PHY2054: Chapter 16 Capacitance 1
Capacitance
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Capacitance: Parallel and Series Connection, Energy Storage, Exams of Physics
The concepts of capacitance in parallel and series connections, including the equations for combining capacitors in these configurations. It also discusses the energy storage capabilities of capacitors and provides examples for calculating charges, voltages, and energies. Concept tests for understanding the behavior of capacitors with dielectric materials.
Typology: Exams
Pre 2010
1 / 18
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PHY2054: Chapter 16 Capacitance
Capacitance
PHY2054: Chapter 16 Capacitance
Capacitors in Parallel
ÎV
= V 1
= V 2
(same potential top and bottom) 3
ÎTotal charge: q
tot^
= q
+ q 1
+ q 2
3
ÎC
V = Ceq
V + C 1
V + C 2
V 3
¾^ Basic law for combiningcapacitors in parallel ¾^ Works for N capacitors
eq^
C^
C^
C^
C
=^
+^
PHY2054: Chapter 16 Capacitance
ConcepTest
ÎTwo identical parallel plate capacitors are shown in anend-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 ^ (d) 0 ^ (e) Need more information
A^
B
Area is doubled
PHY2054: Chapter 16 Capacitance
ConcepTest
ÎEach capacitor is the same in the three configurations.Which configuration has the lowest equivalent capacitance?
^ (1) A ^ (2) B ^ (3) C ^ (4) They all have identical capacitance
C^
C C
C^
C
A^
B^
C
C/2 (series)
PHY2054: Chapter 16 Capacitance
Example: Find q
and Vi
on All Capacitorsi
ÎC
is charged in position A, then S is thrown to B position 1 Initial voltage across C
:^1
V^ = 12^0
^ Initial charge on C
:^1
q^10
= 12 x 4 = 48
μC
ÎAfter switch is thrown to B: V
= V 1
(parallel branches) 23
^ q
and q 2
in series: 3
q
= q 2
= q 3
(C 23
μF)
^ Charge conservation: q
= q 10
- q 1
23
^ 48 = C
V^11
+ C
V 23
(V 1
= V 1
^ Find V
:^1
V= 48 / (C^1
+ C 1
) = 8 V 23
^ Find q
:^1
q= C^1
V^11
μC
^ q
= 48 – 32 = q 23
= q 2
μC
^ V
= q 2
/ C 2
= 2.67 V 2
^ V
= q 3
/ C 3
= 5.33 V 3
4 μF
6 μF 3 μF
12V
A^
B
PHY2054: Chapter 16 Capacitance
Another Example
ÎEach capacitor has capacitance 10
μF. Find the total
capacitance ÎDo it in stages^
μF
^ Add 4
μF
^ Add 5
μF
^ Add 1
μF
PHY2054: Chapter 16 Capacitance
Energy in a Capacitor
ÎCapacitors have energy associated with them
^ Grab a charged capacitor with two hands and find out!
ÎCalculation of stored energy
^ Proof requires calculus derivation ^ Energy = work moving charge from – to + surface
ÎCapacitors store and release energy as they acquire andrelease charge
^ This energy is available to drive circuits
Q^2 U^
CV C =^
=
PHY2054: Chapter 16 Capacitance
Example of Capacitor Energy
ÎC = 5
μF, V = 200
(^
0.1J
U^
CV
=^
=^
×^
×^
×^
PHY2054: Chapter 16 Capacitance
Dielectric Materials and Capacitors
ÎInsulating material that can be polarized in E field ÎInduced charges at dielectric surface partially cancel E field
^ E
E /
κ^
κ^ > 1 is “dielectric constant”
^ V
V /
κ^
(since V = Ed)
^ C
→ κ
C^
(since C = Q / V)
ÎBut “good” dielectric requires more than high
κ^
value
^ Good insulator
(no charge leakage)
^ High breakdown voltage
(no arcing at high voltage)
^ Low cost
(affordable)
Dielectricmaterial
+++++++++++++++++++++++++ ------------------------------------------- +++++++++++++++++++++++++ -------------------------------------------
κ
Inducedcharges
PHY2054: Chapter 16 Capacitance
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
Similar Example
ÎSame capacitor as before, but this time insert dielectricwhile C is in the circuit
^ C
new^
=^ κ
C = 5490
μF
^ V is still 200 volts
ÎCalculate new q and U in this example ÎNotice how q and U are increased by factor
(^
(^) )(
)
new
new
2
2
6
1
1
new
new 2
2 5490
μF
200
1.1 C 5490
10
200
110 J
q^
C^
V
U^
C^
V^
−
=^
=^
×^
=
=^
=^
×^
=
PHY2054: Chapter 16 Capacitance
ConcepTest
ÎTwo identical capacitors are given the same charge Q,then disconnected from a battery.After C
has been charged and disconnected it is filled
with a dielectric. Compare the voltages of the twocapacitors.^
(1) V
< V 2
1
^ (2) V
> V 2
1
^ (3) V
= V 2
1
C^1
C^2
Charge is unchanged, but dielectricreduces E to E/
κ^ and V to V/
κ