1. Physics 2113 Lecture: 16 MON 23 FEB
Jonathan Dowling
Physics Lecture: 16 MON 23 FEB
Capacitance I

2. Capacitors and Capacitance
Capacitor: any two conductors, one with charge +Q, other with charge –Q –Q
Potential DIFFERENCE between conductors = V +Q
Uses: storing and releasing electric charge/energy.
Most electronic capacitors:
micro-Farads (μF),
pico-Farads (pF) — 10–12 F
New technology:
compact 1 F capacitors
Q = CV where C = capacitance
For fixed Q the larger the capacitance the larger the stored voltage.
Units of capacitance:
Farad (F) = Coulomb/Volt

3. Capacitance +Q –Q (We first focus on capacitors
Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors
e.g. Area of conductors, separation, whether the space in between is filled with air, plastic, etc. +Q –Q
(We first focus on capacitors
where gap is filled by AIR!)

4. Capacitance depends only on GEOMETRICAL factors and on the MATERIAL that separates the two conductors. It does not depend on the voltage across it or the charge on it at all.
Same
(b) Same

5. Capacitors Electrolytic (1940-70) Paper (1940-70) Electrolytic (new)
Variable
air, mica
Ceramic (1930 on)
Tantalum (1980 on)
Mica ( )

6. Parallel Plate Capacitor
We want capacitance: C = Q/V 
E field between conducting plates:
Area of each plate = A
Separation = d
charge/area = σ
= Q/A
Relate E to potential difference V: +Q –Q σ d
What is the capacitance C ?

7. Capacitance and Your iPhone!

8. Parallel Plate Capacitor — Example
A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm
What is the capacitance?
C = ε0A/d
= (8.85 x 10–12 F/m)(0.25 m2)/(0.001 m)
= 2.21 x 10–9 F
(Very Small!!)
Lesson: difficult to get large values
of capacitance without special
tricks!

10. Isolated Parallel Plate Capacitor: ICPP
A parallel plate capacitor of capacitance C is charged using a battery.
Charge = Q, potential voltage difference = V.
Battery is then disconnected.
If the plate separation is INCREASED, does the capacitance C:
(a) Increase?
(b) Remain the same?
(c) Decrease? +Q –Q
Q is fixed!
d increases!
C decreases (= ε0A/d)
V=Q/C; V increases.

11. Parallel Plate Capacitor & Battery: ICPP
A parallel plate capacitor of capacitance C is charged using a battery.
Charge = Q, potential difference = V.
Plate separation is INCREASED while battery remains connected. +Q –Q
V is fixed constant by battery!
C decreases (=ε0A/d)
Q=CV; Q decreases
E = σ/ε0 = Q/ε0A decreases
Does the Electric Field Inside:
(a) Increase?
(b) Remain the Same?
(c) Decrease?
Battery does work on capacitor to maintain constant V!

12. Spherical Capacitor Concentric spherical shells:
What is the electric field inside
the capacitor? (Gauss’ Law)
Radius of outer plate = b
Radius of inner plate = a
Concentric spherical shells:
Charge +Q on inner shell,
–Q on outer shell
Relate E to potential difference
between the plates:

13. Spherical Capacitor What is the capacitance? C = Q/V =
Radius of outer plate = b
Radius of inner plate = a
Concentric spherical shells:
Charge +Q on inner shell,
–Q on outer shell
Isolated sphere: let b >> a,

14. Cylindrical Capacitor
What is the electric field in between the plates? Gauss’ Law!
Radius of outer plate = b
Radius of inner plate = a
Length of capacitor = L
+Q on inner rod, –Q on outer shell
Relate E to potential difference
between the plates:
cylindrical Gaussian
surface of radius r

15. Summary Any two charged conductors form a capacitor.
Capacitance : C= Q/V
Simple Capacitors: Parallel plates: C = ε0 A/d Spherical: C = 4πε0 ab/(b-a) Cylindrical: C = 2πε0 L/ln(b/a)]

16. Parallel plates: C = ε0 A/d Spherical:
Cylindrical: C = 2πε0 L/ln(b/a)]
Hooked to battery
V is constant.
Q=VC
(a) d increases -> C decreases -> Q decreases
(b) a inc -> separation d=b-a dec. -> C inc. -> Q increases
(c) b increases -> separation d=b-a inc.-> C dec. -> Q decreases

17. 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
PAR-V (Parallel V the Same)
ΔV=V
Ceq
Qtotal
PARALLEL:
V is same for all capacitors
Total charge = sum of Q

18. 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

19. PARALLEL:
V is same for all capacitors
Total charge = sum of Q
SERIES:
Q is same for all capacitors
Total potential difference =
sum of V
Parallel: Voltage is same V on each but charge is q/2 on each since
q/2+q/2=q.
(b) Series: Charge is same q on each but voltage is V/2 on each since V/2+V/2=V.

20. “Compiling” 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

21. Example: Parallel or Series?
Parallel: Circuit Splits Cleanly in Two (Constant V)
What is the charge on each capacitor?
C1=10 μF
C3=30 μF
C2=20 μF
120V
Qi = CiV
V = 120V = Constant
Q1 = (10 μF)(120V) = 1200 μC
Q2 = (20 μF)(120V) = 2400 μC
Q3 = (30 μF)(120V) = 3600 μC
Note that:
Total charge (7200 μC) is shared between the 3 capacitors in the ratio C1:C2:C3 — i.e. 1:2:3

22. Example: Parallel or Series
Series: Isolated Islands (Constant Q)
What is the potential difference across each capacitor?
C1=10μF
C2=20μF
C3=30μF
Q = CserV
Q is same for all capacitors
Combined Cser is given by:
120V
Ceq = 5.46 μF (solve above equation)
Q = CeqV = (5.46 μF)(120V) = 655 μC
V1= Q/C1 = (655 μC)/(10 μF) = 65.5 V
V2= Q/C2 = (655 μC)/(20 μF) = V
V3= Q/C3 = (655 μC)/(30 μF) = 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)

23. Example: Series or Parallel?
Neither: Circuit Compilation Needed!
10μF
5μF
10V
In the circuit shown, what is the charge on the 10μF capacitor?
10 μF
10μF
10V
The two 5μF capacitors are in parallel
Replace by 10μF
Then, we have two 10μF capacitors in series
So, there is 5V across the 10 μF capacitor of interest
Hence, Q = (10μF )(5V) = 50μC