1. Physics 2102 Lecture: 08 THU 18 FEB
Jonathan Dowling
Physics Lecture: 08 THU 18 FEB
Capacitance II
25.4–7

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
PAR-V (Parallel: V the Same)

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
SERI-Q: Series Q the Same
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 on ALL Capacitors (PAR-V)
Q1 = (10 F)(120V) = 1200 C
Q2 = (20 F)(120V) = 2400 C
Q3 = (30 F)(120V) = 3600 C
Note that:
Total charge (7200 C) 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 (SERI-Q)
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 by symmetry
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. Dielectric Constant C =  A/d DIELECTRIC +Q –Q
If the space between capacitor plates is filled by a dielectric, the capacitance INCREASES by a factor 
This is a useful, working definition for dielectric constant.
Typical values of are 10–200 +Q –Q
DIELECTRIC
C =  A/d
The  and the constant
o are both called dielectric constants. The  has no units (dimensionless).

11. Atomic View
Emol
Molecules set up counter E field Emol that somewhat cancels out capacitor field Ecap.
This avoids sparking (dielectric breakdown) by keeping field inside dielectric small.
Hence the bigger the dielectric constant the more charge you can store on the capacitor.
Ecap

12. Example dielectric slab Capacitor has charge Q, voltage V
Battery remains connected while dielectric slab is inserted.
Do the following increase, decrease or stay the same:
Potential difference?
Capacitance?
Charge?
Electric field?
dielectric
slab

13. Example dielectric slab
Initial values: capacitance = C; charge = Q; potential difference = V; electric field = E;
Battery remains connected
V is FIXED; Vnew = V (same)
Cnew = kC (increases)
Qnew = (kC)V = kQ (increases).
Since Vnew = V, Enew = V/d=E (same)
dielectric
slab
Energy stored? u=e0E2/2 => u=ke0E2/2 = eE2/2

14. Summary Any two charged conductors form a capacitor.
Capacitance : C= Q/V
Simple Capacitors: Parallel plates: C = e0 A/d Spherical : C = 4 e0 ab/(b-a) Cylindrical: C = 2 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=0E2/2
Capacitor with a dielectric: capacitance increases C’=kC