1. Physics 2102 Lecture: 12 MON FEB
Jonathan Dowling
Physics Lecture: 12 MON FEB
Capacitance II
25.4–5

2. Capacitors in Parallel: V=Constant
An ISOLATED wire is an equipotential surface: V=Constant
Capacitors in parallel have SAME potential difference but NOT ALWAYS same charge!
VAB = VCD = V
Qtotal = Q1 + Q2
CeqV = C1V + C2V
Ceq = C1 + C2
Equivalent parallel capacitance = sum of capacitances
V = VAB = VA –VB
V = VCD = VC –VD VA VB VC VD A B C D C1 C2 Q1 Q2
V=V
Ceq
Qtotal

3. Capacitors in Series: Q=Constant
Isolated Wire:
Q=Q1=Q2=Constant
Q1 = Q2 = Q = Constant
VAC = VAB + VBC A B C C1 C2 Q1 Q2
Q = Q1 = Q2
SERIES:
Q is same for all capacitors
Total potential difference = sum of V
Ceq

4. Capacitors in parallel and in series
Cpar = C1 + C2
Vpar = V1 = V2
Qpar = Q1 + Q2 C1 C2 Q1 Q2
Ceq
Qeq
In series :
1/Cser = 1/C1 + 1/C2
Vser = V1  + V2
Qser= Q1 = Q2 C1 C2 Q1 Q2

5. Example: Parallel or Series?
Parallel: Circuit Splits Cleanly in Two (Constant V)
What is the charge on each capacitor?
C1=10 F
C3=30 F
C2=20 F
120V
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

6. Example: Parallel or Series
Series: Isolated Islands (Constant Q)
What is the potential difference across each capacitor?
C1=10mF
C2=20mF
C3=30mF
Q = CserV
Q is same for all capacitors
Combined Cser is given by:
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) = 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)

7. Example: Series or Parallel?
Neither: Circuit Compilation Needed!
10F
5F
10V
In the circuit shown, what is the charge on the 10F capacitor?
10 F
10F
10V
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

8. 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
How much work was needed? dq

9. 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 =
volume = Ad
General expression for any region with vacuum (or air)

10. Example 10mF (C1) Initial charge on 10mF = (10mF)(120V)= 1200mC
10mF capacitor is initially charged to 120V.
20mF capacitor is initially uncharged.
Switch is closed, equilibrium is reached.
How much energy is dissipated in the process?
10mF (C1)
20mF (C2)
Initial charge on 10mF = (10mF)(120V)= 1200mC
After switch is closed, let charges = Q1 and Q2.
Charge is conserved: Q1 + Q2 = 1200mC
Q1 = 400mC
Q2 = 800mC
Vfinal= Q1/C1 = 40 V
Also, Vfinal is same:
Initial energy stored = (1/2)C1Vinitial2 = (0.5)(10mF)(120)2 = 72mJ
Final energy stored = (1/2)(C1 + C2)Vfinal2 = (0.5)(30mF)(40)2 = 24mJ
Energy lost (dissipated) = 48mJ

11. Summary Any two charged conductors form a capacitor.
Capacitance : C= Q/V
Simple Capacitors: Parallel plates: C = e0 A/d Spherical : C = 4p e0 ab/(b-a) Cylindrical: C = 2p e0 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=e0E2/2
Capacitor with a dielectric: capacitance increases C’=kC