Class 07: Outline
Hour 1:
Conductors & Insulators
Expt. 2: Electrostatic Force
Hour 2:
Capacitors
P07 - 1
Last Time:
Gauss’s Law
P07 - 2
Gauss’s Law
ΦE =
w
∫∫
G G Qenc
E ⋅ dA =
closed
surface S
ε0
In practice, use symmetry:
• Spherical (r)
• Cylindrical (r, A)
• Planar (Pillbox, A)
P07 - 3
Conductors
P07 - 4
Conductors and Insulators
A conductor contains charges that are free to
move (electrons are weakly bound to atoms)
Example: metals
An insulator contains charges that are NOT free to
move (electrons are strongly bound to atoms)
Examples: plastic, paper, wood
P07 - 5
Conductors
Conductors have free charges
Æ E must be zero inside the conductor
Æ Conductors are equipotential objects
E
-
Neutral
Conductor
+
+
+
+
P07 - 6
Equipotentials
P07 - 7
Topographic Maps
P07 - 8
Equipotential Curves
All points on equipotential curve are at same potential.
Each curve represented by V(x,y) = constant
P07 - 9
PRS Question:
Walking down a mountain
P07 -10
Direction of Electric Field E
E is perpendicular to all equipotentials
Constant E field
Point Charge
Electric dipole
P07 -11
Properties of Equipotentials
• E field lines point from high to low potential
• E field lines perpendicular to equipotentials
• Have no component along equipotential
• No work to move along equipotential
P07 -12
Conductors in Equilibrium
Conductors are equipotential objects:
1) E = 0 inside
2) Net charge inside is 0
3) E perpendicular to surface
4) Excess charge on surface
E =σ
ε0
P07 -13
Conductors in Equilibrium
Put a net positive charge anywhere inside a
conductor, and it will move to the surface
to get as far away as possible from the
other charges of like sign.
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/34-pentagon/34pentagon320.html
P07 -14
Expt. 2: Electrostatic Force
P07 -15
Expt. 2: Electrostatic Force
http://ocw.mit.edu/ans7870/8/8.02T/f04/visualizations/electrostatics/36-electrostaticforce/36-esforce320.html
P07 -16
Experiment 2:
Electrostatic Force
P07 -17
Capacitors and Capacitance
P07 -
Capacitors: Store Electric Energy
Capacitor: two isolated conductors with equal and
opposite charges Q and potential difference ∆V
between them.
Q
C=
∆V
Units: Coulombs/Volt or
Farads
P07 -
Parallel Plate Capacitor
E =0
+Q = σ A
E =?
d
E =0
−Q = −σ A
P07 -20
Parallel Plate Capacitor
When you put opposite charges on plates, charges
move to the inner surfaces of the plates to get as
close as possible to charges of the opposite sign
http://ocw.mit.edu/ans7870/8/8.02T/f04/visuali
zations/electrostatics/35-capacitor/35capacitor320.html
P07 -21
Calculating E (Gauss’s Law)
G G qin
w
∫∫ E ⋅ dA =
S
ε0
σ AGauss
E ( AGauss ) =
ε0
σ
Q
E= =
ε 0 Aε 0
Note: We only “consider” a single sheet! Doesn’t
the other sheet matter?
P07 -22
Alternate Calculation Method
Top Sheet:
σ
E=−
2ε 0
++++++++++++++
σ
E=
2ε 0
Bottom Sheet:
σ
E=
2ε 0
- - - - - - - -- - - - - -
σ
E=−
2ε 0
σ
σ
σ
Q
E=
+
= =
2ε 0 2ε 0 ε 0 Aε 0
P07 -23
Parallel Plate Capacitor
G G
Q
d
∆V = − ∫ E ⋅ dS = Ed =
Aε 0
bottom
top
ε0 A
Q
C=
=
∆V
d
C depends only on geometric factors A and d
P07 -24
Demonstration:
Big Capacitor
P07 -25
Spherical Capacitor
Two concentric spherical shells of radii a and b
What is E?
Gauss’s Law Æ E ≠ 0 only for a < r < b,
where it looks like a point charge:
G
E=
Q
4πε 0 r
ˆ
r
2
P07 -26
Spherical Capacitor
b
G G
Qrˆ
∆V = − ∫ E ⋅ dS = − ∫
⋅ dr rˆ
2
4πε 0 r
inside
a
outside
Q ⎛1 1⎞
=
⎜ − ⎟
4πε 0 ⎝ b a ⎠
Is this positive or negative? Why?
4πε 0
Q
= −1 −1
C=
∆V
a −b
(
)
For an isolated spherical conductor of radius a:
C = 4πε 0 a
P07 -27
Capacitance of Earth
For an isolated spherical conductor of radius a:
C = 4πε 0 a
ε 0 = 8.85 ×10
−12
Fm
a = 6.4 × 10 m
6
−4
C = 7 × 10 F = 0.7 mF
A Farad is REALLY BIG! We usually use pF (10-12) or nF (10-9)
P07 -28
1 Farad Capacitor
How much charge?
Q = C ∆V
= ( 1 F )( 1 2 V )
= 12C
P07 -29
PRS Question:
Changing C Dimensions
P07 -30
Demonstration:
Changing C Dimensions
P07 -31
Energy Stored in Capacitor
P07 -32
Energy To Charge Capacitor
+q
-q
1. Capacitor starts uncharged.
2. Carry +dq from bottom to top.
Now top has charge q = +dq, bottom -dq
3. Repeat
4. Finish when top has charge q = +Q, bottom -Q
P07 -33
Work Done Charging Capacitor
At some point top plate has +q, bottom has –q
Potential difference is ∆V = q / C
Work done lifting another dq is dW = dq ∆V
+q
-q
P07 -34
Work Done Charging Capacitor
So work done to move dq is:
q 1
dW = dq ∆V = dq = q dq
C C
Total energy to charge to q = Q:
Q
1
W = ∫ dW = ∫ q dq
C0
+q
2
1Q
=
C 2
-q
P07 -35
Energy Stored in Capacitor
Q
Since C =
∆V
2
Q
1
1
U=
= Q ∆V = C ∆V
2C 2
2
2
Where is the energy stored???
P07 -36
Energy Stored in Capacitor
Energy stored in the E field!
Parallel-plate capacitor:
1 εo A
1
2
U = CV =
2 d
2
( Ed )
2
=
C=
εo E 2
2
εo A
d
and V = Ed
× ( Ad ) = uE × (volume)
uE = E field energy density =
εoE
2
2
P07 -37
1 Farad Capacitor - Energy
How much energy?
1
2
U = C ∆V
2
1
2
= ( 1 F )( 1 2 V )
2
= 72 J
Compare to capacitor charged to 3kV:
1
1
2
2
U = C ∆ V = ( 1 0 0 µ F )( 3 k V )
2
2
2
1
3
−4
1× 1 0 F 3 × 1 0 V = 4 5 0 J
=
2
(
)(
)
P07 -38
PRS Question:
Changing C Dimensions
Energy Stored
P07 -39
Demonstration:
Dissectible Capacitor
P07 -40