Transcript Ch. 29 slides

Chapter 29
Electric Potential and Field
Phys 133 – Chapter 29
1
Overview Force, field, energy, potential
Phys 133 – Chapter 29
2
Force, field, energy, potential
or
E =-
dV
dr
(
¶V ˆ ¶V ˆ ¶V ˆ
i+
j+
k)
¶x ¶y
¶z
Phys 133 – Chapter 29
3
Fields and potentials
1. a) E = 0 V/m in throughout some region of space,
can you conclude that the potential V = 0 in this
region?
b) V = 0 V throughout some region of space. Can you
conclude that the electric field E = 0 V/m in this region?
Phys 133 – Chapter 29
4
Graphically convert between E and V
2. The top graph shows Ex vs. x for an electric field parallel to the x-axis.
a) Draw the graph of V vs. x in this region of space. Let V = 0 at x = 0m.
Add an appropriate scale on the vertical axis. (Hint: integration is the area
under the curve)
b) Draw a contour map above the x-axis on a diagram like the one belowright and label your equipotential lines every 20 V. 0V has been drawn
already.
C) Draw several electric field vectors on top of the contour map.
Ex (V/m)
0V
40
20
0
2
4
1
x(m)
2
3
4
x(m)
Phys 133 – Chapter 29
5
Problem: Potential of sphere
--Find the electric potential everywhere for a sphere
(radius R) with charge (Q) uniformly distributed. Take
V=0 at infinity.
--Sketch V vs r and Er vs r.
From Chap 27 E field is:
ì Q
ïK R 3 r rˆ
E =í
ï K Q rˆ
î r2
r<R
R<r
Phys 133 – Chapter 29
6
Problem: Potential of sphere (ans)
Finding Vr
R<r
r
V∞
dr
For all r
For r < R
For R < r
Vr
é KQ ù
= -ê 3 r2 ú
ë 2R û R
KQ KQ
=
r
¥
=-
Defined V¥ = 0
KQ
r
KQ
(R < r)
r
∆V
r
é KQ ù
= - êë r úû ¥
r
Vr =
r<R
∆V
VR E
Vr Vr
definition
Vr - V¥ =
ì Q
ïK R 3 r rˆ
E =í
ï K Q rˆ
î r2
æ KQ 2 KQ 2ö
r - 3R
è 2R 3
ø
2R
KQ KQ 2
From R < r : VR =
= 3 R
R
R
Vr -
KQ
æ KQ 2 KQ 2ö
=r - 3R
è 2R 3
ø
R
2R
3KQ 2 KQ 2
V
=
R r (r < R)
Phys 133 -- Chapter
r
2R 3 30 2R 3
7
Problem: Potential of sphere (ans)
Find the electric potential everywhere for a sphere
(radius R) with charge (Q) uniformly distributed.
ì 3KQ KQ 2
ï 2R - 2R 3 r (r < R)
ï
V =í
ï
Q
(R < r)
K
ïî
r
ì Q
ïK R 3 r
Er = í
Q
ï K 2
î r
(r < R)
(R < r)
Phys 133 – Chapter 29
8
Problem: Field from Potential
Find the x,y and z components of the electric field,
given that the electric potential of a disk is given by
Vdisk
Q
=
2pR 2e0
(
z 2 + R2 - z
)
Phys 133 – Chapter 29
9
Problem: Field from Potential (Answer)
Find the z component of the electric field, given that
the electric potential of a disk is given by
Vdisk
Q
=
2pR 2e0
( Edisk )z
(
z 2 + R2 - z
)
ö
dV
h æ
z
==
1ç
2
2 ÷
dz 2e 0 è
z +R ø
Phys 133 – Chapter 29
10
Geometry of potential/field
 is perp to equipotential surfaces
points downhill (decreasing V)
--strength proportional to spacing equipotentials
Phys 133 – Chapter 29
11
Kirchhoff’s loop rule
(Conservation of energy)
Phys 133 – Chapter 29
12
Conductor in equilibrium: field and potential
--field is zero inside conductor
--field is perpendicular at surface
--conductor is at equipotential (no work to move)
Phys 133 – Chapter 29
13
Conductor in equilibrium: equipotentials
--equipotentials are parallel
to nearby conductor
Phys 133 – Chapter 29
14
Do Workbook 30.4, 6, 11, & 12a
Phys 133 – Chapter 29
15
Problem: Finding Potential
--Find the electric potential everywhere for a
point charge (q) at the center of a hollow metal
sphere (inner radius a, outer radius b) with
charge Q. (Take V=0 at infinity.)
a
--Sketch V vs r and Er vs r.
b
ì
Kq
, r<a
ï
2
r
ï
E r (r) = í
0, a < r < b
ï K(Q + q)
ï
, b<r
î r2
Phys 133 – Chapter 29
16
Problem: Finding Potential (ans)
Finding Vr
∆V
Va
For all r
é Kq ù
= - ê- ú
ë r ûa
r
K(Q + q) K(Q + q)
=
r
¥
K(Q + q)
Vr - V¥ =
r
K(Q + q)
(R < r)
r
E
Kq Kq
r
a
K(Q + q)
From before Va = Vb =
b
∆V
For a < r < b
=
Vr -
K(Q + q) Kq Kq
=
b
r
a
Vr =
K(Q + q) æ Kq Kqö
+
(r < a)
è r
b
aø
Phys 133 -- Chapter 30
r
Vr dr V∞
r
é K(Q + q) ù
= - êúû
r
ë
¥
Defined V¥ = 0
Vr
For r < a
For b < r
Vr =
ì
Kq
, r<a
ï
2
r
ï
E r (r) = í
0, a < r < b
ï K(Q + q)
ï
, b<r
î r2
from b < r; Vb =
Vr =
K(Q + q)
b
K(Q + q)
(a < r < b)
b
17
Problem: finding Potential (Answer)
ì
q+Q
K
ï
r
ïï
q+Q
V =í
K
b
ï
ïK q + Q + Kqæ 1 - 1 ö
ïî
è r aø
b
b< r
a< r<b
r<a
ì
Kq
, r<a
ï
2
r
ï
E r (r) = í
0, a < r < b
ï K(Q + q)
ï
, b<r
î r2
not origin
Phys 133 -- Chapter 30
18
Sources of potential: Capacitor
-charge separation
-not sustained
Phys 133 -- Chapter 30
19
Sources of potential: Battery
--sustained
--
Phys 133 -- Chapter 30
20
Capacitors
charge separation on capacitor
Apply to capacitor
E capcitor
Q
=(constant)
e0 A
Q
=(
)d
e0 A
where C =
Phys 133 -- Chapter 30
e0 A
d
21
Circuit geometry
--generic circuit elements
--imagine battery connection
1
2
1
2
parallel
series
Phys 133 -- Chapter 30
22
Capacitors: series equivalent
Q1 = Q2 = Qequivalent
Q1 Q2 Qequivalent
+
=
C1 C2 Cequivalent
1
1
1
+
=
C1 C2 Cequivalent
Phys 133 -- Chapter 30
23
Capacitors: parallel equivalent
Q1 + Q2 = Qequivalent
C1 + C2 = Cequivalent
Phys 133 -- Chapter 30
24
Capacitors
1
1
2
2
series
parallel
V
V1+V2= Veq
V1=V2
Q
Q1=Q2
Q1+Q2= Qeq
1
1
1
= +
+…
Ceq C1 C2
Ceq = C1 + C2 +…
Phys 133 -- Chapter 30
25
Do Workbook 30.26 & 27
Phys 133 -- Chapter 30
26
Energy in capacitor
--work done charge separation
--or in electric field
Phys 133 -- Chapter 30
27
Problem
--Find the charge on (Q) and potential difference (V)
across each capacitor.
--What is the total energy stored in the system?
5 mF
2 mF
8 mF
12 V
2 mF
Phys 133 -- Chapter 30
28
Cequ = C2 + C8 (parallel)
Problem (ans)
= 2mF +8mF
5 mF
2 mF
8 mF
=10mF
12 V
2 mF
5 mF
1
1
1
1
=
+
+
(series)
Cequ C5 C10 C2
10 µF
12 V
2 mF
1
1
1
=
+
+
5mF 10mF 2mF
8
=
10mF
5mF
Þ Cequ =
4
5/4 µF
12 V
Phys 133 -- Chapter 30
29
Problem (ans)
5/4 µF
12 V
Q5 = Q10 = Q2 = Qtot = 15mC (in series)
5 mF
10 µF
12 V
2 mF
5 mF
2 mF
8 mF
12 V
2 mF
Phys 133 -- Chapter 30
30
Dielectrics change the potential difference
• The potential between to
parallel plates of a capacitor
changes when the material
between the plates changes.
It does not matter if the
plates are rolled into a tube
as they are in Figure 24.13 or
if they are flat as shown in
Figure 24.14.
C
K=
C0
V0
V=
K
E0
E=
K
Table 24.1—Dielectric constants
Field lines as dielectrics change
• Moving from part (a)
to part (b) of Figure
24.15 shows the change
induced by the
dielectric.
In vacuum, energy density is
1
u = e0 E 2
2
In dielectric
e = Ke 0
C = Ke0
A
A
=e
d
d
1 2
u = eE
2
Examples to consider, capacitors with and without dielectrics
• If capacitor is disconnected from circuit, inserting a
dielectric changes decreases electric field, potential and
increases capacitance, but the amount of charge on the
capacitor is unchanged.
• If the capacitor is hooked up to a power supply with
constant voltage, the voltage must remain the same, but
capacitance and charge increase
Q24.8
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the charges on the
plates remain constant.
What effect does adding the dielectric have on the potential
difference between the capacitor plates?
A. The potential difference increases.
B. The potential difference remains the same.
C. The potential difference decreases.
D. not enough information given to decide
A24.8
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the charges on the
plates remain constant.
What effect does adding the dielectric have on the potential
difference between the capacitor plates?
A. The potential difference increases.
B. The potential difference remains the same.
C. The potential difference decreases.
D. not enough information given to decide
Q24.9
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the charges on the
plates remain constant.
What effect does adding the dielectric have on the energy
stored in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide
A24.9
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the charges on the
plates remain constant.
What effect does adding the dielectric have on the energy
stored in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide
Q24.10
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the potential
difference between the plates remains constant.
What effect does adding the dielectric have on the amount
of charge on each of the capacitor plates?
A. The amount of charge increases.
B. The amount of charge remains the same.
C. The amount of charge decreases.
D. not enough information given to decide
A24.10
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the potential
difference between the plates remains constant.
What effect does adding the dielectric have on the amount
of charge on each of the capacitor plates?
A. The amount of charge increases.
B. The amount of charge remains the same.
C. The amount of charge decreases.
D. not enough information given to decide
Q24.11
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the potential
difference between the plates remains constant.
What effect does adding the dielectric have on the energy
stored in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide
A24.11
You slide a slab of dielectric between the plates of a
parallel-plate capacitor. As you do this, the potential
difference between the plates remains constant.
What effect does adding the dielectric have on the energy
stored in the capacitor?
A. The stored energy increases.
B. The stored energy remains the same.
C. The stored energy decreases.
D. not enough information given to decide