capacitance
TRANSCRIPT
25-2. Capacitance25-2. Capacitance
charge storage, energy storage
CVq
Capacitor: Any two isolated conductors
Given
As a Result
25-2. Capacitance25-2. Capacitance
00
1
A
qEEE
EdVV if
if VVV Potential (or Voltage)d
A
qEdV
0
Capacitanced
AC 0
FV
CUnits of capacitance: Farad (F) = Coulomb/Volt
25-2. Capacitance25-2. Capacitance
+q -q
E field between the plates: (Gauss’ Law)
Relate E to potential difference V:
What is the capacitance C ?
Area of each plate = ASeparation = dcharge/area = = q/A
dA
qEdV
0
00
1
A
qE
d
A
V
QC 0
FV
C
25-2. Capacitance25-2. Capacitance
d
A
V
QC 0
Capacitance Capacitance and Your iPodand Your iPod
25-2. Capacitance25-2. Capacitance
• A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm
• What is the capacitance?
C = 0 A/d
= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)
= 2.21 x 10–9 F
(Very Small!!)
FV
CUnits of capacitance: Farad (F) = Coulomb/Volt
25-2. Capacitance25-2. Capacitance
Does the capacitance of a capacitor increase, decrease, or remain the same?
(a)When the charge q on it is doubled
(b)When the potential difference across it tripled
CVq
Determine what is NOT changed.
25-2. Capacitance25-2. Capacitance
• A parallel plate capacitor of capacitance C is charged using a battery.
• Charge = Q, potential difference = V.• Plate separation is INCREASED while
battery remains connected.
+Q –Q
• V is fixed by battery!• C decreases (=0A/d)• Q=CV; Q decreases• E = Q/ 0A decreases
Does the Electric Field Inside:
(a) Increase?
(b) Remain the Same?
(c) Decrease?
25-3. Calculating the Capacitance25-3. Calculating the Capacitance
Cylindrical Capacitor
abh
Cln
2 0
a
b
h
qdrrh
qdrEV
b
a
b
aln
2
1
2 00
0
)2(
qhrE
25-3. Calculating the Capacitance25-3. Calculating the Capacitance
Spherical Capacitor
0
2 )4(
qrE
abh
qdr
r
qdrEV
b
a
b
a
11
2
1
4 02
0
baa
C1
4 0
25-3. Calculating the Capacitance25-3. Calculating the Capacitance
Spherical Capacitor
baa
C1
4 0
aC 04
ab b
Capacitance of a single conducting sphere
25-4. C in Parallel and in Series 25-4. C in Parallel and in Series
V: the same
Parallel Series Q: the same
25-4. C in Parallel and in Series 25-4. C in Parallel and in Series
V: the sameParallel
• A wire is an equipotential surface.
• VAB = VCD = V
• Qtotal = Q1 + Q2
• CeqV = C1V + C2V
• Equivalent parallel capacitance = sum of capacitances
A B
C D
C1
C2
Q1
Q2
CeqQtotal
21 CCCeq
25-4. C in Parallel and in Series 25-4. C in Parallel and in Series
Q: the sameSeries
• Q1 = Q2 = Q (WHY??)
• VAC = VAB + VBC
AB C
C1 C2
Q1 Q2
Ceq
Q
21 C
Q
C
Q
C
Q
eq
21
111
CCCeq
25-4. Example 125-4. Example 1
10 F
30 F
20 F
120V
• Q = CV; V = 120 V
• Q1 = (10 F)(120V) = 1200 C • Q2 = (20 F)(120V) = 2400 C• Q3 = (30 F)(120V) = 3600 C
Note that:• Total charge (7200 C) is shared between the 3
capacitors in the ratio C1:C2:C3 — i.e. 1:2:3
What is the charge on each capacitor?
25-4. Example 225-4. Example 2
10 F 30 F20 F
120V
• Q = CV; Q is same for all capacitors• Combined C is given by:
)30(
1
)20(
1
)10(
11
FFFCeq
• Ceq = 5.46 F• Q = CV = (5.46 F)(120V) = 655 C• V1= Q/C1 = (655 C)/(10 F) = 65.5 V• V2= Q/C2 = (655 C)/(20 F) = 32.75 V• V3= Q/C3 = (655 C)/(30 F) = 21.8 V
What is the potential difference across each capacitor?
25-4. Example 325-4. Example 3
What is the charge on the 10F capacitor?
10 F
5 F5 F 10V
10 F
10 F10V
• Then, we have two 10F capacitors in series
• So, there is 5V across the 10F capacitor of interest
• Hence, Q = (10F )(5V) = 50C
25-4. Example 425-4. Example 4
(N-1) Capacitors with A and d: Parallel Connection
+
–
25-4. Example 525-4. Example 5
What is the charge on each capacitor?
10F (C1)
20F (C2)
• 10F capacitor is initially charged to 120V.• 20F capacitor is initially uncharged.• Switch is closed, equilibrium is reached.
Initial charge on 10F = (10F)(120V)= 1200C
After switch is closed, let charges = Q1 and Q2.
Charge is conserved: Q1 + Q2 = 1200C
And, Vfinal is same:
2
2
1
1
C
Q
C
Q
22
1
• Q1 = 400C• Q2 = 800C• Vfinal= Q1/C1 = 40 V
25-5. Energy Stored25-5. Energy Stored
2
2
1CVU
• Start out with uncharged capacitor• Transfer small amount of charge dq from
one plate to the other until charge on each plate has magnitude Q
• How much work was needed?dq
Q
VdqU0
Q
C
Qdq
C
q
0
2
2 2
2CV
25-5. Energy Stored in an E-Field25-5. Energy Stored in an E-Field
• Energy stored in capacitor:U = Q2/(2C) = CV2/2 • View the energy as stored in ELECTRIC FIELD• For example, parallel plate capacitor: • Energy DENSITY = energy/volume = u
General expression for any region with
vacuum (or air)
Add
AQ
AdC
Q
Volume
Uu
1
2
1
2 0
22
20
0
2
0 2
1
2E
A
Qu
25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric
Dielectric: Noncoducting material
d
AC
d
AC 00
HW5-8
Breakdown Potential: Dielectric Strength (kV/mm) X
Length (mm)
00
25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric
00
d
AC
d
AC 00
2
2
1CVU C
QU
2
2
1or ??
25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric
00 d
AC
d
AC 00
25-6. Capacitor with a Dielectric25-6. Capacitor with a Dielectric
00 d
AC
d
AC 00
25-6. Example25-6. Example
• Capacitor has charge Q, voltage V• Battery remains connected while
dielectric slab is inserted.
• Do the following increase, decrease or stay the same:– Potential difference?– Capacitance?– Charge?– Electric field?
dielectric slab
25-6. Example25-6. Example
• Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E
• Battery remains connected• V is FIXED; Vnew = V (same)
• Cnew = C (increases)• Qnew = (C)V = Q (increases).• Since Vnew = V, Enew = E (same)
dielectric slab