Capacitors Phy 1161: PreLecture 06 Today’s lecture will cover Textbook Sections 20-5 – 20-6.
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Transcript of Capacitors Phy 1161: PreLecture 06 Today’s lecture will cover Textbook Sections 20-5 – 20-6.
Comparison:Electric Potential Energy vs. Electric Potential
VAB : the difference in electric potential between points B and A
UAB : the change in electric potential energy of a charge q when moved from A to B
UAB = q VAB
q
A B
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 (–) terminalPositive charges tend to go “downhill”, from + to -Negative charges go in the opposite direction, from - to +
UAB = q VAB
Important Special CaseUniform 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=4kQ/A
zero everywhere else
This result is independent of plate separation
This is call a parallel plate capacitor
Parallel Plate CapacitorPotential Difference
A B
E=E0
d
V = VA – VB = +E0 d
+
+
+
+
+
-
-
-
-
-
A B
d
+
+
+
+
+
-
-
-
-
-
+
+
+
+
+
-
-
-
-
-
E=
V = VA – VB = +2E0d
Potential difference is proportional to charge: Double Q Double V
E0 = 4πkQ/A
Charge Q on plates Charge 2Q on plates
Capacitance
• The ability to store separated charge
• Definition:
• Units: Farad (F) – named in honor of Michael Faraday– 1 F = 1C/V
QC
V
From Faraday’s notebook
Capacitor
• Any pair of conductors separated by a small distance. (e.g. two metal plates)
• Capacitor stores separated charge – Positive Q on one conductor, negative Q on other– Net charge is zero
E
d
+
+
+
+
+
-
-
-
-
-
Q = CV
U = (½) Q V • Stores Energy
Capacitance of Parallel Plate Capacitor
V = Ed AND E = Q/(0A) (Between two large plates)
So: V = Qd/ /(0A) Remember: CQ/V
So:
Equation based on geometry of capacitor
A
d
AE+
-
V
0= 8.85x10-12 C2/Nm2
If there is adielectric (κ>1) between plates C = κ C0
0 AC
d
Dielectric• Placing a dielectric between
the plates increases the capacitance.
C = C0
Capacitance with dielectric
Dielectric constant ( > 1)
Capacitance without dielectric
Material Constant Material ConstantVacuum 1 Germanium 16
Polyvinyl chloride
3.18 Strontium titanate
310
Mica 3 - 6 Water 80.4
Mylar 3.1 Glycerin 42.5
Neoprene 6.70 Benzene 2.284
Plexiglass 3.40 Glass 5 – 10
Polyethylene 2.25 Air (1 atm) 1
Liquid ammonia
(-78oC)
25 Titanium dioxide (rutile)
173 perp
86 para
Dielectrics
Voltage in Circuits• Elements are connected by wires. • Any connected region of wire has the same
potential.
Vwire 1= 0 V Vwire 2= 5 V Vwire 3= 12 V Vwire 4= 15 V
VC1= _____ V VC3
= _____ VVC2= _____ V
C1 C2 C3
• The potential difference across an element is the element’s “voltage.”
Voltage in Circuits• Elements are connected by wires. • Any connected region of wire has the same
potential.
Vwire 1= 0 V Vwire 2= 5 V Vwire 3= 12 V Vwire 4= 15 V
VC1= 5 V VC3
= 3 VVC2= 7 V
C1 C2 C3
• The potential difference across an element is the element’s “voltage.”
Capacitors in Parallel• Both ends connected together by wire
C1 C2 Ceq
• Share Charge: Qeq = Q1 + Q2
• Total Cap: Ceq = (Q1 + Q2)/V = C1 + C2
= Veq • Same voltage: V1 = V2
Capacitors in Parallel• Both ends connected together by wire
C1 C2 Ceq
15 V
10 V
15 V
10 V
15 V
10 V
• Share Charge: Qeq = Q1+ Q2
• Total Cap: Ceq = (Q1+ Q2)/V = C1+ C2
= Veq • Same voltage: V1 = V2
Capacitors in Series• Connected end-to-end with NO other exits
• Same Charge: Q1 = Q2 = Qeq
Ceq
C1
C2
++++
++++
++++
+
-+-
+
-+-
+
-
+Q
-Q +Q
-Q
+Q
-Q
21
111CCCeq
• Share Voltage: V1+V2=Veq