, a charged
From before, the capacitance of a sphere is
C = 4πε!R = 4π × 8.85 × 10 −12 × 6400 ×103
= 7.1 × 10−4 F.
object has potential energy (U); a device that is
specifically designed to hold or store charge is called a
capacitor.
Earlier, question 22.4, we found the charge on the
Earth was
Q = −9.11 ×105 C.
So, what is the corresponding potential? By definition
Question 24.1: The Earth is a conductor of radius
6400km. If it were an isolated sphere what would be its
capacitance?
V=
Q 9.11 × 105
9
=
−4 = 1.28 × 10 V
C 7.1× 10
which is what we found in chapter 23 (question 23.7).
Two parallel plates (Practical considerations):
Capacitance of two parallel plates:
−Q
+Q
Area = A
+ −
Example: A = 5.0cm × 5.0cm with d = 0.5cm.
A
∴C = ε! = 4.4 × 10 −12 F = 4.4pF.
d
Maximum possible value of E in air (from earlier)
d
We can use a battery or a
≈ 3 × 106 V/m. Therefore, the maximum potential
generator to move an amount
difference (voltage) we can get between this pair of
of charge Q from one plate to
plates (in air) is:
the other. The electric field between the plates is:
σ
Q
E= =
.
(From ch. 22)
ε! ε! A
Also, the potential difference (voltage) between the two
plates is:
σ
V = E.d = d.
ε!
Vmax = E max .d ≈ 3 × 106 × 0.005 ≈ 15,000V.
Note: Vmax depends only on the spacing.
Also, the maximum charge we can achieve is
Qmax = CVmax = 4.4 × 10 −12 × 15 ×103 = 66nC.
A pair of parallel plates is a useful capacitor. Later, we
(From ch. 23)
So the capacitance of this pair of plates is
Q σA
A
C= =
= ε! .
V V
d
will find an expression for the amount of energy stored.
_________________________________
To have a 1F capacitance the area would have to be ~ 5.6 × 10 8 m2 ,
i.e., the length of each side of the plates would be ~ 23.8km (i.e.,
about 14 miles!) with a spacing of 0.5cm.
Two parallel plates:
C = ε!
A
d
How can we increase the capacitance, i.e, get more
A cylindrical (coaxial) capacitor:
charge per unit of potential difference?
• increase A
-
• decrease d
-
Metal foil
The capacitor is rolled up into a cylindrical shape.
increase ε!
by changing the medium between the plates, i.e.,
ε! ⇒ ε = κε! (later).
-
Insulating spacer
(dielectric)
•
−Q
-
++
- ++
+ +
- +Q
-
-
-
L
ra
rb
-
The capacitance of a coaxial capacitor of length L is:
2πε!L
C=
r
ln b r
a
( )
A cylindrical (coaxial) capacitor (continued):
Example: coaxial (antenna) wire.
copper
wire
insulation
copper
braid
Assume an outer conductor (braid) radius rb ≈ 2.5mm,
and an inner conductor (wire) radius ra ≈ 0.5mm, with
Question 24.2: What is the relationship between the
neoprene insulation ( ε! ⇒ κε ! = 6.9ε! ).
charge density on the inner and outer plates of a
The capacitance per meter is:
C 2πκε!
=
L ln rb
ra
cylindrical capacitor? Is it
( )
2 × π × 6.9 × 8.85 × 10 −12
=
= 2.384 × 10 −10 F/m
1.609
= 238.4 pF/m
A: larger on the outer plate?
B: larger on the inner plate?
C: the same on both plates.
Storing energy in a capacitor
+Q
−Q
When an uncharged capacitor is connected to a source
EFM10VD2.MOV
of potential difference (like a battery), charges move
from one plate to the other. Therefore, the magnitude of
Because work is done to move charges onto the plates of
the charge ( ±Q) on each plate is always the same. But
the charge density depends on area ( σ = Q A ); because
a capacitor, the capacitor stores energy, electrostatic
the inner plate has a smaller area than the outer plate,
the charge density on the inner plate is greater than the
charge density on the outer plate.
Therefore, the answer is B (larger on the inner plate).
potential energy. The energy is released when the
capacitor is discharged.
Where is the energy stored?
... in the electric field (between the plates), which
has been produced during the charging process!
Storing energy in a parallel plate capacitor:
+q + + + + + + +
B
+ dq
−q - - - - - - -
A
Energy stored in a capacitor (continued)
To store energy in a
So, in charging a capacitor from 0 → Q the total increase
capacitor we “charge” it,
in potential energy is:
producing an electric field.
We do work moving
charges from plate A to plate B. If the plates already
Q⎛ q ⎞
1 2 Q 1 Q2
U = ∫ dU = ∫ ⎜ ⎟ dq =
q
=
⎝
⎠
0
C
2C
2 C
0
[ ]
⎛ 1
1
⎞
⎜ = QV = CV 2 ⎟ .
⎝ 2
⎠
2
have charge ±q and dq is then moved from A to B, the
incremental work done is
dW = −dq(VA − VB ),
where VA and VB are the potentials of plates A and B,
respectively. This, then is the incremental increase in
potential energy, dU, of the capacitor system.
If V = VB − VA ( VB > VA ), then dU = Vdq .
q
But, by definition: V = ,
C
so the incremental increase in energy when dq of charge
is taken from A → B is:
⎛ q⎞
∴dU = ⎜ ⎟ dq.
⎝ C⎠
Potential
difference
Slope = V Q = 1C
V
V = qC
Area = dU = Vdq
dq
q
Charge
Q
Note, U is the area under the V − Q plot. This potential
energy can be recovered when the capacitor is
discharged, i.e., when the stored charge is released.
(Note also, this is the same expression we obtained earlier
for a charged conducting sphere.)
−Q
+Q
+
−
+
−
D
d
[1] Algebraically: Put C1 = ε!
D
d
Question 24.3: A parallel plate capacitor, with a plate
separation of d, is charged by a battery. After the battery
is disconnected, the capacitor is discharged through two
wires producing a spark. The capacitor is re-charged
exactly as before. After the battery is disconnected, the
plates are pulled apart slightly, to a new distance D
(where D > d). When discharged again, the energy of the
spark is ___________ it was before the plates were
pulled apart.
−Q
+Q
A
A
and C2 = ε!
d
D
But D > d, so C2 < C1. Since the charge must be the
same in each case (where can it go or come from??)
1 Q2
1 Q2
∴U1 =
and U2 =
,
2 C1
2 C2
i.e., U2 > U1
The stored energy increases when the plates are pulled
apart so the spark has more energy. Answer A.
[2] Conceptually: You do work to separate the plates
(why? ... because there is an attractive force between
A: greater than
B: the same as
C: less than
them). Where does that work go? Into the capacitor
system! So, when the plates are pulled apart, the stored
energy increases. Answer A.
Combining capacitors (parallel):
Energy density of an electric field ...
Assume we have a parallel plate capacitor, then stored
VB
VBA = VB − VA
energy is:
A
and V = E.d,
d
1 A
1
∴U = ε! (E.d)2 = ε! (A.d)E2 .
2 d
2
But A.d ⇒ volume of
the electric field. So
Area = A
d
the energy density is:
1
u e = U volume = ε!E2 .
2
This result is true for all electric fields.
C2
VA
1
U = CV 2 .
2
But C = ε!
C1
VBA
Ceq
Equivalent
capacitor
The potential difference is the same on each capacitor.
The charges on the two capacitors are:
Q1 = C1VBA and Q2 = C2 VBA
and the total charge stored is:
Q = Q1 + Q2 = (C1 + C2 )VBA = Ceq VBA
where Ceq = C1 + C2 .
So, this combination is equivalent to a single capacitor
with capacitance
Ceq = C1 + C2 .
When more than two capacitors are connected in
parallel:
Ceq = ∑i Ci .
Combining capacitors (series):
VB
VBA = VB − VA
VA
C1
Vm
C2
+Q
−Q
+Q
−Q
VBA
Ceq
200V
C1
C2
Equivalent
capacitor
C1 = 4.0 µF
C3
C 2 = 15.0 µF
C 3 = 12.0 µF
Question 24.4: In the circuit shown above, the capacitors
The charge on the two capacitors is the same: ±Q.
were completely discharged before being connected to
The individual potential differences are:
Q
Q
V1 = (VB − Vm ) =
and V2 = (Vm − VA ) = .
C1
C2
the voltage source. Find
Therefore, the total potential difference is:
⎛ 1
Q Q
1⎞
Q
∴VBA = V1 + V2 =
+
= Q⎜ + ⎟ =
,
C1 C2
⎝ C1 C2 ⎠ Ceq
1
1
1
providing
=
+ .
Ceq C1 C2
With more than two capacitors:
1
1
= ∑i .
Ceq
Ci
(a) the equivalent capacitance of the combination,
(b) the charge on the positive plate of each capacitor,
(c) the potential difference (voltage) across each
capacitor, and
(d) the energy stored in each capacitor.
200V
C1
C2
C3
C12
C3
Ceq
(a) Note that C1 and C2 are in series:
1
1
1
1
1
∴
=
+
=
+
= 0.3167 × 106 .
C12 C1 C2 4µF 15µF
∴C12 = 3.16µF
But C12 and C3 are in parallel:
∴Ceq = C12 + C3 = 3.16µF + 12µF = 15.16µF.
(b) We have: Q1 = Q2 (= Q)
⎛ 1
Q Q
1⎞
and V = V1 + V2 =
+
= Q⎜ + ⎟
C1 C2
⎝ C1 C2 ⎠
∴200 = Q × 0.3167 × 106 ,
200
−3
i.e., Q =
6 = 0.632 × 10 C ( = Q1 = Q2 )
0.3167 × 10
Also
Q3 = C3V = 12 × 10 −6 × 200 = 2.4 × 10 −3 C.
200V
C1
C2
C3
C12
C3
Ceq
Q
(c) By definition Vi = i .
Ci
Q
0.632 × 10 −3
∴V1 = 1 =
= 158V,
C1
4 × 10−6
Q2 0.632 × 10 −3
V2 =
=
= 42V,
C2
15 × 10 −6
200V
and V3 = 200V.
1
(d) By definition Ui = Qi Vi ,
2
∴U1 = 0.05J: U2 = 0.013J: U3 = 0.24J.
Check total stored energy using the equivalent ...
1
1
U = Ceq V2 = × 15.16 ×10 −6 × 2002 = 0.303J,
2
2
i.e., the same.
d
d
d
3
d
d
3
(a)
separation d. A slab of metal is placed between the
plates as shown. In case (a) the slab is not connected to
either plate; in case (b) it is connected to the upper plate.
d
3
(b)
(a)
(b)
Question 24.5: Two capacitors each have a plate
3
Case (a) is equivalent to two capacitors in series each
with spacing d 3 and capacitance C:
1
1 1
= +
C
Ceq C C
Which arrangement produces the higher capacitance ?
C
∴Ceq =
C
.
2
A: Case (a)
Case (b) is a single capacitor: Ceq = C.
B: Case (b)
Therefore, (b) has the higher capacitance. So, the answer
C: Neither, they are the same.
is B.
+Q
−Q
Question 24.6: A 2.00 µF capacitor is energized to a
potential difference of 12.0 V. The wires connecting the
2µF
12V
+Q1
−Q1
2µF
4V
C2 +Q2
−Q2
(a) The initial charge on the 2µF capacitor is
Q = C1V = 2 × 10 −6 × 12 = 24 × 10 −6 C.
capacitor to the battery are then disconnected from the
battery and connected across of second, initially
When connected across the second capacitor, this charge
uncharged capacitor. The potential difference across the
is redistributed (none is lost!!). The “new” charges are
Q1 = C1V′ = 2 × 10 −6 × 4 = 8 × 10 −6 C,
2.00 µF capacitor then drops to 4.00 V.
and
(a) What is the capacitance of the second capacitor?
(b) What is the energy of the system before and after?
Q2 = C2 V′ = C2 × 4.
But total charge is conserved.
So, Q1 + Q2 = Q.
∴Q2 = Q − Q1 = C 2 V′ .
i.e., C2 × 4 = (24 × 10 −6 ) − (8 × 10−6 ) = 16 ×10 −6 .
∴C2 = 4 ×10 −6 F = 4µF.
General case ...
+Q
−Q
2µF
12V
+Q1
2µF
−Q1
4V
C2 +Q2
−Q2
+Q
−Q
C1
V
+Q1
−Q1
C1
V′
C2 +Q2
−Q2
(b)
•
•
1
Energy before ⇒ C1V2
2
1
= × 2 × 10 −6 × 122 = 1.44 × 10 −4 J
2
1
1
Energy after ⇒ C1V′2 + C2 V′ 2
2
2
1
1
= × 2 × 10 −6 × 42 + × 4 ×10 −6 × 42
2
2
= 0.48 × 10−4 J
i.e, a loss of 0.96 × 10 −4 J!
What? Where
has it gone?
Does it remind you of something in Physics 1?
If C2 is uncharged initially ...
1
• Initial energy Ui = QV.
2
1
1
• Final energy Uf = Q1V′ + Q2 V′
2
2
1
1
= (Q1 + Q2 ) V′ = QV′ .
2
2
U
V′
∴ f = .
Ui V
But Q1 + Q2 = C1V′ + C2 V′ = (C1 + C2 )V′ .
V′
C1
Since Q1 + Q2 = Q = C1V then,
=
.
V C1 + C2
U
C1
∴ f =
Ui C1 + C2
Like an inelastic collision!
i.e., always < 1.
Dielectrics:
Dielectrics (continued):
+Q
+Q
Dielectric
d
(a)
−Q
−Q
(b)
C! = Q V
!
C = QV
(a) The electric field in an isolated charged parallel plate
capacitor (in vacuum) is: E! = σ ε .
!
σ
∴V! = E!d = d.
ε
!
(b) When a dielectric is inserted, ε! ⇒ κε ! , where κ is
σ
the dielectric constant, so E =
.
κε !
σ
V
∴V = Ed =
d = !,
κε!
κ
i.e., the potential difference is reduced.
Q κQ
⎛
A⎞
∴C = =
= κC! ⎜ = κε! ⎟ .
⎝
V V!
d⎠
Thus, the capacitance increases by a factor of κ.
Three advantages:
• maintains plate separation when small,
• increased capacitance for a given size.
• dielectric increases the max. electric field possible,
and hence potential difference (voltage), between
plates before breakdown (dielectric strength).
Material
κ
Dielectric
Strength (V/m)
Air
1.00059
3 × 106
Paper
3.7
16 × 106
Neoprene
6.9
12 × 106
Polystyrene
2.55
24 × 106
How does a dielectric work? ...
−σin +σin
+σ −σ
+
+
+
-
+
d
no dielectric
σ
E! =
ε!
+
-
+
-
-
+
+
-
+
-
-
+
+
E! − Ein
+
-
σ
Ein = in
ε!
- +
+ -+ -+ -+
+
-+ -+ -+ -+
+
polarized dielectric
Because the dielectric is polarized, the electric field in
the presence of a dielectric is reduced to:
σ σ
E = E! − Ein = − in
ε ! ε!
=
σ − σin E!
= .
ε!
κ
Therefore, the potential difference V (= Ed) is reduced
by a factor of κ also.
∴C > C! .
d
+σ
−σ
−σin +σin
+
+
-
+
+
-
+
σin
Ein =
ε!
+
-
σ
E! =
ε!
-
E! − Ein
Hence, the induced electric field is:
E
⎛ 1⎞
Ein = E! − ! = ⎜1 − ⎟ E! .
κ ⎝ κ⎠
Substituting for Ei and E!, we obtain
⎛ 1⎞
σin = ⎜1 − ⎟ σ.
⎝ κ⎠
Hence σ ≥ σin . Note: σin = 0 if κ = 1 (free space, i.e.,
no dielectric).
For a conducting slab κ = ∞ and σin = σ.
∴E = E! − Ein = 0,
i.e., there is no electric field inside a conductor.
+
+
X
−
Y
−
(a) When the dielectric is inserted, the capacitance of Y
+
+
X
−
Y
−
increases from C to κC, where κ is the dielectric
constant. But the capacitance of X is unchanged. The
potential difference across both capacitors remains the
same ( = V) and since the charge on a capacitor is given
Question 24.7: Two, identical capacitors, X and Y, are
connected across a battery as shown. A slab of dielectric
is then inserted between the plates of Y. Which capacitor
has the greater charge or do the charges remain the
same?
by Q = CV, if C increases to κC, then Q increases to κQ.
Where does the extra charge come from? ...
... from the battery!
(b) If the battery is disconnected before the dielectric is
inserted, the charge on each capacitor is unchanged. But,
since the capacitance of Y increases to κC, the potential
difference across Y changes from
Q to Q , i.e., it gets smaller.
C
κC
Two ways to answer part (a) ...
Question 24.8: A dielectric is placed between the plates
of a capacitor. The capacitor is then charged by a
battery. After the battery is disconnected, the dielectric
is removed. (a) Does the energy stored by the capacitor
increase, decrease or remain the same after the dielectric
is removed? (b) If the battery remains connected when
the dielectric is removed, does the energy increase,
decrease or remain the same?
[1] Algebraically: Let C → C! and V → V!
1
With the dielectric: U = CV 2 .
2
1
Without the dielectric: U! = C! V!2 .
2
V
C
But C = κC! and V = ! . ∴C! =
and V! = κV.
κ
κ
1C
1
(κV)2 = κ CV 2 = κU.
∴U! =
2κ
2
Therefore, the energy increases.
[2] Conceptually: You must do work to remove the
dielectric therefore the
+++ +++
- - - - +++++
- - - - - -
stored (potential) energy
increases!
(b) If the battery remains connected, the potential
difference remains constant ( = V). But the capacitance
changes from C → C! , with C = κC!, i.e., C > C ! .
The energy changes from
1
1
U = CV 2 to U! = C! V2 .
2
2
Since C > C ! , then U > U! , i.e., the energy decreases.
Question 24.9: You are required to fabricate a 0.01 µF
parallel plate capacitor that can withstand a potential
difference of 2.0 kV. You have at your disposal a
dielectric with a large dielectric constant of 24 and a
dielectric strength of 4.0 × 10 7 V/m.
(a) What is the minimum plate separation required?
(b) What is the area of each plate at this separation?
V
V
⇒d = .
d
E
V
2000
∴d min =
=
= 50 ×10 −6 m.
7
Emax 4 ×10
(a) Since E =
A variety of (fixed value) capacitors
εA
Cd
(b) C = κ ! ⇒ A =
.
d
κε!
Cd 0.10 × 10 −6 × 50 ×10 −6
∴A =
=
κε !
24 × 8.854 × 10 −12
= 2.35 × 10−2 m2 ⇒ 235 cm2 .
A simple variable (air) capacitor
+
-
+
-
“charged”
1
“uncharged”
0
+
-
+
-
+
-
“charged”
+
-
1
“charged”
1
Binary [1 0 1 1] ⇒ Decimal 11
(a)
(b)
Question 24.10: A parallel plate capacitor is charged by
a generator. The generator is then disconnected (a). If
the spacing between the plates is decreased (b), what
happens to:
(i) the charge on the plates,
Dynamic random access memory (DRAM) is composed
(ii) the potential across the plates, and
of banks of capacitors. A “charged” capacitor represents
(iii) the energy stored by the capacitor.
the binary digit “1” and “uncharged” capacitor represents
the binary digit “0”.
A 32MB DRAM chip contains 256 million capacitors.
(a)
(b)
A parallel plate capacitor is charged by a generator. The
generator is then disconnected (a). If the spacing between
the plate is decreased (b), what happens to:
(i) the charge on the plates remains the same, where
could it go or where could charge come from?
(ii) the potential across the plates decreases, because the
electric field remains constant - it’s independent of d
(chapter 22) - but as spacing decreases, the potential
(V= E.d) decreases.
(iii) the energy stored by the capacitor decreases,
because the system does work to reduce spacing.
Conversely, you would have do work in order increase
the spacing.
(a)
(b)
Question 24.11: A parallel plate capacitor is charged by
a battery (a). When fully charged, and while the battery
is still connected, the spacing between the plate is
decreased (b), what happens to:
(i) the potential across the plates,
(ii) the charge on the plates, and
(iii) the energy stored by the capacitor.
(a)
(b)
A parallel plate capacitor is charged by a generator (a).
When fully charged, and while the generator is still
connected, the spacing between the plate is decreased (b).
(i) the potential across the plates remains the same since
the source of potential difference is still connected!
(ii) the charge on the plates increases, because the
capacitance increases and so, if V is unchanged, Q
increases.
(iii) the energy stored by the capacitor increases,
1
1
because Q and C increase ( U = QV = CV 2 ).
2
2