Capacitance
PHY2054: Chapter 16 Capacitance
1
Purpose of Capacitors
ÎStorage
‹ Used
in DC and AC circuits
ÎStorage
‹ Can
ÎUsed
of charge: Q = CV
of energy
provide energy to circuits
in DC and AC circuits
Q2 1
U=
= 2 CV 2
2C
‹ Timing
in DC circuits
‹ Resonance in AC circuits
‹ (Later in course)
PHY2054: Chapter 16 Capacitance
2
Capacitors in Parallel
Î V1
= V2 = V3 (same potential top and bottom)
ÎTotal
ÎCeqV
charge: qtot = q1 + q2 + q3
= C1V + C2V + C3V
V
Ceq = C1 + C2 + C3
V
¾ Basic law for combining
capacitors in parallel
¾ Works for N capacitors
PHY2054: Chapter 16 Capacitance
3
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
PHY2054: Chapter 16 Capacitance
V
V
4
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
PHY2054: Chapter 16 Capacitance
B
5
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
PHY2054: Chapter 16 Capacitance
C
6
Capacitors in Circuits
ÎFind
total capacitance Ceq between (a) and (b)
‹ Use
ÎC23
multi-step process
= 2 + 3 in series
‹ 1/C23
‹ C23
ÎC1
= 1/C + 1/C = 2/C
= C/2
+ C23 in parallel
‹ Ceq
(a) o
Ceq
C
2
1 C
3 C
(b) o
= C + C/2 = 3C/2
PHY2054: Chapter 16 Capacitance
7
Another Example
ÎAssume
all capacitors = 10 μF. Find total capacitance
and C4 in parallel: C34 = 10 + 10 = 20 μF
‹ C1, C34, C2 in series
‹ C3
1
1 1 1
5
= + + =
Ceq 10 20 10 20
Ceq = 4.0μF
ÎHow
much charge provided by battery to fully charge
capacitors? Assume V = 20.
‹Q
= Ceq x V = 4 x 20 = 80 μC
PHY2054: Chapter 16 Capacitance
8
Another Example
ÎAssume
all capacitors = 10 μF. Find total capacitance
= 30 μF (parallel)
‹ C56 = 20 μF (parallel)
‹ C1, C234, C56 (series)
‹ C234
1
1 1
1 11
= + +
=
Ceq 10 30 20 60
Ceq = 5.45μF
ÎHow
much charge provided by battery to fully charge
capacitors? Assume V = 20.
‹Q
= Ceq x V = 5.45 x 20 = 109 μC
PHY2054: Chapter 16 Capacitance
9
Find Charges on Series Capacitors
ÎLet
V = 10, C1 = 6μF, C2 = 12μF
‹ Find
charges, voltages on C1, C2
ÎCombine
‹ This
ÎFind
is what battery “sees”!
Ceq =
C1C2
72
=
= 4μF
C1 + C2 18
qeq, then use qeq = q1= q2
‹ qeq
ÎFind
series capacitances
= Ceq V = 4 x 10 = 40 μC
V1, V2
‹ V1
= q1 / C1 = 40 / 6 = 6.67 V
‹ V2 = q2 / C2 = 40 / 12 = 3.33 V
‹ V1 + V2 = 10, as expected
PHY2054: Chapter 16 Capacitance
10
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:
‹ Charge
flows from C1 to C2 and C3
‹ V1 = V23 (parallel branches)
Îq2
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
‹ q23 = 48 – 32 = q2 = q3 = 16μC
‹ V2 = q2 / C2 = 2.67 V
‹ V3 = q3 / C3 = 5.33 V
PHY2054: Chapter 16 Capacitance
A
B
6μF
12V
4μF
3μF
11
Another Example
ÎEach
ÎDo
capacitor has C = 10μF. Find the total capacitance
it in stages
‹2
&3
‹ Add 4
‹ Add 5
‹ Add 1
ÎCharge
‹ qtot
⇒
⇒
⇒
⇒
C23 = 5 μF
C234 = 15 μF
C2345 = 6 μF
C12345 = 16 μF
supplied by battery (20V)
2
1
4
3
5
= C12345 x V = 16 x 20 = 320 μC
PHY2054: Chapter 16 Capacitance
12
Find Charges, Voltages on All Capacitors
ÎEach
‹ q1
ÎC2345
capacitor has capacitance 10μF. V = 20 volts
= C1V = 10 x 20 = 200μC
= 6μF, q2345 = 6 x 20 = 120 μC
‹ q2345
= q234 = q5 = 120μC (series)
‹ V5 = q5 / C5 = 120 / 10 = 12
ÎFind
q4, V4 = V23
= V4 = 20 – 12 = 8
‹ q4 = C4V4 = 10 x 8 = 80μC
2
1
4
3
‹ V234
ÎFind
5
q2, q3, V2, V3 (C23 = 5μF)
‹ q2
= q3 = q23 = C23 x V23 = 5 x 8 = 40μC
‹ V2 = q2 / C2 = 40 / 10 = 4
Check: V2 + V3 = 8 (= V23)
‹ V3 = q3 / C3 = 40 / 10 = 4
PHY2054: Chapter 16 Capacitance
13
Capacitor Monster
All voltages = 4V, all capacitors = 2μF. What is the charge on C?
Can you find the charge on all capacitors?
PHY2054: Chapter 16 Capacitance
14
Capacitor Monster
All voltages = 4V, all capacitors = 2μF. What is the charge on C?
q = CV = 2 x 4 = 8μC
4
4
4
4
4
4
4
4
4
0
0
0
0
0
4
4
0
PHY2054: Chapter 16 Capacitance
15
Energy in a Capacitor
ÎCapacitors
‹ Grab
have energy associated with them
a charged capacitor with two hands and find out!
ÎCalculation
of stored energy
‹ Proof
requires simple calculus derivation
‹ Energy = work moving charge from – to + surface
Q2 1
U=
= 2 CV 2
2C
ÎCapacitors
store and release energy as they acquire and
release charge
‹ This
energy is available to drive circuits
PHY2054: Chapter 16 Capacitance
16
Example of Capacitor Energy
ÎC
= 5 μF, V = 200
(
)
U = 12 CV 2 = 12 × 5 × 10−6 × 2002 = 0.1J
ÎChange
U=
V to 20000 (as in demo of large HV capacitor)
1 CV 2
2
=
1×
2
(5 ×10 ) × 20000
−6
PHY2054: Chapter 16 Capacitance
2
= 1000J
17
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
PHY2054: Chapter 16 Capacitance
18
Dielectric Materials and Capacitors
ÎInsulating
Induced
charges
ÎInduced
material that can be polarized in E field
+++++++++++++++++++++++++
-------------------------------------------
κ
Dielectric
material
+++++++++++++++++++++++++
-------------------------------------------
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
ÎBut
“good” dielectric requires more than high κ value
‹ Good
insulator
‹ High breakdown voltage
‹ Low cost
(no charge leakage)
(no arcing at high voltage)
(affordable)
PHY2054: Chapter 16 Capacitance
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
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/dielec.html
PHY2054: Chapter 16 Capacitance
20
Dielectric Constants of Some Materials
Material
Vacuum
Air
Bakelite
Strontium titanate
Water
Ethanol
Mica
Barium Titanate
Paper
Beeswax
Silica glass
Dielectric constant κ
1.0
1.00059
4.9
310
80
24
5.4
100 – 1250
3.7
2.7 – 3.0
3.8
PHY2054: Chapter 16 Capacitance
21
Example of Dielectric Use
ÎSimple
capacitor: A = 2m2, d = 1μm (no dielectric)
‹ Place
200 volts across C
A
2
= 8.85 × 10−12 −6 = 1.77 × 10−5 = 17.7 μF
d
10
q = CV = 17.7 μF × 200 = 3540 μC
C = ε0
U=
ÎNow
1 CV 2
2
=
1
2
(1.77 ×10 ) ( 200)
−5
2
= 0.35 J
disconnect C from circuit
‹ Insert
strontium titanate dielectric (κ = 310) into capacitor
‹ Charge is conserved, calculate new C, V and U
Cnew = κ C = 310 × 17.7 = 5490 μF
Vnew = V / κ = 200 / 310 = 0.65 volts
U new = U / κ = 0.35 / 310 = 0.0011 J
(
2
= 12 CnewVnew
PHY2054: Chapter 16 Capacitance
)
22
Similar Example
ÎSame
capacitor as before, but this time insert dielectric
while C is in the circuit
= κC = 5490 μF
‹ V is still 200 volts (maintained by battery)
‹ Cnew
ÎCalculate
new q and U in this example
qnew = CnewV = 5490 μF × 200 = 1.1 C
U new =
ÎNotice
2
1C
V
2 new
=
1
2
(5490 ×10 ) ( 200)
−6
2
= 110 J
how q and U are increased by factor κ
PHY2054: Chapter 16 Capacitance
23
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.
‹ (1)
V2 < V1
‹ (2) V2 > V1
‹ (3) V2 = V1
C1
C2
Charge is unchanged, but dielectric
reduces E to E/κ and V to V/κ
PHY2054: Chapter 16 Capacitance
24
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
C1
C2
Energy is reduced from U
to U/κ, so work is negative
If U is total energy in capacitor
¾Positive work:
One “pushes in” dielectric
¾Negative work:
Capacitor “sucks in” dielectric
PHY2054: Chapter 16 Capacitance
ΔU > 0
ΔU < 0
25