Unformatted text preview: Capacitance PHY2049: Chapter 25 1 What You Know: Electric Fields
Coulomb’s law
Electric fields
Equilibrium
Gauss’ law
Electric 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
Electric potential energy
Electric potential
Equipotential surfaces
Potential of point charge
Potential of charge distribution
Special cases: dipole, line, ring, disk, sphere Relationship of potential and electric field
Calculating the potential from the field
Calculating the field from the potential Potential energy from a system of charges PHY2049: Chapter 25 3 Capacitance: Basic Idea
Capacitance: 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
Electric field lines start on + charges, terminate on −
+ V+ − V− = − ∫ E ⋅ ds
− Follow 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
From Gauss’ law (conducting sheet) E =
=
ε 0 Aε 0
⎛q⎞
So V = Ed = ⎜
⎟d
⎝ Aε 0 ⎠
ε0 A
Therefore q =
V ≡ CV
d
A
C = ε 0 = ε 0 × "length"
d
Depends only on geometry
of device ++++++++++++++++ E d  PHY2049: Chapter 25 6 Example
A = 1m2, d = 1 μm A
C = ε 0 = 8.85 × 1012 × 106 = 8.85 × 106 F
d
C = 8.85 μ F Largish, but somewhat typical value PHY2049: Chapter 25 7 Cylindrical Capacitor
Inner = a, outer = b, length = L
Gauss’ law:
Using λ=q/L + V = ∫ E ds
− V = −∫ a b q1
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
Outer shell “very close” to inner shell: b − a = d (d small)
Use 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
Inner radius = a, outer radius = b
Coulomb’s 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
Isolated sphere: corresponds to b = ∞ 4π ab
4π a
= ε0
→ ε 0 4π a
C = ε0
b−a
1− a / b
Outer 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
V1 = V2 = V3 (same potential top and bottom)
Total charge: Qtot = Q1 + Q2 + Q3
CeqV = C1V + C2V + C3V Ceq = C1 + C2 + C3
Basic law for combining
capacitors in parallel
Works for N capacitors
PHY2049: Chapter 25 12 Capacitors in Series
q1 = q2 = q3 (same current charges all capacitors)
Total potential: V = V1 + V2 + V3
q/Ceq = q/C1 + q/C2 + q/C3 1
1
1
1
=
+
+
Ceq C1 C2 C3
Basic law for combining
capacitors in series
Works for N capacitors PHY2049: Chapter 25 13 ConcepTest
Two identical parallel plate capacitors are shown in an
endview 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
Each capacitor is the same in the three configurations.
Which configuration has the lowest equivalent capacitance?
(1)
(2)
(3)
(4) A
C/2 (series)
B
C
They all have identical capacitance C C C C C
A B
PHY2049: Chapter 25 C
15 Energy in a Capacitor
Capacitors have energy associated with them
Grab a charged capacitor with two hands and find out! Calculate energy
Continually move charge from – to + surface
dU = V dQ
Q
Q2 1
2
V=Q/C U =
Q / C dQ =
= CV ∫0 ( ) 2C 2 Q2 1
U=
= 2 CV 2
2C
So 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?
Answer: Energy is stored in the electric field itself!!
Example: Find energy density of two plate capacitor
E field is constant ε 0 ( A / d ) ( Ed )
U
CV
=
=
= 1 ε0E 2
u=
2
2 Ad
Ad 2 Ad
2 2 Energy density depends only on E field!
A general result, independent of geometry
Can be shown more generally by Maxwell’s equations u = 1 ε0E 2
2
PHY2049: Chapter 25 17 Example: Spherical Charged Conductor
Capacitance: C = ε 0 4π R Q2
Q2
=
Total energy of spherical conductor: U =
2C 8πε 0 R
Calculate directly: integrate energy density over volume u= 1 ε E2
20 U =∫ U = ∫ u dV
R dV = 4π r 2dr 2 ⎛Q⎞
Q2
ε0 ⎜
4π r 2dr =
2
⎜ 4πε r 2 ⎟
⎟
8πε 0 R
0
⎝
⎠ ∞1 R ∞ PHY2049: Chapter 25 Checks! 18 Dielectric Materials and Capacitors
Insulating material that can be polarized in E field
Induced
charges +++++++++++++++++++++++++
 κ Dielectric
material +++++++++++++++++++++++++
 Induced charges at dielectric surface partially cancel E field
E → E / κ κ > 1 is “dielectric constant”
V → V / κ (since V = Ed)
C → κC
(since C = Q / V) “Good” dielectric requires more than high κ value
Good 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 E applied, partially aligns dipoles
Aligned dipoles induce surface charges
Surface charges partially cancel E field
Yow! http://hyperphysics.phyastr.gsu.edu/hbase/electric/dielec.html
PHY2049: Chapter 25 20 ConcepTest
Two 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
When 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
Positive work: One “pushes in” dielectric
ΔU > 0
Negative work: Capacitor “sucks in” dielectric ΔU < 0
PHY2049: Chapter 25 22 Capacitors in Circuits
Multistep process
2 + 3 in series C o Combine 2 & 3 ⇒ C/2 Ceq 1 + (2+3) in parallel
Combine 1 & (2 & 3) ⇒ 3C/2 2
1C 3C o PHY2049: Chapter 25 23 Example: Find qi and Vi on All Capacitors
C1 is charged in position A, then S is thrown to B position
Initial voltage across C1:
Initial charge on C1: V0 = 12
q10 = 12 x 4 = 48μC After 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
Each capacitor has capacitance 10μF. Find the total
capacitance
Do 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
Each capacitor has capacitance 10μF. V = 10 volts
q1 = 10 x 10 = 100μC C2345 = 6μF
q2345 = q234 = q5 (series)
q2345 = q234 = q5 = 10 x 6 = 60μC
V5 = q5 / C5 = 6 Find q4, V4
V234 = V4 = 10 – 6 = 4
q4 = C4 x 4 = 40μC 2
1 4 3 5 Find 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 ...
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This note was uploaded on 07/31/2011 for the course PHY 2049 taught by Professor Any during the Summer '08 term at University of Florida.
 Summer '08
 Any
 Physics, Capacitance, Charge, Electric Fields

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