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Physics 2102
Jonathan Dowling
Physics 2102
Lecture: 12 MON FEB
Capacitance II
25.4–5
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V=Constant
• An ISOLATED wire is an
equipotential surface:
V=Constant
• Capacitors in parallel have
SAME potential difference but
NOT ALWAYS same charge!
V = VAB = VA –VB
Q1
A
C
• VAB = VCD = V
• Ceq = C1 + C2
VA
Q2
VB
C2
VC
VD
B
D
V = VCD = VC –VD
• Qtotal = Q1 + Q2
• CeqV = C1V + C2V
C1
Cparallel  C1  C2
• Equivalent parallel
capacitance = sum of
capacitances

Qtotal
V=V
Ceq
Capacitors in Series: Q=Constant
Isolated Wire:
Q=Q1=Q2=Constant
• Q1 = Q2 = Q = Constant
• VAC = VAB + VBC
Q
Q Q
 
Ceq C1 C2
1
Cseries
Q1
B
A
1
1
 
C1 C2
SERIES:
• Q is same for all capacitors
• Total potential difference = sum of V
Q2
C1
C
C2
Q = Q1 = Q2
Ceq
Capacitors in parallel and in
series
• In parallel :
Cpar = C1 + C2
Vpar = V1 = V2
Qpar = Q1 + Q2
Q1
C1
Qeq
Q2
C2
Ceq
• In series :
1/Cser = 1/C1 + 1/C2
Vser = V1 + V2
Qser= Q1 = Q2
Q1
Q2
C1
C2
Example: Parallel or Series?
Parallel: Circuit Splits Cleanly in Two (Constant V)
What is the charge on each capacitor?
• Qi = CiV
• V = 120V = Constant
• Q1 = (10 F)(120V) = 1200 C
• Q2 = (20 F)(120V) = 2400 C
• Q3 = (30 F)(120V) = 3600 C
Note that:
• Total charge (7200 mC) is shared
between the 3 capacitors in the ratio
C1:C2:C3 — i.e. 1:2:3
C1=10 F
C2=20 F
C3=30 F
120V
Cpar  C1  C2  C3  10  20  30F  60F
Example: Parallel or Series
Series: Isolated Islands (Constant Q)
What is the potential difference across each capacitor?
• Q = CserV
• Q is same for all capacitors
• Combined Cser is given by:
1
1
1
1



Cser (10F) (20F) (30F)
C1=10F
C2=20F
C3=30F
120V
• Ceq = 5.46 F (solve above equation)
• Q = CeqV = (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
Note: 120V is shared in the
ratio of INVERSE
capacitances i.e.
(1):(1/2):(1/3)
(largest C gets smallest V)
Example: Series or Parallel?
Neither: Circuit Compilation Needed!
In the circuit shown,
what is the charge on
the 10F capacitor?
• The two 5F capacitors are in
parallel
• Replace by 10F
• Then, we have two 10F
capacitors in series
• So, there is 5V across the 10 F
capacitor of interest
• Hence, Q = (10F )(5V) = 50C
5F
10F
5F
10V
10 F
10F
10V
Energy U Stored in a Capacitor
• 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
dq
• How much work was needed?
Q
Q
2
2
q
Q
CV
U   Vdq   dq 

C
2C
2
0
0
Energy Stored in Electric Field of Capacitor
• 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 =
2
Q
Q
0  Q  0E 2
Q



u



2


2CAd 2  0 A  Ad 2 0 A
2

A
2
0




2
2
2
 d 
volume = Ad
General
expression for
any region with
vacuum (or air)
Example
•
•
•
•
10F capacitor is initially charged to 120V.
20F capacitor is initially uncharged.
Switch is closed, equilibrium is reached.
How much energy is dissipated in the process?
10F (C1)
Initial charge on 10F = (10F)(120V)= 1200C
After switch is closed, let charges = Q1 and Q2.
20F (C2)
Charge is conserved: Q1 + Q2 = 1200C
• Q1 = 400C
Q2
Also, Vfinal is same: Q1  Q2
• Q2 = 800C
Q1 
C1 C 2
2
• Vfinal= Q1/C1 = 40 V
Initial energy stored = (1/2)C1Vinitial2 = (0.5)(10F)(120)2 = 72mJ
Final energy stored = (1/2)(C1 + C2)Vfinal2 = (0.5)(30F)(40)2 = 24mJ
Energy lost (dissipated) = 48mJ
Summary
• Any two charged conductors form a capacitor.
• Capacitance : C= Q/V
• Simple Capacitors:
Parallel plates: C = 0 A/d
Spherical : C = 4p 0 ab/(b-a)
Cylindrical: C = 2p 0 L/ln(b/a)
• Capacitors in series: same charge, not necessarily equal
potential; equivalent capacitance 1/Ceq=1/C1+1/C2+…
• Capacitors in parallel: same potential; not necessarily same
charge; equivalent capacitance Ceq=C1+C2+…
• Energy in a capacitor: U=Q2/2C=CV2/2; energy density u=0E2/2
• Capacitor with a dielectric: capacitance increases C’=kC