jan25.pdf

1
19.1 Electrical Potential Energy
r
Special Case 2
EPE of test charge q2 in the
electric field of source
charge q1:
Electric field created by point
charge q1:
(diagram drawn assuming q1>0)
(1) Direction: pointing directly
away from q1 at any point.
(2) Magnitude given by
Coulomb’s Law
(1)
)
( 2
1
1
r
q
k
r
E =
(2)
)
(
EPE 2
1
21
r
q
q
k
r =
***Derivation of (2) from (1) requires integral calculus
This formula works
even if one or both
charges are negative
***
“(r)” means “function
of distance r”

Example: An electron (q=−e) is sent, from very far away, towards a
charged Q = −0.25 nC. It gets to a distance of 1.10 mm from the
center of the sphere before turning around.
What is the initial speed of the electron?
Q = −0.25 nC electron
?
=
∞
=
i
i
v
r
0
m
10
1
.
1 3
=
×
= −
f
f
v
r
e=1.6x10-19C, me=9.11x10-31 kg,
k=8.99x109 Nm2/C2

Example: An electron (q=−e) is sent, from very far away, towards a
charged Q = −0.25 nC. It gets to a distance of 1.10 mm from the
center of the sphere before turning around.
What is the initial speed of the electron?
Q = −0.25 nC electron
?
=
∞
=
i
i
v
r
0
m
10
1
.
1 3
=
×
= −
f
f
v
r
i
e
i
f
e
f
i
i
f
f
i
f
r
kQq
mv
r
kQq
mv
E
E
+
=
+
+
=
+
→
=
2
2
1
2
2
1
EPE
KE
EPE
KE
2
2
1
2
2
1
2
2
1
)
(
)
(
)
(
)
0
( i
f
i
f
mv
r
e
kQ
e
kQ
mv
r
e
kQ
m =
−
→
∞
−
+
=
−
+
( )( )
( )( )
2
2
14
3
31
19
10
2
2
9
2
s
m
10
18
.
7
m
10
1
.
1
kg
10
11
.
9
C
10
6
.
1
C
10
5
.
2
C
Nm
10
99
.
8
2
)
(
2
×
=
×
×
×
−
×
−








×
=
−
=
⇒ −
−
−
−
f
i
mr
e
kQ
v
s
m
vi
7
10
68
.
2 ×
=
⇒
e=1.6x10-19C, me=9.11x10-31 kg,
k=8.99x109 Nm2/C2

The overall potential energy is the amount of work required of an external agent to
assemble this set of charges—this can be done in different ways, for example:
(a) Bring in q1 from ∞ (no cost), W1=0
(b) Bring in q2 from ∞: W2=U12
(c) Bring in q3 from ∞: W3=U13+U23
Total W1+ W2+ W2  U=U12+U13+U23
Can do this in any order and obtain the SAME result
4
Total Potential energy of a system of point charges
q2
q3
q1 r12
r13
r23
ij
j
i
ij
r
q
kq
U =
12
2
1
12
r
q
kq
U =
13
3
1
13
r
q
kq
U =
23
3
2
23
r
q
kq
U =
23
13
12 U
U
U
U +
+
=
∑
∑∑ ≠
=
−
=
=
=
i
j ij
j
i
N
i
i
j ij
j
i
r
q
kq
r
q
kq
U 2
1
1
1
1
There are N(N-1)/2 distinct pairs
Extending to N point charges
We start with three point
charges: q1, q2, q3
EPE of qi, qj

19.2 The Electric Potential Difference
DEFINITION OF ELECTRIC POTENTIAL
The electric potential at a given point is the electric potential energy of a small test charge
divided by the charge itself:
o
q
V
EPE
= SI Unit of Electric Potential:
joule/coulomb = volt (V)
( )
o
o
o q
W
q
q
V
V
V
AB
B
B
A
B
AB
EPE
EPE −
=
−
=
−
=
∆
5
Electrical Potenetial Difference from A to B
=difference, per unit charge, of the EPE of
a test charge WOULD HAVE from location A
to location B
rA
rB
A
B
A
1
A
A
2
1
A ,
EPE
r
kq
V
r
q
kq
=
=
B
1
B
B
2
1
B ,
EPE
r
kq
V
r
q
kq
=
=
r
kq
r
V 1
)
( =
r

19.3 The Electric Potential Difference Created by Point Charges
Example The Potential of a Point Charge
Using a zero reference potential at infinity***,
determine the amount by which a point charge of 4.0x10-8C
alters the electric potential at a spot 1.2m away when the charge
is (a) positive and (b) negative.
6

Example The Potential of a Point Charge
Using a zero reference potential at infinity***,
determine the amount by which a point charge of
4.0x10-8C alters the electric potential at a spot 1.2m
away when the charge is (a) positive and (b) negative.
7
( )( )
V
300
m
2
.
1
C
10
0
.
4
C
m
N
10
99
.
8 8
2
2
9
+
=
×
+
⋅
×
=
=
−
r
kq
V
(a)
( )( )
V
300
m
2
.
1
C
10
0
.
4
C
m
N
10
99
.
8 8
2
2
9
−
=
×
−
⋅
×
=
=
−
r
kq
V
(b)
*** In Ch. 6 you learned that the choice of the
reference zero for gravitational potential energy
is arbitrary. This is also true for EPE and hence
for the electric potential V.
The choice of zero at infinity is only valid for a
collection of point charges, and not for parallel
plate capacitors or infinite wires

Example Where is the Potential equal to Zero?
Two point charges are fixed in place. The positive
charge is +2q and the negative charge is –q. On the
line that passes through the charges, how many
places are there at which the total potential is zero?
Where are they?
8
a
0
x

Example Where is the Potential
equal to Zero?
Two point charges are fixed in place.
The positive charge is +2q and the
negative charge is –q. On the line that
passes through the charges, how many
places are there at which the total
potential is zero? Where are they?
9
a
0
x
2
2
2
)
(
2
|
|
,
|
|
3
2
2
3
2
2
)
(
2
|
|
,
|
|
0
2
2
2
)
(
2
|
|
,
|
|
0
0
0
:
cases
3
|
|
1
|
|
2
0
|
|
1
|
|
2
0
|
|
|
|
2
0
0
2
2
1
1
2
2
1
1
a
x
x
a
x
x
a
x
a
x
a
x
x
x
a
x
a
x
a
x
x
x
a
x
x
a
x
a
a
x
x
x
a
x
a
x
x
x
a
x
x
a
x
a
a
x
x
x
x
a
x
a
x
x
a
x
x
a
x
x
a
x
q
x
q
r
q
r
q
r
kq
r
kq
V
=
→
=
−
→
=
−
→
−
=
−
=
→
>
=
→
=
→
=
−
→
=
−
→
−
=
−
=
→
<
<
=
→
−
=
−
→
−
=
−
→
−
=
−
−
=
→
<





>
<
<
<
−
=
→
=
−
−
→
=
−
−
+
+
→
=
+
→
=
+
=
Extraneous root:
x<0 amd x=2a are
not compatible

+q
−q
19.5 Capacitors and Dielectrics
Charge q (-q) is proportional to the
potential difference between the plates.
Italy
France
Britain
V
q
C
:
volt
coulomb
farad =
=
SI Unit of Capacitance:
coulomb/volt = farad (F)
1 F=1 C/V=1 C2/J
10
CV
q =
Memory Aid
A capacitor is a device designed to store energy and
charge. E.g. defibrillators consist of a large capacitor
that discharges current (flow of charges) when the
paddles are applied to a patient
In electrical circuits they often are used to stabilize
electric potential

THE CAPACITANCE OF A PARALLEL PLATE CAPACITOR
Simplest case: air/vacuum gap (no “dielectric”)
V = potential difference from the − to the + plate
(by this definition, V for a capacitor is always positive)
q = charge on the + plate (−q on the − plate)
A = area of the plates
d = plate separation/gap
Note the different notation used by the textbook for this section
11
+q
A
A
d
V
d
A
A
qd
q
V
q
C
C
A
q
E
E
A
qd
d
E
Ed
V
Ey
y
V
0
0
0
0
0
0
0
0
:
(downward)
r
vacuum/ai
in
field
Electric
gap
r
vacuum/ai
of
case
special
the
indicates
0
subscript
the
Here
then
is
plate
positive
at
Potential
)
(
plate;
negative
from
y
distance
at
Potential
ε
ε
ε
ε
σ
ε
=
÷
=
=
=
=
=
=
=
=
=
=
Parallel plate capacitor
filled with vacuum/air
d
A
C
C 0
0
ε
=
=
−q
0

12
Example: A parallel plate capacitor have the following attributes:
Area of plates: A=4.50m2 Plate separation: d=2.00cm
The gap is filled with air
Find its capacitance C
+
q
-q
A
A
d
V

13
Example: A parallel plate capacitor have the
following attributes:
Area of plates: A=4.50m2
Plate separation: d=2.00cm
Find its capacitance C
filled with vacuum/air
d
A
C
C 0
0
ε
=
=
nF
99
.
1
F
10
99
.
1
)
m
0200
.
0
(
)
m
50
.
4
)](
m
N
/(
C
10
85
.
8
[
9
2
2
2
12
0
=
×
=
⋅
×
=
=
−
−
C
C
Note that a farad (F) is a very
large capacitance
This 4.50 m2 plates at 2 .00 cm
separation gives only ~2 nF
+q
-q
A
A
d
V

14
Capacitors as an energy storage device:
It takes work (i.e. energy) to charge up a capacitor from zero charge to q (zero
potential to V).
The figure shows a capacitor at charge q, potential difference V (between the − plate
and the + plate).
To increase q and V, we move a small amount of charge ∆q from the − plate to the +
plate. This requires work done BY an external agent, AGAINST the electric field:
V
q
∆q
∆W
= (∆q)V
V=q/C
+V 0
+
+
+
+
+
+
+
+
−
−
−
−
−
−
−
−
∆q
+q −q
graph
vs.
in
stripe
shaded
the
of
area
)
(
)
(
field)
electric
by the
(done
q
V
V
q
Ed
q
W
W
EPE EXT
=
∆
=
∆
=
−
=
=
∆ +
−

15
Capacitors as an energy storage device:
(continued)
To charge a capacitor to (q, V) from (0,0), the total amount of work = area enclosed
by the blue triangle, which is the energy stored in the capacitor.
V
q
∆q
∆W
= (∆q)V
V=q/C
+V 0
+
+
+
+
+
+
+
+
−
−
−
−
−
−
−
−
∆q
+q −q
2
2
1
2
Energy
but
2
Energy
CV
U
CV
q
C
q
U
=
=
=
=
=
Energy =
U = area of triangle
=½ qV = ½ q2/C

16
+q
-q
A
A
d
Example: Parallel plate capacitor
Area of plates: A=4.50m2 Plate separation: d=2.00cm
Plate charge: q=22.5µC The gap is filled with air
(a) Find the voltage (potential difference) of the capacitor.
(b) Find the energy stored in this capacitor

17
+q
-q
A
A
d
Example: Parallel plate capacitor
Area of plates: A=4.50m2
Plate separation: d=2.00cm
Plate charge: q=22.5µC
(a) Find the voltage (potential difference) of the capacitor.
(b) Find the energy stored in this capacitor
2
2
1
2
2
Energy CV
C
q
U =
=
=
J
127
.
0
)
F
10
99
.
1
/(
C)
10
22.5
(
/
V)
10
13
.
1
)(
J
/
C
10
99
.
1
(
Energy
(b)
V
10
13
.
1
J
/
C
10
99
.
1
C
10
22.5
F
10
99
.
1
C
10
22.5
(a)
9
2
9
2
1
2
2
1
2
4
2
9
2
1
2
2
1
4
2
9
6
9
9
=
×
×
=
=
×
×
=
=
=
×
=
×
×
=
×
×
=
=
−
−
−
−
−
−
−
C
q
CV
U
C
q
V
Application:
Typical defibrillators have banks
of capacitors charged up to
~1000 V and usually an energy
of up to ~200 J is delivered to the
patient
This means C = 2U/V2 ≈ 0.4 mF
So how do we make capacitance
much bigger (approx. ×105)?
Resident
Evil?

Dielectric constant
E
Eo
=
κ
18
Dielectrics are materials that can be electrically
polarized by the application of an external electric field.
Filling the gap of a capacitor with a dielectric: the
molecules align such that “bound” charges (that are not
free to move around as charges in a conductor) develop
at the surface.
The induced surface charges (opposite in sign to the
charges on the plates) reduce the actual electric field
inside the material.
Note: dielectric = dia-electric; “dia” means “against”
(Greek root)
The factor by which the applied field E0 is reduced (to
E), is called:
THE DIELECTRIC CONSTANT Field in absence
of dielectric
Field in presence
of dielectric
κ
o
E
E =
A
q
Eo
0
ε
=

The Effect of a Dielectric When a
Capacitor Has a Constant Charge
An empty capacitor is connected to a battery and
charged up. The capacitor is then disconnected from the
battery, and a slab of dielectric material is inserted
between the plates. Does the voltage across the plates
increase, remain the same, or decrease?
19
fold
-
increases
of
factor
a
by
reduced
is
)
(
)
(
:
)
out
(dividing
plates
between
difference
Potential
reduced
is
constant
held
is
:
1)
(
dielectric
in
field
Electric
0
0
0
0
0
0
κ
κε
κ
κε
κε
κε
σ
κ
κ
C
d
A
V
q
C
V
A
qd
Ed
y
E
V
V
q
A
q
E
E
E
q
=
=
=
=
∆
=
∆
=
=
=
=
→
>
+
−
+
−
A
q
Eo
0
ε
=
κ
o
E
E =
filled with dielectric
d
A
C
C 0
0
κε
κ =
=

20
+q
-q
A
A
d
Example: Parallel plate capacitor in a defibrillator
(note this is both a high capacitance AND a precision application)
Area of plates: A=0.0100m2 (4”x4”: fits in the enclosure)
Maximum potential: V=1.00×103 V
Maximum energy stored: U=200 J
The gap is filled with BaTiO3 (barium titanate) : κ=5.00 ×103
Find the required plate separation d and the capacitance C
mF
0.398
F
10
98
.
3
10
11
.
1
)
m
0100
.
0
)](
m
N
/(
C
10
85
.
8
)[
10
00
.
5
(
μm
11
.
1
m
10
11
.
1
J)
200
(
2
J/C)
10
00
.
1
)(
m
0100
.
0
)](
m
N
/(
C
10
85
.
8
)[
10
00
.
5
(
2
2
4
6
2
2
2
12
3
0
6
2
3
2
2
2
12
3
2
0
2
0
2
0
2
1
2
2
1
=
×
=
⋅
×
×
=
=
=
×
=
×
⋅
×
×
=
=
→
=






=
=
−
−
−
−
−
C
x
d
A
C
d
U
AV
d
d
AV
V
d
A
CV
U
κε
κε
κε
κε

19.4 Equipotential Surfaces and Their Relation to the Electric Field
An equipotential surface is a surface on which the electric
potential is the same everywhere. In this case they are for the
point chareg potential
r
kq
V =
The net electric force does no work on a charge as
it moves on an equipotential surface.
21
The equipotential surfaces
are analogous to the
“contours” (curves on the
boundaries between
different colors instead of
surfaces in this picture) on
an elevation map such as
this (V is analogous to
height)

19.4 Equipotential Surfaces and Their Relation to the Electric Field
The electric field created by any charge or
group of charges is everywhere perpendicular
to the associated equipotential surfaces and
points in the direction of decreasing potential.
Equipotential surfaces are typically drawn at
equal intervals
The electric field is always perpendicular to the
surface pof a conductor: Conducting
surfaces are equipotential surfaces!
(the whole conductor is at the same potential)
22
r
kq
V =

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