Capacitance
PHY2049: Chapter 25
1
What You Know: Electric Fields
ÎCoulomb’s
ÎElectric
law
fields
ÎEquilibrium
ÎGauss’
law
ÎElectric
fields for several charge configurations
‹ Point
‹ Dipole
(along axes)
‹ Line
‹ Plane
(nonconducting)
‹ Plane (conducting)
‹ Ring (along axis)
‹ Disk (along axis)
‹ Sphere
‹ Cylinder
PHY2049: Chapter 25
2
What You Know: Electric Potential
ÎElectric
potential energy
ÎElectric
potential
ÎEquipotential
surfaces
ÎPotential
of point charge
ÎPotential
of charge distribution
‹ Special
cases: dipole, line, ring, disk, sphere
ÎRelationship
of potential and electric field
‹ Calculating
the potential from the field
‹ Calculating the field from the potential
ÎPotential
energy from a system of charges
PHY2049: Chapter 25
3
Capacitance: Basic Idea
ÎCapacitance:
Capacity to store charge
‹ Like
a tank
‹ Capacitor is electrically neutral (equal + and − charge regions)
‹ q = CV (C is a property of the device, independent of q, V)
‹ Units: [C] = “Farad” = Coul/Volt
PHY2049: Chapter 25
4
Calculating Potential Difference
ÎElectric
field lines start on + charges, terminate on −
+
V+ − V− = − ∫ E ⋅ ds
−
ÎFollow
E field line during integration (note cosθ = −1)
+
V ≡ V+ − V− = ∫ E ds
−
E, ds positive
++++++++++++++++
E
d
--------------------PHY2049: Chapter 25
5
Parallel Plate Capacitor
σ
q
ÎFrom Gauss’ law (conducting sheet) E =
=
ε 0 Aε 0
⎛ q ⎞
ÎSo V = Ed = ⎜
⎟d
⎝ Aε 0 ⎠
ε0 A
ÎTherefore q =
V ≡ CV
d
A
C = ε 0 = ε 0 × "length"
d
ÎDepends
of device
++++++++++++++++
E
only on geometry
d
--------------------PHY2049: Chapter 25
6
Example
ÎA
= 1m2, d = 1 μm
A
C = ε 0 = 8.85 × 10-12 × 106 = 8.85 × 10-6 F
d
C = 8.85 μ F
Largish, but somewhat typical value
PHY2049: Chapter 25
7
Cylindrical Capacitor
ÎInner
= a, outer = b, length = L
ÎGauss’
‹
Î
Î
law:
Using λ=q/L
+
V = ∫ E ds
−
V = −∫
a
b
q 1
E=
2π Lε 0 r
(ds = −dr)
b
-
-
+
+
+
-
-
a
+
-
+
-
q
ln ( b / a )
Edr =
2π Lε 0
C
2π
= ε0
L
ln ( b / a )
-
-
+
+
-
+
-
-
Capacitance per unit length,
e.g. “coaxial” cable (RF frequencies)
PHY2049: Chapter 25
8
Special Case for Cylinder
ÎOuter
ÎUse
shell “very close” to inner shell: b − a = d (d small)
ln(1+x) ≅ x (for x small)
⎛b⎞
⎛a+d⎞ d
ln ⎜ ⎟ = ln ⎜
⎟
⎝a⎠
⎝ a ⎠ a
Asurface
2π L
2π aL
≅ ε0
= ε0
C = ε0
ln ( b / a )
d
d
A
‹ Just like parallel plate capacitor: C = ε 0
d
‹ Always
true if surfaces are close together
PHY2049: Chapter 25
9
Spherical Capacitor
ÎInner
radius = a, outer radius = b
ÎCoulomb’s
Î
Î
+
law:
V = ∫ E ds
−
V = −∫
a
b
E=
q
b
-
1
4πε 0 r
2
+
-
-
a
+
q b−a
Edr =
4πε 0 ab
-
+
+
-
+
-
(ds = −dr)
-
-
+
+
-
+
-
-
4π ab
C = ε0
b−a
PHY2049: Chapter 25
10
Two Special Cases
ÎIsolated
sphere: corresponds to b = ∞
4π ab
4π a
= ε0
→ ε 0 4π a
C = ε0
b−a
1− a / b
ÎOuter
shell “very close” to inner shell: b − a = d (d small)
Asurface
4π ab
4π a
≅ ε0
= ε0
C = ε0
b−a
d
d
A
C = ε0
d
2
‹ Again,
just like parallel plate:
PHY2049: Chapter 25
11
Capacitors in Parallel
Î V1
= V2 = V3 (same potential top and bottom)
ÎTotal
ÎCeqV
charge: Qtot = Q1 + Q2 + Q3
= C1V + C2V + C3V
Ceq = C1 + C2 + C3
¾ Basic law for combining
capacitors in parallel
¾ Works for N capacitors
PHY2049: Chapter 25
12
Capacitors in Series
Îq1
= q2 = q3 (same current charges all capacitors)
ÎTotal
potential: V = V1 + V2 + V3
Îq/Ceq
= q/C1 + q/C2 + q/C3
1
1
1
1
=
+
+
Ceq C1 C2 C3
¾ Basic law for combining
capacitors in series
¾ Works for N capacitors
PHY2049: Chapter 25
13
ConcepTest
ÎTwo
identical parallel plate capacitors are shown in an
end-view in Figure A. Each has a capacitance of C.
If the two are joined together at the edges as in Figure B,
forming a single capacitor, what is the final capacitance?
‹ (a)
C/2
‹ (b) C
‹ (c) 2C
Area is doubled
‹ (d) 0
‹ (e) Need more information
A
PHY2049: Chapter 25
B
14
ConcepTest
ÎEach
capacitor is the same in the three configurations.
Which configuration has the lowest equivalent capacitance?
‹ (1)
A
C/2 (series)
‹ (2) B
‹ (3) C
‹ (4) They all have identical capacitance
C
C
C
C
C
A
B
PHY2049: Chapter 25
C
15
Energy in a Capacitor
ÎCapacitors
‹ Grab
have energy associated with them
a charged capacitor with two hands and find out!
ÎCalculate
energy
‹ Continually
‹ dU
= V dQ
‹V = Q / C
move charge from – to + surface
U =∫
Q
0
Q2 1
2
( Q / C ) dQ = = 2 CV
2C
Q2 1
U=
= 2 CV 2
2C
ÎSo
capacitors store and release energy as they acquire
and release charge
‹ This
energy is available to drive circuits
PHY2049: Chapter 25
16
Where is the Energy Stored?
ÎAnswer:
Energy is stored in the electric field itself!!
ÎExample:
‹E
Find energy density of two plate capacitor
field is constant
ε 0 ( A / d )( Ed )
U
CV
=
=
= 12 ε 0 E 2
u=
2 Ad
Ad 2 Ad
2
2
ÎEnergy
density depends only on E field!
‹A
general result, independent of geometry
‹ Can be shown more generally by Maxwell’s equations
u = 12 ε 0 E 2
PHY2049: Chapter 25
17
Example: Spherical Charged Conductor
ÎCapacitance:
C = ε 0 4π R
Q2
Q2
=
ÎTotal energy of spherical conductor: U =
2C 8πε 0 R
ÎCalculate
u=
directly: integrate energy density over volume
1 ε E2
2 0
U =∫
U = ∫ u dV
R
dV = 4π r 2dr
2
2
⎛ Q ⎞
Q
2
ε
4
π
r
dr =
2 0⎜
⎜ 4πε r 2 ⎟⎟
8πε 0 R
0
⎝
⎠
∞1
R
∞
PHY2049: Chapter 25
Checks!
18
Dielectric Materials and Capacitors
ÎInsulating
material that can be polarized in E field
+++++++++++++++++++++++++
-------------------------------------------
Induced
charges
κ
Dielectric
material
+++++++++++++++++++++++++
-------------------------------------------
ÎInduced
charges at dielectric surface partially cancel E field
→ E / κ κ > 1 is “dielectric constant”
‹ V → V / κ (since V = Ed)
‹ C → κC
(since C = Q / V)
‹E
ΓGood”
‹ Good
dielectric requires more than high κ value
insulator
‹ High breakdown voltage
‹ Low cost
(no charge leakage)
(no arcing at high voltage)
(affordable)
PHY2049: Chapter 25
19
Dielectric Mechanism is Due to Polarization
E = 0, Dipoles randomly aligned
¾ E applied, partially aligns dipoles
¾ Aligned dipoles induce surface charges
¾ Surface charges partially cancel E field
¾ Yow!
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html
PHY2049: Chapter 25
20
ConcepTest
ÎTwo
identical capacitors are given the same charge Q,
then disconnected from a battery.
After C2 has been charged and disconnected it is filled
with a dielectric. Compare the voltages of the two
capacitors.
Voltage lowered
to V/κ
‹ (1)
V1 > V2
‹ (2) V1 < V2
‹ (3) V1 = V2
C1
PHY2049: Chapter 25
C2
21
ConcepTest
ÎWhen
we fill the capacitor with the dielectric, what is the
amount of work required to fill the capacitor?
‹ (1)
W>0
‹ (2) W < 0
‹ (3) W = 0
Energy lowered
to U/κ
C1
C2
If U is total energy in capacitor
¾Positive work: One “pushes in” dielectric
ΔU > 0
¾Negative work: Capacitor “sucks in” dielectric ΔU < 0
PHY2049: Chapter 25
22
Capacitors in Circuits
ÎMulti-step
Î2
+ 3 in series
‹ Combine
Î1
process
2 & 3 ⇒ C/2
Ceq
+ (2+3) in parallel
‹ Combine
C
o
1 & (2 & 3) ⇒ 3C/2
2
1 C
3 C
o
PHY2049: Chapter 25
23
Example: Find qi and Vi on All Capacitors
ÎC1
is charged in position A, then S is thrown to B position
‹ Initial
voltage across C1:
‹ Initial charge on C1:
ÎAfter
V0 = 12
q10 = 12 x 4 = 48μC
switch is thrown to B: V1 = V23 (parallel branches)
‹ q2
and q3 in series: q2 = q3 = q23 (C23 = 2μF)
‹ Charge conservation: q10 = q1 + q23
‹ 48 = C1V1 + C23V1 (V1 = V23)
‹ Find V1: V1 = 48 / (C1 + C23) = 8 V
‹ Find q1: q1 = C1V1 = 32μC
12V
‹ q23 = q2 = q3 = 48 – 32 = 16μC
‹ V2 = q2 / C2 = 2.67 V
‹ V3 = q3 / C3 = 5.33 V
PHY2049: Chapter 25
A
B
6μF
4μF
3μF
24
Another Example
ÎEach
capacitor has capacitance 10μF. Find the total
capacitance
ÎDo
it in stages
‹2
&3
‹ Add 4
‹ Add 5
‹ Add 1
⇒
⇒
⇒
⇒
5 μF
15 μF
6 μF
16 μF
2
1
4
3
5
PHY2049: Chapter 25
25
Find Charges on All Capacitors
ÎEach
‹ q1
ÎC2345
capacitor has capacitance 10μF. V = 10 volts
= 10 x 10 = 100μC
= 6μF
‹ q2345
= q234 = q5 (series)
‹ q2345 = q234 = q5 = 10 x 6 = 60μC
‹ V5 = q5 / C5 = 6
ÎFind
q4, V4
‹ V234
= V4 = 10 – 6 = 4
‹ q4 = C4 x 4 = 40μC
ÎFind
2
1
4
3
5
q2, q3, V2, V3 (C23 = 5μF)
‹ q2
= q3 = q23 = C23 x 4 = 5 x 4 = 20μC
‹ V2 = 20 / C2 = 2
‹ V3 = 20 / C3 = 2
PHY2049: Chapter 25
26