Phy 1161: PreLecture 06
Capacitors
• Today’s lecture will cover Textbook
Sections 20-5 – 20-6
Comparison:
Electric Potential Energy vs. Electric Potential
q
A
B
DVAB : the difference in electric potential between
points B and A
DUAB : the change in electric potential energy of a
charge q when moved from A to B
DUAB = q DVAB
Electric Potential: Summary
• E field lines point from higher to lower potential
• For positive charges, going from higher to lower
potential is “downhill”
• For a battery, the (+) terminal is at a higher potential
than the (–) terminal
Positive charges tend to go “downhill”, from + to Negative charges go in the opposite direction, from
- to +
DUAB = q DVAB
Important Special Case
Uniform Electric Field
+
-
+
-
+
-
+
-
+
-
Two large parallel conducting plates of area A
+Q on one plate
-Q on other plate
Then E is
uniform between the two plates: E=4kQ/A
zero everywhere else
This result is independent of plate separation
This is call a parallel plate capacitor
Parallel Plate Capacitor
Potential Difference
Charge Q on plates
V = VA – VB = +E0 d
E=E0
+
-
+
+A
B
-
+
-
+
d
Charge 2Q on plates
V = VA – VB = +2E0d
+ E=
+
+
+
+
+A
+
+
+
+
d
B -
Potential difference is proportional to charge: Double Q  Double V
E0 = 4πkQ/A
Capacitance
• The ability to store separated charge
C 
• Definition:
Q
V
• Units: Farad (F) – named in honor of Michael
Faraday
– 1 F = 1C/V
From Faraday’s notebook
Capacitor
• Any pair of conductors separated by a small
distance. (e.g. two metal plates)
• Capacitor stores separated charge Q = CV
– Positive Q on one conductor, negative Q on
other
+
+
– Net charge is zero
-
E
-
• Stores Energy U = (½) Q V
+
-
+
-
+
d
Capacitance of
Parallel Plate Capacitor
V = Ed AND E = Q/(e0A)
V
(Between two large plates)
So: V = Qd/ /(e0A)
Remember: CQ/V
So:
C 
+E
e0A
A
d
Equation based on
geometry of capacitor
If there is adielectric (κ>1)
between plates C = κ C0
d
e0= 8.85x10-12 C2/Nm2
-A
Dielectric
• Placing a dielectric between
the plates increases the
capacitance.
Dielectric
constant (k > 1)
C = k C0
Capacitance with
dielectric
Capacitance
without dielectric
Dielectrics
Material
Constant
Material
Constant
Vacuum
Polyvinyl
chloride
Mica
Mylar
Neoprene
Plexiglass
Polyethylene
Liquid
ammonia
(-78oC)
1
3.18
Germanium
Strontium
titanate
Water
Glycerin
Benzene
Glass
Air (1 atm)
Titanium
dioxide
(rutile)
16
310
3-6
3.1
6.70
3.40
2.25
25
80.4
42.5
2.284
5 – 10
1
173 perp
86 para
Voltage in Circuits
• Elements are connected by wires.
• Any connected region of wire has the same
potential.
• The potential difference across an element
is the element’s “voltage.”
Vwire 1= 0 V
C1
VC1= _____ V
Vwire 2= 5 V
Vwire 3= 12 V
C2
VC2= _____ V
Vwire 4= 15 V
C3
VC3= _____ V
Voltage in Circuits
• Elements are connected by wires.
• Any connected region of wire has the same
potential.
• The potential difference across an element
is the element’s “voltage.”
Vwire 1= 0 V
Vwire 2= 5 V
Vwire 3= 12 V
C1
C2
VC1= 5 V
VC2= 7 V
Vwire 4= 15 V
C3
VC3= 3 V
Capacitors in Parallel
•
•
•
•
Both ends connected together by wire
Same voltage: V1 = V2 = Veq
Share Charge: Qeq = Q1 + Q2
Total Cap: Ceq = (Q1 + Q2)/V = C1 + C2
C1
C2
Ceq
Capacitors in Parallel
•
•
•
•
Both ends connected together by wire
Same voltage: V1 = V2 = Veq
Share Charge: Qeq = Q1+ Q2
Total Cap: Ceq = (Q1+ Q2)/V = C1+ C2
15 V
15 V
C1
C2
10 V
10 V
15 V
Ceq
10 V
Capacitors in Series
• Connected end-to-end with NO other exits
• Same Charge: Q1 = Q2 = Qeq
• Share Voltage: V1+V2=Veq
1
C eq
+Q
-Q
+Q
-Q
+
+
+
++
--++
+
+
++
- +
-+
+
+
+
C1

1
C1

1
C2
+Q
+
Ceq
C2
-Q
-
Electromotive Force
• Battery
–
–
–
–
+
Maintains potential difference V
Not constant power
Not constant current
Does NOT produce or supply charges, just
“pushes” them.