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2. electronic components

Physics 121 - Electricity and Magnetism
Lecture 06 - Capacitance
Y&F Chapter 24 Sec. 1 - 6
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Overview
Definition of Capacitance
Calculating the Capacitance
Parallel Plate Capacitor
Spherical and Cylindrical Capacitors
Capacitors in Parallel and Series
Energy Stored in an Electric Field
Atomic Physics View of Dielectrics
Electric Dipole in an Electric Field
Capacitors with a Dielectric
Dielectrics and Gauss Law
Summary
Copyright R. Janow – Fall 2014
1
What does Capacitance measure?
How much charge an arrangement of conductors holds when a
given voltage is applied.
 The charge needed per volt depends on
a geometrical factor called capacitance.
Example:
•
•
•
•
Q
Q
C
 V 
V
C
Two charged, conducting spheres: Radii R1 and R2 = 2R1.
Spheres touch and come to the same potential V1 = V2 between R and infinity
Different final charges Q1 and Q2.
Apply point charge potential formula, V(infinity) = 0
V1 
1
4 0
Q1
Q
 1
R 1 C1
and also
V2 
1
4 0
Q2
Q
 2
R2
C2
Capacitance of a single isolated sphere:
Example: A primitive capacitor
• The right ball’s potential is the same as the +
side of the battery. Similarly for the – ball.
• How much charge flows onto each ball to
produce a potential difference of 1.5 V ?
• The answer depends on the capacitance.
Q1
C
R
1
 1  1 
Q 2 C2 R 2 2
C  4 0R
V = 1.5 V
_
+ charges
+
+
1.5 V
battery
_
+
charges
Copyright
R. Janow – Fall 2014
Definition of
CAPACITANCE :
Q
C
V
or

Q  C V
Coulombs
Volt

• Measures the charge needed per volt of potential difference between plates
• Does not depend on applied V or charge Q. Always positive.
• Depends on geometry (and on dielectric materials) only
• Units: 1 FARAD = 1 Coulomb / Volt. - Farads are very large
1 mF = 10-6 F.
1 pF = 1 pico-Farad = 10-12 F = 10-6 mF = 1 mmF
Thought experiment:
Geometry effect on Capacitance
+ charges
Move the balls closer together while still connected to the _
battery
• The potential difference V cannot change.
 


• But:
+
+
1.5 V
battery
V    E  ds  Eav  s
• The distance s between the balls decreased
so the E field had to increase as did the stored energy. + charges
• Charge flowed from the battery to the balls to increase E.
• The two balls now hold more charge for the same potential
difference: i.e. the capacitance increased.
_
constant
Q  CV
increases
increases
Copyright R. Janow – Fall 2014
Capacitors are charge storage devices
• Two conductors not in electrical contact
• Electrically neutral before & after being charged
Qenc= Qnet=0
• Current can flow from + plate to – plate if
there is a conducting path
(complete circuit but not across the plates)
• Capacitors store charge and potential energy
+
-
+Q
d
V
• power supplies
• radio circuits
• memory bits
• Common type: “parallel plate”, often tubular
Method for calculating capacitance from geometry:
• Assume two conducting plates (equipotentials) with


qenc
equal and opposite charges +Q and –Q
E 
 E  dA
0
• Possibly use Gauss’ Law to find E between the plates
S
• Calculate V between plates using a convenient path
f
 
• Capacitance C = Q/V
Vfi   E  ds
• Insulating materials (“dielectrics”) can reduce the E field
i
between plates by “polarizing” - capacitance increasesCopyright R. Janow – Fall 2014


-Q
EXAMPLE: CALCULATE C for a PARALLEL PLATE CAPACITOR
d
Find E between plates
•
•
•
•
A = plate area. Treat plates as infinite sheets
|s+| = |- s| = s = Q / A = uniform surface charge density
E is uniform between the plates (d << plate size)
Use Gaussian surface S (one plate). Flux through ends
and attached conductors is zero. Total flux is EA
• Qenc = sA = 0f  0 EA
S
+Q
 E  s/  0
-Q
path
i.e.,
E  Q/  0 A
(infinite conducting sheet)
Find potential difference V:
• Choose Vi = 0 on negative plate (grounded)
• Choose path from – plate to + plate, opposite to E field


s
Qd Q
Vfi    E  ds  ()( )Ed  + d 

0
0 A C
path
0A
Q
 C

V
d
• C DEPENDS ONLY ON GEOMETRY
• C  infinity as plate separation d  0
• C directly proportional to plate area A
• Other formulas for other geometries
Copyright R. Janow – Fall 2014
Charge on a capacitor - no battery attached
6-1: Suppose that we charge an ideal parallel plate capacitor to
20 V, then disconnect the battery.
What happens to the charge and voltage on it?
A.
The charge stays on the plates indefinitely, and the voltage stays
constant at 20 V.
B. The charge leaks out the bottom quickly,
and the voltage goes to 0 V.
C. The charge jumps quickly across the air gap,
and the voltage goes to 0 V.
D. The charge stays on the plates, but the voltage drops to 0 V.
E. The charge instantly disappears,
but the voltage stays constant at 20 V.
Copyright R. Janow – Fall 2014
EX 24.03: FIND C for a SPHERICAL CAPACITOR
• 2 concentric spherical, conducting shells, radii a & b
• Charges are +q (inner sphere), -q (outer sphere)
• All charge on the outer sphere is on its inner
surface (by Gauss’s Law)
• Choose spherical Gaussian surface S as shown
and find field using Gauss’s Law:
 
 0  E  dA  q
q   0EA   0E(4r 2 )
S
• As before: E  q /( 4 0r )
2
• To find potential difference use outward radial
integration path from r = a to r = b.
r b
 
V  Vb  Va    E  ds 
r a
V 
r b
b
q
dr
 q ()1

4 0 r  a r 2
4 0 r a
q  1 1
q  ab
  


4 0  b a  4 0  ba 
4 0 ab
q
 C

| V |
ba
Negative
For b > a
Vb < Va
Let b  infinity. Then a/b  0 and
result becomes the earlier formula for
the isolated sphere:
C
4 0ab
b
Copyright
 4 R.aJanow – Fall 2014
0
EX 24.04: Find C for a CYLINDRICAL CAPACITOR
EXTRA
•
•
•
•
2 concentric, long cylindrical conductors
Radii a & b and length L >> b => neglect end effects
Charges are +q (inner) and -q (outer),  is uniform
All charge on the outer conductor is on its inner
surface (by Gauss’s Law)
• Choose cylindrical Gaussian surface S between plates
and find field at radius r.
• E is perpendicular to endcaps => zero flux contribution
 0  cyl
• So:
 
  0  E  dA  q
S
q   0EA   0E(2rL)
E  q /(2 0rL)  /(2  0r )
• To find potential difference use outward radial
integration path from r = b to r = a.
r b
 
V  Vb  Va    E  ds 
r a
C  q / V  2 0
L
ln(b / a)
C  0 as b/a  inf
C  inf as b/a  1
r b
q
dr
q
q
b

ln(
r
)

ln(b / a)

a
2 0L r  a r
2 0L
2 0L
Vb < Va
For b > a
C depends only on
geometrical parameters
Copyright R. Janow – Fall 2014
Examples of Capacitance Formulas
•
Capacitance for isolated Sphere
C  4 0R
C
•
Parallel Plate Capacitor
•
Concentric Cylinders Capacitor
•
Concentric Spheres Capacitor
•
Units: F (Farad) = C2/Nm = C/ Volt = 0length
- named after Michael Faraday. [note: 0 = 8.85 pF/m]
0A
d
L
ln(b / a)
ab
C  4 0
ba
C  2 0
All of these formulas depend only on geometrical factors
Copyright R. Janow – Fall 2014
Capacitors in circuits
CIRCUIT SYMBOLS:
CIRCUIT DEFINITIONS:
Current
+
-
 i  dq/dt
 rate of + charge
flow past a point in the circuit
Open Circuit:
NO closed path. No current. Conductors are equi-potentials
Closed Circuit: Current can flow through completed paths somewhere.
Loop Rule:
Potential field is conservative
 Potential CHANGE around ANY closed path = 0
Example: CHARGING A CAPACITOR
i
+
+
E
-
-
• Current begins flowing as switch S is CLOSED, completing circuit
• Battery (EMF) maintains V (= EMF E) & supplies energy. It moves free
+ charges inside battery from – to + terminal.
C • Convention: i flows from + to – outside of battery
• When switch closes, current (charge) flows until V across
capacitor equals battery voltage E.
Then current stops as E field in wire  0
S
DEFINITION: EQUIVALENT CAPACITANCE
• Capacitors can be connected in series, parallel, or more complex combinations
• The “equivalent capacitance” is the capacitance of a SINGLE capacitor that would
have the same capacitance as the combination.
Copyright R. Janow – Fall 2014
• The equivalent capacitance can replace the original combination in analysis.
Parallel capacitors - Equivalent capacitance
The actual parallel circuit...
Qi  Ci V
Q2
Q3
C1
C2
C3
E
V is the same for each branch
...and the equivalent circuit:
Q tot  CeqV
Q1
....
V ....
Qtot
V
E
Ceq
The parallel capacitors are just like a
single capacitor with larger plates so....
Q tot   Qi
(parallel)
Charges on parallel capacitors add
 Q tot  C1V + C2V + C3V + ...  (C1 + C2 + C3 + ...) V
Parallel capacitances
add directly
Question: Why is V
the same for all
elements in parallel?
Ceq   Ci
Answer: Potential is
conservative field, for ANY
closed loop around circuit:
(parallel)
Vi  0 (Kirchoff Loop Rule)
Copyright
R. Janow – Fall 2014
Series capacitors - equivalent capacitance
The actual series circuit...
Vtot
V1
V2
V3
C1
C2
C3
The equivalent circuit...
Qtot
Vtot
Vtot
Ceq
Vi may be different for capacitors in series
Vtot   Vi 
V1 + V2 + V3 + .....
Q tot  CeqVtot
Gaussian surface
Qenc = 0
neutral..so
Q1 = - Q2
Vi  Qi / Ci BUT All Qi  same Q
Charges on series capacitors are all
equal - here’s why.....
Q1
Q2
 Vtot  Q/C1 + Q/C2 + Q / C3 + ...  Q 1/Ci  Q/C eq
Reciprocals of series
capacitances add
For two capacitors
in series:
1

Ceq

1
Ci
(series)
1
1
1
C + C1
C2C1

+
 2
Copyright
Ceq R. Janow
– Fall 2014
Ceq C1 C2
C1C2
C1 + C2
Example 1:
A 33mF and a 47 mF capacitor are connected in parallel
Find the equivalent capacitance
Solution:
Example 2:
Same two capacitors as above, but now in series connection
Solution:
Example 3:
Cpara  C1 + C2  80 mF
C ser 
C1 C2
C1 + C2

33 x 47
 19.4 mF
33 + 47
A pair of capacitors is connected as shown
• C1 = 10 mF, charged initially to 100V = Vi
• C2 = 20 mF, uncharged initially
Close switches. Find final potentials across C1 & C2.
Solution:
• C1 & C2 are in parallel  Same potential Vf for each
• Total initial charge:
Q tot  C1 Vi  10-3 C.
• Charge is conserved – it redistributes on both C1 & C2
Ceq  Q tot / Vf  C1 + C2  Vf 
• Final charge on each:
10-3
30 x 10
-6
 33 V.
Copyright R. Janow – Fall 2014
Q1f  C1 Vf  3.3 x 10 C. Q2f  C2 Vf  6.7 x 10-4 C.
-4
Three Capacitors in Series
6-2: The equivalent capacitance for two
capacitors in series is:
1
C1C2
Ceq 
1
C1
+
1
C2

C1 + C2
Which of the following is the equivalent
capacitance formula for three capacitors
in series?
C1 + C 2 + C 3
C1C 2C 3
D.
Ceq 
A.
Ceq 
C1C 2C 3
C1 + C 2 + C 3
B.
Ceq
C1C2 + C2C3 + C1C3

C1 + C2 + C3
Ceq
C1C2 + C2C3 + C3C1

C1C2C 3
C.
E.
Ceq 
C1C2C 3
C1C2 + C2C3 + C3C1
Apply formula for Ceq twice
Copyright R. Janow – Fall 2014
Example: Reduce circuit to find Ceq=C123
for mixed series-parallel capacitors
After Step 1
C1
C2
parallel
V
V
C123
C123 
C12C 3
C12 + C 3
series
C3
C12  C1 + C2
Values:
C12
V
C3
After Step 2
1
1
1

+
C123 C12 C 3
C1 = 12.0 mF, C2 = 5.3 mF, C3 = 4.5 mF
C123 = (12 + 5.3) 4.5 / (12 + 5.3 + 4.5) mF = 3.57 mF
Copyright R. Janow – Fall 2014
Series or Parallel?
6-3: In the circuits below, which ones show
capacitors 1 and 2 connected in series?
C2
A.
B.
C.
D.
E.
I, II, III
I, III
II, IV
III, IV
None
I
C3
II
C1
E
E
C3
C1
C2
C1
III
IV
C3
E
C2
C1
E
C2
C3
Copyright R. Janow – Fall 2014
Energy Stored in a Capacitor
When charge flows in the sketch, energy stored in the
battery is depleted. Where does it go?
•
•
Charge distributions have
potential energy. Charges that
are separated in a neutral body
store energy.
The electric potential is defined
to be
V = U/q,
•
_
+
U = qV
A small element of charge
added to each plate of a
capacitor stores potential
energy:
+ charges
dq
dU = V dq
•
V=1.5 V
V
q
C
+
The energy stored by charging a
capacitor from charge 0 to Q is
the integral:
1.5 V
battery
_
1 Q
Q2 1
U   dU   q dq 
 2 CV2
0
C 0
2C
Q
+ charges
Copyright R. Janow – Fall 2014
Capacitors Store Energy in the
Electrostatic Field
•
The total energy in a parallel plate capacitor is
U  12 CV 2 
•
0A 2
V
2d
The volume of space filled by the electric field
in the capacitor is = Ad, so the energy density
u is
 A
U
V
u
 0 V 2  12  0  
vol 2dAd
d
•
But for a parallel plate capacitor,
•
so
2
 
V    E  ds  Ed
u  12  0E 2
Energy is stored in
the electric field
Copyright R. Janow – Fall 2014
What Changes?
6-4: A parallel plate capacitor is connected to a
battery of voltage V. If the plate separation is
decreased, which of the following increase?
A. II, III and IV.
B. I, IV, V and VI.
C. I, II and III.
D. All except II.
E. All increase.
q  CV
U  12 CV 2
C
0A
d
u  12  0E 2
I.
Capacitance of capacitor
II.
Voltage across capacitor
III.
Charge on capacitor
IV.
Energy stored on capacitor
V.
Electric field magnitude
between plates
VI.
Energy density of E field
E  V/d
Copyright R. Janow – Fall 2014
Model for Polarization in Dielectrics
A dipole in a uniform external field..
..feels torque, stores electrostatic potential energy


p  qd
  
  px E
• |torque| = 0 at q = 0 or q = 
• |torque| = pE at q = +/- /2
• RESTORING TORQUE: (q)  (+q)
 
UE  p  E
See Lecture 3
Polarization: An external field aligns dipoles in an insulator,
causing a polarization field that reduces the external, applied field

E0



Ediel  E0 + Epol
-

p

Epol
+
+
+
+
+
Copyright R. Janow – Fall 2014
Dielectric materials in capacitors
• Insulators POLARIZE when an external electric field is applied
• The NET field inside the material is lowered below vacuum level.
MOLECULAR VIEW
Non-polar Material Polar Material
induced dipole permanent dipole
NO EXTERNAL FIELD
WITH EXTERNAL FIELD Eo
E0= Evac is field due to free charge
Response to Evac is the polarization
field Epol
Actual weakened net field inside is
Ediel = E0 – Epol = Enet

Ediel

Epol
Polarization surface charge density
reduces free surface charge density
snet= sfree – spol
(sfree = sext = svac)
Dielectrics increase capacitance
Dielectric    Cdielectric  1
For a given V = Enetd, more
C vacuum
constant
movable charge sfreeis needed
Copyright R. Janow – Fall 2014
Inside conductors, polarization reduces Enet to zero
The Dielectric Constant 
•
All materials (water, paper, plastic, air) polarize to some extent and
 = 0
is the free space permittivity.  is called the dielectric constant -
have different permittivities
•
•
•
0
a dimensionless number.

Wherever you see 0 for a vacuum, you can substitute
considering dielectric materials.
The capacitance increases when the space between plates is filled with
a dielectric. For a parallel plate capacitor:
Cdiel
•
0 when
0 A

 C vac
d
A dielectric weakens the field, compared to what it would be for
vacuum



Ediel  Evac /  D/ 0 
•
The induced polarization charge density on the dielectric surface is
proportional to the “free” surface charge density created by the
external circuit:
1
spol  s free [1  ]
κ
Copyright R. Janow – Fall 2014
Copyright R. Janow – Fall 2014
Copyright R. Janow – Fall 2014
What happens as you insert a dielectric?
Initially, charge capacitor C0 to voltage V, charge Q, field Enet.
•
•
•
•
With battery detached insert dielectric
Q remains constant, Enet is reduced
Voltage (fixed Q) drops to V’.
Dielectric reduced Enet and V.
Q
V
C0
•
•
•
With battery attached, insert dielectric.
Enet and V are momentarily reduced
but battery maintains voltage E
Charge flows to the capacitor as
dielectric is inserted until V and Enet
are back to original values.
OR
Q  C0 V
E
V 
Q'  C0 V
Q
C0
E
Q  a constant
 
V    E  ds  Ed
V  a constant
Copyright
QR.
' Janow
Q – Fall 2014
Gauss’ Law with a dielectric
0 
S


Enet  dA  qfree  qpol  qnet
OPTIONAL
TOPIC
S
{
Alternatively:






 0Evac  dA  qfree   0Enet  dA   D  dA
S
S
free charge on plates
field not counting polarization = 0Evac
The “Electric Displacement” D measures field that would be present due to
the “free” charge only, i.e. without polarization field from dielectric
Evac  Enet
D  0Evac  0Enet  Enet
•  could vary over Gaussian surface S. Usually it is constant and factors
• Flux is still measured by field without dielectric:



d  Evac  dA  Enet  dA
• Only the free charge qfree (excluding polarization) is counted as qenc
above. Ignore polarization charges inside the Gaussian surface
Copyright R. Janow – Fall 2014