Download Lecture 7

Lecture 7 Capacitors & Energy Storage
(Finish electric potential and gradient)
• Capacitors: what they are and how to
calculate their capacitance
Application: flash device for a
camera (stores charge and energy)
Charge up a flash device in a
camera and then short out the
two leads in the 18.5 µ F cap
CAPACITOR
• A capacitor is device formed with two or
more separated conductors that store charge
and electric energy.
• Consider any two conductors and we put
+Q on a and –Q on b. Conductor a has constant
Va and conductor br has constant Vb , then
rb r r
Va − Vb = ∫ E ⋅ dl
Y&F fig. 24.1
r
ra
• The electric field is proportional to the charges ±Q. If we double the
charges ±Q, the electric field doubles. Then the voltage difference is
Va-Vb proportional to the charge. This proportionality constant
depends on size, shape and separation of the conductors.
Q = const × (Va − Vb )
CAPACITOR, continued
• If we call this constant, Capacitance, C,
and the voltage difference, V= Va-Vb, then,
Q = CV
Or
Q
C=
V
• Capacitance, depends on the geometry of the
two conductors (size, shape, separation) and capacitance is always
a positive quantity by its definition (voltage difference and charge
of + conductor) [N.B. C does not depend on voltage or charge]
• UNITs of capacitance, Coulomb/Volts or Farads, after Michael Faraday
Example:
Parallel Plate Capacitor
• Calculate the capacitance. We
assume +σ, -σ charge densities
on each plate with potential
difference V:
A
++++
d
-----
Q
C ≡
V
• Need Q:
Q = σA
• Need V:
from def’n:
r r
Vb −Va = −∫ E • dl
b
a
– Use Gauss’ Law to find E
CLICKER QUESTION
What is the electric potential difference, Va-Vb ?
r
a) + E d
r
b) − E d
r
c) + q0 E d
r
d ) − q0 E d
Test charge moves to lower potential
Recall:
Two Infinite Sheets
(into screen)
+
-
σ
• Field outside the sheets is zero
• Gaussian surface encloses
zero net charge
• Field inside sheets is not zero:
• Gaussian surface encloses
non-zero net charge Q = σA
r r
∫ E • dS = AEinside
E =
σ
ε0
σ
- E=0
-
+
+
E=0
+
+
A +
+
+
+
+
A
+
+
E
Example: Parallel Plate Capacitor
•
•
Calculate the capacitance:
A
Assume +Q, -Q on plates with
potential difference V.
++++
d
σ
Q
E=
=
ε 0 Aε 0
r r
Q
Vb −Va = −∫ E • dl = Ed =
d
Aε0
a
b
⇒
C≡
-----
Q Aε 0
=
V
d
• As expected, the capacitance of this capacitor
depends only on its GEOMETRY (A,d). DOES NOT
DEPEND on Q or on V !!!!!
• Note that C ~ length; this will always be the case!
Cylindrical Capacitor Example
• Calculate the capacitance:
• Assume +Q, -Q on surface of
r
cylinders with potential difference V.
a
b
• Gaussian surface is cylinder of
radius r (a < r < b) and length L
r r
Q
• Apply Gauss' Law: ∫ E • dS = 2πrLE =
ε0
⇒
L
E =
Q
2 πε 0 Lr
If we assume that inner cylinder has +Q, then the potential V is
positive if we take the zero of potential to be defined at r = b:
a
b
r r
V = −∫ E • dl = −∫ Edr = ∫
a
b
b
b
dr =
ln 
2πε 0 rL
2πε 0 L  a 
a
Q
Q
⇒
C ≡
Q
2 πε 0 L
=
V
 b 
ln 

a


Spherical Capacitor Example
•
Suppose we have 2 concentric spherical shells
of radii a and b and charges +Q and –Q.
-Q
• Question: What is the capacitance?
+Q
b
• E between shells is same as a point
charge +Q. (Gauss’s Law):
a
• What is the capacitance of the earth?
1 Q
Er =
4πεo r 2
C ≡
r r br r
Vab = Va −Vb = −∫ E • dl = ∫ E • dl
a
b
b
b
4πε0r
a
a
=−
Q
b
4πε0r a
=
Q
4 πε
o
Q
1
1
( − )
a
b
a
Q
= ∫ Er dr = ∫
Q
=
V ad
dr
2
1 1
( − )
4πε0 a b
Q
⇒
4 πε 0 ab
C =
b−a
Summary
•
A Capacitor is an object with two spatially separated conducting
surfaces used to store charge or energy.
•
The definition of the capacitance of such an object is:
C ≡
Q
V
• The capacitance depends on the geometry :
-Q
A
++++
d
----Parallel Plates
C=
Aε 0
d
r
+Q
-Q
+Q
a
a
b
b
L
Cylindrical
Spherical
2 πε 0 L
 b 
ln 

 a 
4 πε 0 ab
C =
b−a
C =
CAPACITORs in parallel
To find the equivalent capacitance Ceq:
• Voltage is same across each capacitor
Q1 Q2
Q1
Q2
Vab =
=
; C1 =
; C2 =
C1 C2
Vab
Vab
• Total charge and voltage ratios for
parallel capacitor,
Q1 + Q2
= C parallel = C1 + C2
Vab
For more parallel
capacitors:
C parallel = ∑ Ci
i
Note; 2 parallel capacitors C, doubles capacitance.
CAPACITORs in series
Want to find the equivalent capacitance Ceq:
• If a voltage is applied across a and b, then
a +Q appears on upper plate and –Q on lower
plate.
• A –Q charge is induced on lower plate of C1
and a +Q charge is induced on upper plate of
C2. The total charge in circuit c is neutral.
Q
Q
V1 =
; V2 =
C1
C2
Q Q
V = V1 + V2 =
+
C1 C2
V
1
1
1
=
= +
Q
Cseries C1 C2
Note; 2 series capacitors C, halves capacitance.
Examples:
Combinations of Capacitors
a
≡
C3
b
C1
a
C2
C
• How do we start??
• Recognize C3 is in series with the parallel
combination on C1 and C2. i.e.,
1 1
1
= +
C C3 C1 + C2
⇒
b
C=
C 3 (C1 + C 2 )
C1 + C 2 + C 3
UI7PF7: CLICKER PROBLEM
Two identical parallel plate
capacitors are shown in an endview in A) of the figure. Each
has a capacitance of C.
4) If the two are joined together as shown in B), forming a
single capacitor, what will be the final capacitance?
a) C/2
b) C
c) 2C
UI7PF7: CLICKER PROBLEM
C
C
C
C
C
Configuration A
Configuration B
Configuration C
Three configurations are constructed using identical capacitors
6) Which of these configurations has the lowest overall capacitance?
a) Configuration A
b) Configuration B
c) Configuration C
UI7ACT 3: CLICKER PROBLEM
• What is the equivalent capacitance, Ceq, of the
combination shown?
o
Ceq
C
C
C
o
(a) Ceq = (3/2)C
(b) Ceq = (2/3)C
(c) Ceq = 3C
UI7ACT 3
• What is the equivalent capacitance, Ceq, of the
combination shown?
o
Ceq
C
C
C
o
(b) Ceq = (2/3)C
(a) Ceq = (3/2)C
C
C
1 1 1
= +
C1 C C
C
≡
C
C1 =
2
(c) Ceq = 3C
C1
C
C 3
Ceq = C + = C
2 2
CLICKER QUESTION
Which Circuit has the LARGEST capacitance between connections
a and b? assume all capacitors have the same value.
(A)
(B)
(C)
(D)
Not a clicker problem
A circuit consists of three unequal capacitors C1, C2, and C3 which are
connected to a battery of emf ε. The capacitors obtain charges Q1 Q2,
Q3, and have voltages across their plates V1, V2, and V3. Ceq is the
equivalent capacitance of the circuit.
8) Check all of the following
that apply:
a) Q1= Q2
b) Q2= Q3
e) V1 < V2
f) Ceq > C1
c) V2= V3
d) ε = V1
• Energy storage in capacitors
• Improving capacitors by adding dielectrics
Hints for Homework (remind me)
UI8ACT1/Clicker Problem
• What is the relationship between V0 and V in the
systems shown below?
+Q
V0
(Area A)
+Q
d
-Q
(a) V = (2/3)V0
(Area A)
d/3
V
-Q
(b) V = V0
d/3
conductor
(c) V = (3/2)V0
UI8ACT1/Clicker Problem
• What is the relationship between V0 and V in the
systems shown below?
+Q
+Q
(Area A)
V0
d/3
V
d
d/3
-Q
-Q
(b) V = V0
(a) V = (2/3)V0
•
(Area A)
conductor
(c) V = (3/2)V0
The arrangement on the right is equivalent to capacitors (each with
separation = d/3) in SERIES!!
+Q
+Q
d/3
≡
d/3
-Q
C eq
1
C eq = C
2
1 Aε 0
3 Aε 0 3
=
=
= C0
2 (d / 3) 2 d
2
(Area A)
d/3
V
d/3
-Q
V =
conductor
Q
Q
2
=
= V0
C eq (3 / 2 )C0 3
Energy storage in CAPACITORs
Charge capacitor by transferring bits of
charge dq at a time from bottom to top plate.
Can use a battery to do this. Battery does
work that increases potential energy of
capacitor.
+q
-q
q is magnitude of charge on plates
V= q/C
V across plates
dU = V dq
increase in potential energy
(
q
Q
CV )
1
U = ∫ dU = ∫ dq =
=
= CV 2
C
2C
2C
2
0
0
U
Q
2
2
two ways to express the result
V
dq
Question!—constant Q
A
++++
• Suppose the capacitor shown here is
charged to Q and then the battery is
disconnected.
d
-----
• Now suppose I pull the plates further apart so that the final
separation is d1.
• How do the quantities Q, C, E, V, U change?
•
•
•
•
•
Q:
C:
E:
V:
U:
remains the same.. no way for charge to leave.
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?.. exercise for student!!
Answers:
C1 =
d
C
d1
V1 =
d1
V
d
U1 =
d1
U
d
Related Question—constant V
• Suppose the battery (V) is kept
attached to the capacitor.
A
++++
d
-----
V
• 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
must decrease ( E =
V
σ
, E= )
D
E0
must decrease (U = 12 CV 2 )
• How much do these quantities change?.. exercise for student!!
Answers:
d
C1 =
C
d1
E1 =
d
E
d1
U
1
=
d
U
d1
Clicker questions:
Two parts
Two identical parallel plate capacitors are connected to a battery, as
shown in the figure. C1 is then disconnected from the battery, and the
separation between the plates of both capacitors is doubled.
2) What is the relation between the charges on the two capacitors ?
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
3) How does the electric field between the plates of C2 change as
separation between the plates is increased ? The electric field:
a) increases
b) decreases
c) doesn’t change
Clicker problem
• Two identical parallel plate capacitors are connected to a battery.
– C1 is then disconnected from the battery and the separation
between the plates of both capacitors is doubled.
d
V
2d
d
C1
V
C2
2d
– What is the relation between the U1, the energy stored in C1, and
the U2, energy stored in C2?
(a) U1 < U2
(b) U1 = U2
(c) U1 > U2
UI8ACT2
• Two identical parallel plate capacitors are connected to a battery.
– C1 is then disconnected from the battery and the separation
between the plates of both capacitors is doubled.
d
V
2d
d
C1
V
C2
2d
– What is the relation between the U1, the energy stored in C1, and
the U2, energy stored in C2?
(a) U1 < U2
(b) U1 = U2
(c) U1 > U2
• What is the difference between the final states of the two capacitors?
• The charge on C1 has not changed.
• The voltage on C2 has not changed.
• The energy stored in C1 has definitely increased since work must be
done to separate the plates with fixed charge, they attract each other.
• The energy in C2 will actually decrease since charge must leave in order
to reduce the electric field so that the potential remains the same.
2
Q
1
1
1
0
Later: U 1 =
= 2U 0
U 2 = C 2V02 = U 0
Initially: C1 = C2
2 C1
2
2
Where is the Energy Stored?
• Energy is stored in the electric field itself ! Think of
the energy needed to charge the capacitor as being
the energy needed to create the field.
• To calculate the energy density in the field, first consider the
constant field generated by a parallel plate capacitor, where
-Q
-------- ------
++++++++ +++++++
+Q
1 Q2 1 Q2
U=
=
2 C 2 ( Aε 0 / d )
• The electric field is given by:
Q
σ
E= =
ε0 ε0 A
⇒
This is the energy
density, u, of the
electric field….
1
U = ε0 E2 Ad
2
• The energy density u in the field is given by:
U
U 1 2
u=
= = ε0E
volume Ad 2
Units:
J
m3
Energy Density
Claim: the expression for the energy density of the
electrostatic field
1
u = ε0E 2
2
is general and is not restricted to the special case of the
constant field in a parallel plate capacitor.
• Example (and another exercise for the student!)
– Consider E- field between surfaces of cylindrical
capacitor:
– Calculate the energy in the field of the capacitor by
integrating the above energy density over the volume of
the space between cylinders.
U=
1
1
ε 0 ∫ E 2 dV = ε 0 ∫ ∫ E 2π r dr dl = etc.
2
2
– Compare this value with what you expect from the
general expression:
1
W
U = CV 2
2
Clicker problem
• Consider two cylindrical capacitors,
each of length L.
C1
1.1
1
– C1 has inner radius 1 cm and outer radius 1.1cm.
– C2 has inner radius 1 cm and outer radius 1.2cm.
If both capacitors are given the same amount of
charge, what is the relation between U1, the energy
stored in C1, and U2, the energy stored in C2?
(a) U2 < U1
(b) U2 = U1
C2
1.2
1
(c) U2 > U1
Lecture 8, ACT 3
• Consider two cylindrical capacitors,
each of length L.
C1
1.1
1
– C1 has inner radius 1 cm and outer radius 1.1cm.
– C2 has inner radius 1 cm and outer radius 1.2cm.
If both capacitors are given the same amount of
charge, what is the relation between U1, the energy
stored in C1, and U2, the energy stored in C2?
C2
1.2
1
Hint: what is C for a cylindrical capacitor ?
(a) U2 < U1
(b) U2 = U1
(c) U2 > U1
The magnitude of the electric field from r = 1 to 1.1 cm is the same
for C1 and C2. But C2 also has electric energy density in the volume
1.1 to 1.2 cm. In formulas:
2πε o L
2
1
1
U
Q
/ 2C2 C1
1.2
C=
C2 ~
2
C1 ~
= 2
=
= ln(
)
 1.2 
 router 
 1.1
1.1
U1 Q / 2C1 C2
ln 
ln 


ln

 1 
 1 
r

inner

DIELECTRICS
Consider parallel plate
capacitor with vacuum
separating plates (left)
Suppose we place a material
called a dielectric in between
the plates (right)
The charge on the plates
remain the same, but a
Y&F Figure 24.13
dielectric has a property of
having induced charges on its surface that REDUCE
the electric field in between and the voltage difference.
Since C = Q/V, the resulting capacitance will INCREASE.
DIELECTRICS
Suppose the charges on the plate and the
dielectric are, σ and σi. The electric
Fields before and after are
σ
σ − σi
E0
σ
E0 =
; E=
; K≡
=
ε0
ε0
E σ − σi
We define the ratio of the original field over
the new field as the dielectric constant, K.
Hence, the voltage difference changes by
1/K and the capacitance, Co=Q/V, changes
by C=KQ/V=K Co
For same Q:
But
C = KCo
C = KCo
E = Eo/K
General
V = Vo/K
DIELECTRICS Materials
Glass, mica, plastics are very good dielectrics
DIELECTRICS and permittivity
Define ε, called permittivity, as
ε = Κ ε0
Consider a parallel plate capacitor with no dielectric
A
C0 = ε 0
d
A capacitor with a dielectric becomes simply,
A
A
C = KC 0= Kε 0 = ε
d
d
The change in capacitance can be accounted for
by changing permittivity.
Y&F Problems 24.72 and 24.71
A
A parallel plate capacitor has two
dielectrics, side by side, find the
capacitance
A K1 + K 2
C = ε0
d
2
A
A parallel plate capacitor has two
dielectrics, stacked, find the
capacitance
A 2 K1K 2
C = ε0
d K1 + K 2
EXAMPLE
Two parallel plate capacitors, C1 = C2 = 2 µF, are connected across
a 12 V battery in parallel.
a.) What energy is stored?
1
U1 = U 2 = CV 2 = 144µJ U T = 288µJ
2
b.) A dielectric (K = 2.5) is inserted between the plates of C2. Energy?
C2' = KC2 = 2.5 × 2µF = 5µF
1 ' 2
U = C2V = 360µJ U T = 504µJ
2
'
2
Note: a dielectric increases amount of energy stored in C2.
EXAMPLE of parallel plate capacitor problem
A parallel plate capacitor is made by placing polyethylene (K = 2.3)
between two sheets of aluminum foil. The area of each sheet is
400 cm2, and the thickness of the polyethylene is 0.3 mm. Find the
capacitance.
C =K εo A/d = (2.3) (8.85 x 10-12 C2/Nm2) (400 cm2)(1m2/104 cm2)
0.3 x 10-3 m
= 2.71 nF
UI8PF 8: TWO PART CLICKER PROBLEM
Two identical parallel plate capacitors are connected to a battery.
Remaining connected, C2 is filled with a dielectric.
7) Compare the voltages of the two capacitors.
a) V1 > V2
b) V1 = V2
c) V1 < V2
8) Compare the charges on the plates of the capacitors.
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
Note: Unlike constant Q case, here V and E remain the same
but C = K Co still.
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.
CLICKER PROBLEM:
7) Compare the voltages of the two capacitors.
a) V1 > V2
b) V1 = V2
c) V1 < V2
8) Compare the electric fields between the plates of both
capacitors.
a) E1 > E2
b) E1 = E2
c) E1 < E2
UI8ACT1/Clicker Problem
• What is the relationship between V0 and V in the
systems shown below?
+Q
V0
(Area A)
+Q
d
-Q
(a) V = (2/3)V0
(Area A)
d/3
V
-Q
(b) V = V0
d/3
conductor
(c) V = (3/2)V0
UI8ACT1/Clicker Problem
• What is the relationship between V0 and V in the
systems shown below?
+Q
+Q
(Area A)
V0
d/3
V
d
d/3
-Q
-Q
(b) V = V0
(a) V = (2/3)V0
•
(Area A)
conductor
(c) V = (3/2)V0
The arrangement on the right is equivalent to capacitors (each with
separation = d/3) in SERIES!!
+Q
+Q
d/3
≡
d/3
-Q
C eq
1
C eq = C
2
1 Aε 0
3 Aε 0 3
=
=
= C0
2 (d / 3) 2 d
2
(Area A)
d/3
V
d/3
-Q
V =
conductor
Q
Q
2
=
= V0
C eq (3 / 2 )C0 3