Phys102
General Physics II
Capacitance
Capacitance
•Capacitance definition and examples.
•Discharge a capacitor
•Circular parallel plate capacitior
•Cylindrical capacitor
•Concentric spherical capacitor
•Dielectric Slabs
1
Capacitance
ƒ Definition of capacitance
A capacitor is a useful device in electrical circuits
that allows us to store charge and electrical energy in a
controllable way. The simplest to understand consists
of two parallel conducting plates of area A separated
by a narrow air gap d. If charge +Q is placed on one
plate, and -Q on the other, the potential difference
between them is V, and then the capacitance is
defined as C=Q/V. The SI unit is C/V, which is called
the Farad, named after the famous and creative
scientist Michael Faraday from the early 1800’s.
ƒ Applications
– Radio tuner circuit uses variable capacitor
– Blocks DC voltages in ac circuits
– Act as switches in computer circuits
– Triggers the flash bulb in a camera
– Converts AC to DC in a filter circuit
Parallel Plate Capacitor
q = CV
2
Charging a Capacitor
q = CV
Electric Field of Parallel Plate Capacitor
Gauss Law
E=
σ
ε0
V = Ed =
C=
σ=
q
A
⇒
E=
q
ε0 A
q = ε0 EA
qd
ε0 A
ε A
q
q
=
= 0
V qd ε 0 A
d
Coulomb/Volt = Farad
3
Circular parallel plate capacitor
r
r = 10 cm
r
A = πr2 = π(.1)2
A = .03 m 2
s
C=
S = 1 mm = .001 m
ε0A
S
C = (10 −11 )
.03 Coulomb
.001 Volt
}Farad
C = 3 ×10 −10 F
C = 300 pF
p = pico = 10-12
Model of coaxial cable for calculation of
capacitance
Outer metal braid
Signal wire
4
Find the capacitance of a ordinary piece of coaxial cable (TV cable)
cable)
For a long wire we found that
Er =
where r is radial to the wire.
• r
a
a
b
b
Va − Vb = − ∫ E. ds = −2kλ ∫
2kλ
r
metal braid
with - q
outer insulator
dr
= −2kλ ln r
r
a
radius b
b
E. ds = Edscos180 = −Eds = Edr
ds = - dr because path of integration is radially inward
a
Va −Vb = −2kλ ln
b
b
a
Va is higher than Vb
λ =Q
k=
V=
4 πε 0 ∼ air
Q
2πε 0 L
C=Q
L
1
a = 0.5 mm
C 2πε 0
=
L ln ba
or
V = 2kλ ln
C=
signal
wire
radius a
with + q
Insulator
(dielectric ∼ε)
•
V
=
ln
b = 2.0 mm
κ≅2
−11
C 6 ×10
=
L
ln 4
b
a
Q2πε 0 L
Qln ba
=
6 ×10 −11
1.38
C
= 43 PF m
L
C
= 86 PF m
L
2πε 0 L
ln ba
ε0 (air)
κ=2
Capacitance of two concentric spherical shells
a
a
b
b
Va − Vb = − ∫ E. ds = − ∫ Edr
E. ds = Edscos180 = −Eds = Edr
ds = - dr
a
a
a
b
b
b
Va − Vb = − ∫ Edr = − ∫ kq /r 2 dr = −kq ∫
dr
r2
a
V = kq
1
1 1
b − a)
= kq( − ) = kq(
rb
a b
ab
C = q /V =
ab
ab
= 4 πε0
b−a
k(b − a)
5
Spherical capacitor or sphere
Recall our favorite example for E and V is spherical symmetry
Q
The potential of a charged sphere is V = (kQ)/R with V = 0 at r = ∞
.
The capacitance is
R
C=
Q
Q
R
=
= = 4πε 0 R
V kQ R k
Where is the other plate (conducting shell)?
It’s at infinity where it belongs, since that’s where the electric lines of flux terminate.
k = 1010 and R in meters we have
Earth: C = (6x108 cm)PF = 600µF
R
C = 10 = 10−10 R(m) = 10−12 R(cm) Marble: 1 PF
10
Basketball: 15 PF
C = R(cm) PF
You: 30 PF
Parallel Combination of Capacitors
Typical electric circuits have several capacitors in them. How do they
combine for simple arrangements? Let us consider two in parallel.
V
+
-
C1
Q1
C2
Q2
V
+
-
C1
C = Q/V
We wish to find one equivalent capacitor to replace C1 and C2.
Let’s call it C.
The important thing to note is that the voltage across each is the
same and equivalent to V. Also note what is the total charge
stored by the capacitors? Q.
Q = Q1 + Q 2 = C1V + C 2V = (C1 + C 2)V
Q
= C1 + C 2 ⇒ C = C1 + C 2
V
6
Series Combination of Capacitors
+
+
-
V
Q
-
+
C1
V1
Q
-
+
C2
V2
V
+
-
Q
-
C
V
C=
Q
V
V=
Q
C
What is the equivalent capacitor C?
Voltage across each capacitor does not have to be the same.
The charges on each plate have to be equal and opposite in sign by
charge conservation.
The total voltage across each pair is:
V = V 1 +V 2 =
So
1
1
1
Q Q
+
= Q( + ) = Q( )
C1 C 2
C1 C 2
C
1
1
1 ; Therefore,
= +
C1 C1 C 2
C=
C1C 2
C1 + C 2
Sample problem
C1 = 10 µF
C1
V
+
-
C3
C2
C2 = 5.0 µF
C3 = 4.0 µF
a) Find the equivalent capacitance of the entire combination.
C1 and C2 are in series.
1
1
1
C1C 2
= +
⇒ C12 =
C12 C 1 C 2
C1 + C 2
10 × 5 50
C12 =
=
= 3.3µF
10 + 5 15
C12 and C3 are in parallel.
Ceq = C12 + C 3 = 3.3 + 4.0 = 7.3µF
7
Sample problem (continued)
C1 = 10 µF
C1
V
C2 = 5.0 µF
C3
+
-
C3 = 4.0 µF
C2
b) If V = 100 volts, what is the charge Q3 on C3?
C = Q/V
Q 3 = C 3V = 4.0 ×10 −6 ⋅100
Q 3 = 4.0 ×10 −4 Coulombs
c) What is the total energy stored in the circuit?
1
1
U = CeqV 2 = ⋅1.3 ×10 −6 ⋅10 4 = 3.6 × 10 − 2 J
2
2
U = 3.6 ×10 −2 J
Electric Potential Energy of Capacitor
ƒ
+q
As we begin charging a capacitor, there is initially no potential
potential difference
between the plates. As we remove charge from one plate and put itit on the
other, there is almost no energy cost. As it charges up, this changes.
changes.
-q
+ -
At some point during the charging, we
have a charge q on the positive plate.
The potential difference between the plates is V = q/C. As
we transfer an amount dq of positive charge from the
negative plate to the positive one, its potential energy
increases by an amount dU.
dU = Vdq = q dq.
C
The total potential energy increase is
Also
Q
q
q2
U = ∫ dq =
C
2C
0
1
1
1 Q2
U = QV = CV 2 =
2
2
2 C
=
Q2
2C
using C = Q/ V
8
Graphical interpretation of integration
dU = Vdq
Q
U = ∫ Vdq where V = q/C
0
V
V = q/c
q/c
q
dq
U=
Q
1 N
∑ qi∆qi = Area under the triangle
C i =1
Q
Area under the triangle is the value of the integral
q
∫ C dq
0
Area of the triangle is also = 1/2 bh
1
1
Q 1 Q2
Area = (b)(h) = (Q)( ) =
C 2 C
2
2
Q
q
∫ C dq =
0
q2 Q Q2
=
2
2C
0
Where is the energy stored in a capacitor?
• Find energy density for parallel plate capacitor. When we charge a
capacitor we are creating an electric field. We can think of the work
done as the energy needed to create that electric field. For the
parallel plate capacitor the field is constant throughout, so we can
evaluate it in terms of electric field E easily.
Use U = (1/2)QV
E=
σ σ Q and V = Ed
= =
κε 0 ε εA
We are now
including dielectric
effects: ε
Solve for Q = εAE, V = ES and substitute in
1
1
1
U = QV = (ε AE )( Ed ) = ε E 2 ( Ad )
2
2
2
U
1
= ε E2 = u
Ad 2
1
u = ε E2
2
volume occupied by E
Electrostatic energy density general result for all
geometries.
To get total energy you need to integrate over volume.
9
How much energy is stored in the Earth’s atmospheric
electric field?
(Order of magnitude estimate)
atmosphere
20 km
h
Earth
R
R = 6x106 m
E = 100 Vm = 10 2
1
U = ε 0 E 2 ⋅ Volume
2
Volume = 4πR 2 h
Volume = 4π (6 ×10 6 ) 2 (2 × 10 4 ) = 8.6 × 1018 m3
1
U = (10 −11 )(10 2 )(8.6 ×1018 )
2
U = 4.3 ×1011 J
This energy is renewed daily by the sun. Is this a lot?
The total solar influx is 200 Watts/m2
Usun = 200 ⋅ 3.14(6 × 106 ) ≅ 2 × 1016 J s = 2 × 10 21 J day
U Usun ≅ 2 ×10 −10
World consumes
about 1018
J/day. This is
1/2000 of the
solar flux.
Only an infinitesimal fraction gets converted to electricity.
Dielectrics
ƒ A dielectric is any material that is not a conductor, but
polarizes well. Even though they don’t conduct they
are electrically active.
– Examples. Stressed plastic or piezopiezo-electric crystal will
produce a spark.
– When you put a dielectric in a uniform electric field (like in
between the plates of a capacitor), a dipole moment is
induced on the molecules throughout the volume. This
produces a volume polarization that is just the sum of the
effects of all the dipole moments. If we put it in between the
plates of a capacitor, the surface charge densities due to the
dipoles act to reduce the electric field in the capacitor.
10
Dielectrics
ƒ The amount that the field is reduced defines the
dielectric constant κ from the formula E = E0 / κ, where
E is the new field and E0 is the old field without he
dielectric.
ƒ Since the electric field is reduced and hence the
voltage difference is reduced (since E = Vd),
Vd), the
capacitance is increased.
– C = Q / V = Q / (V
(V0 / κ) = κ C0
– κ is typically between 2 – 6 with water equal to 80
Permanent
dipoles
Induced
dipoles
_ ++ _
E0 = the applied field
E’ = the field due to
induced dipoles
E = E0 - E’
11
C=q
V=
V
V = E0 d
d
σ
ε0
E0 =
C=
qS
ε0A
ε0A
d
σ =qA
q
E0 =
E=
V=
V=
ε0A
E0
C=q
κ
E0 d
C=
κ
V0
V
κq
V0
C = κC 0
κ
What is the electric field in a sphere of uniform
distribution of positive charge. (nucleus of protons)
ρ=
R
r
E
∫
Q
4 3
πR
3
EdA =
E 4 πr 2 =
E=
ρ
q enc
ε0
4πr3
3
ε0
ρr
Q
=
r
3ε0 4 πε0 R 3
12