Physics 202, Lecture 8Today’s Topics
Capacitance Review Energy storage in capacitors Dielectric materials, electric dipoles Dielectrics and Capacitance
Reminder: Midterm I, Oct 5, 5:30-7:00pm
Exam 1 Reminder
Exam time: Wednesday, October 5, 5:30-7PMCovering: Chapters 21-24, lecture material, lab material(E1 & EC2), and homework.
About 20 Multiple Choice Questions
Bring: Pen/pencil Calculator (no programming functionality!)
1 single-sided formula sheet, self-prepared --no photocopying!
Exam room assignment, special arrangements, otherlogistics, etc will be spelled out in a forthcoming email,which contains also a practice exam.
Capacitors: Summary
• Definition:
• Capacitance depends on geometry:
d
A
- - - - -+ + + +
Parallel Plates
a
b L
r
+Q
-Q
Cylindrical
a
b
+Q-Q
Spherical
C has units of “Farads” or F (1F = 1C/V) ε o has units of F/m
C ≡QΔV
C =εoAd
C = 4πεoabb − a
C =2πεoL
ln ba
⎛⎝⎜
⎞⎠⎟
Capacitors in Series and Parallel
Parallel:
Series:
ΔV
C2
C1
ΔV
Q=Q1+Q2
Cp ΔV=Q
ΔV1 = ΔV2 = ΔV
ΔV1 ΔV2ΔV=ΔV1+ΔV2
Q
ΔV
C1 C2CSΔV=Q
Q1 = Q2 = Q
Cp = C1 + C2
Cs =1C1
+1C2
⎛⎝⎜
⎞⎠⎟
−1
ΔV
Energy of a Capacitor• How much energy is stored in a charged capacitor?
– Calculate the work provided (usually by a battery) to charge acapacitor to +/- Q:
Incremental work dW needed to add charge dq to capacitor at voltage V:- +
• In terms of the voltage V:
• The total work W to charge to Q (W = potential energy of the chargedcapactior) is given by:
Two ways to write W
dW = V (q)idq = q
C⎛⎝⎜
⎞⎠⎟idq
W =1C
qdq =0
Q
∫12Q2
C
W =12CV 2
Capacitor Variables
• In terms of the voltage V:
• The total work to charge capacitor to Q equals the energy Ustored in the capacitor:
You can do one of two things to a capacitor :
• hook it up to a battery specify V and Q follows
• put some charge on it specify Q and V follows
U =
12
CV 2
U =
1C
qdq =0
Q
∫12
Q2
C
Q = CV
V =QC
Example (I)
d
A
- - - - -+ + + +• Suppose the capacitor shown here is charged
to Q. The battery is then disconnected.
• Now suppose the plates are pulled further apart to a finalseparation d1.
• How do the quantities Q, C, E, V, U change?
• Q:• C:• E:• V:• U:
remains the same.. no way for charge to leave.
increases.. add energy to system by separating
decreases.. capacitance depends on geometry
increases.. since C ↓, but Q remains same (or d ↑ but E the same)remains the same... depends only on charge density
C1 =dd1C V1 =
d1dV
U1 =
d1
dU
• Suppose the battery (V) is keptattached to the capacitor.
• Again pull the plates apart from d to d1.
• Now what changes?
• C:• V:• Q:• E:• U:
decreases (capacitance depends only on geometry)must stay the same - the battery forces it to be Vmust decrease, Q=CV charge flows off the plate
d
A
- - - - -+ + + +V
U1 =
dd1
U C1 =
dd1
C E1 =
dd1
E
must decrease ( , ) E =
σE0
E =VD
must decrease ( ) U = 1
2 CV 2
Example (II)
Where is the Energy stored?• Claim: energy is stored in the electric field itself.
• The electric field is given by:
⇒• The energy density u in the field is given by:
Units:
• Consider the example of a constant field generated by a parallelplate capacitor:
++++++++ +++++++
- - - - - - - - - - - - - -- Q
+Q
u =
Uvolume
=UAd
=12ε0 E2
E =σε0
=Qε0A
U =12ε0 E2 Ad
U =
12
Q2
C=
12
Q2
( Aε0 / d)
Jm3
Dielectrics• Empirical observation:
Inserting a non-conducting material (dielectric) between theplates of a capacitor changes the VALUE of the capacitance.
• Definition:The dielectric constant κ of a material is the ratio of thecapacitance when filled with the dielectric to that without it:
κ values are always > 1 (e.g., glass = 5.6; water = 80)Dielectrics INCREASE the capacitance of a capacitorMore energy can be stored on a capacitor at fixed voltage:
κ =CC0
′U =
CV 2
2=κC0V
2
2=κU
ε ≡ κε0permittivity:
Dielectric MaterialsDielectrics are electric insulators:
Charges are not freely movable, but can still have smalldisplacements in an external electric field
Atomic view: composed of permanent (or induced)electric dipoles:
- -
++
Permanent dipole
O--
H+ H+
P
+ - + -
+ - + -
Induced dipole E>0
E=0
+q-qp (dipole moment vector)(see Chapter 21)
Electric Dipole in External E FieldElectric dipole moment p.
Dipole in constant E field:
+
-
d-q
+q
F∑ = 0
τ = p ×
E
U = − piE
Net Force
Net Torque
Potential energy
p = q
d
dipole tends to become aligned with external field!
Dielectrics In External FieldAlignment of permanent dipoles in external field (or alignment
of non-permanent dipoles)
Note: induced field always opposite to the external field E0
Applyingexternal E field
Zeroexternal field Equilibrium
E =E0 −
Eind =
E0κ
E0 =
σε0
Eind =
σ ind
ε0σ ind = σ κ −1
κ
+++++++++++++++
Parallel Plate Example (I)• Deposit a charge Q on parallel plates
filled with vacuum (air)—capacitance C0
• Disconnect from battery• The potential difference is V0 = Q / C0.
Now insert material with dielectric constant κ .
Charge Q remains constantCapacitance increases C = κ C0
Voltage decreases from V0 to:
Q+++++++++++++++
- - - - - - - - - - - - - - -
E0V0
Electric field decreases also:
- - - - - - - - - - - - - - -
Q
V E+- +
-
+- +
-
+-
+-+-
κ
Note: The field onlydecreases when thecharge is held constant!
V =QC
=Q
κC0
=Vκ
E =Vd=V0dκ
=E0κ
++++++++++++++++++
Parallel Plate Example (II)• Deposit a charge Q0 on parallel plates
filled with vacuum (air)—capacitance C0
• The potential difference is V = Q / C0.• Leave battery connected.
Now insert material with dielectric constant κ .
Voltage V remains constantCapacitance increases C = κ C0
Charge increases from Q0 to:
Q0+++++++++++++++
- - - - - - - - - - - - - - -
E0V
Net Electric field stays same (V, d constant)
- - - - - - - - - - - - - - - - - -
Q
V E+- +
-
+- +
-
+-
+-+-
κ
Q = CV =κC0V =κQ0