Download 4.1 The Concepts of Force and Mass

19.1 Electrical Potential Energy
Special Case 2
r
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
E1 ( r ) = k
“(r)” means “function
of distance r”
q1
r
2
(1)
***
EPE of test charge q2 in the
electric field of source
charge q1:
q1q2
EPE 21 ( r ) = k
r
(2)
This formula works
even if one or both
charges are negative
1
***Derivation of (2) from (1) requires integral calculus
19.1 Electrical Potential Energy
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.
Q = −0.25 nC
What is the initial speed of the electron?
electron
rf = 1.1 × 10 −3 m
ri = ∞
vf = 0
vi = ?
e=1.6x10-19C, me=9.11x10-31 kg,
k=8.99x109 Nm2/C2
19.1 Electrical Potential Energy
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.
Q = −0.25 nC
What is the initial speed of the electron?
E f = Ei → KE f + EPE f = KE i + EPE i
1
2
1
2
mv 2f +
m ( 0) 2 +
kQqe 1 2 kQqe
= 2 mvi +
rf
ri
kQ ( −e) 1 2 kQ ( −e)
= 2 mvi +
rf
∞
rf = 1.1 × 10 −3 m
ri = ∞
vf = 0
vi = ?
e=1.6x10-19C, me=9.11x10-31 kg,
k=8.99x109 Nm2/C2
→
kQ ( − e) 1 2
= 2 mvi
rf
2

9 Nm 
−10
−19

(
)(
2 8.99 × 10
−
2
.
5
×
10
C
−
1
.
6
×
10
C)
2 
C 
2kQ ( −e)

2
⇒ vi =
=
(9.11 × 10 −31 kg)(1.1 × 10 −3 m )
mrf
m2
= 7.18 × 10
s2
14
⇒ vi = 2.68 × 10 7 m
s
electron
19.1 Electrical Potential Energy
EPE of qi, qj
Total Potential energy of a system of point charges
We start with three point
charges: q1, q2, q3
U 12 =
kq1q2
r12
q1
U ij =
q2
r12
U = U 12 + U 13 + U 23
U 13 =
kq1q3
r13
r13
r23
U 23 =
kqi q j
rij
kq2 q3
r23
q3
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
Extending to N point charges
N
i −1
U = ∑∑
i =1 j =1
kqi q j
rij
=
1
2
∑
j ≠i
kqi q j
rij
There are N(N-1)/2 distinct pairs
4
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:
EPE SI Unit of Electric Potential:
V=
qo
joule/coulomb = volt (V)
(∆V )AB = VB − VA
=
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
EPE B EPE B − WAB
−
=
qo
qo
qo
A
rA
r
EPE A =
kq1q2
kq
, VA = 1
rA
rA
kq1
V (r) =
r
rB
B
EPE B =
kq1q2
kq
, VB = 1
rB
rB
5
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
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.
(a)
kq
=
r
8.99 ×109 N ⋅ m 2 C 2 + 4.0 ×10 −8 C
1.2 m
= +300 V
V=
(
(b)
)(
)
kq
=
V =
r
(8.99 × 109 N ⋅ m 2 C 2 )(− 4.0 × 10 −8 C)
1.2 m
= −300 V
*** 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
7
19.3 The Electric Potential Difference Created by Point Charges
a
Example Where is the Potential equal to Zero?
x
0
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
19.3 The Electric Potential Difference Created by Point Charges
a x
Example Where is the Potential
equal to Zero?
0
V =
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?
kq1 kq2
q q
+ 2q
−q
+
=0 → 1 + 2 =0 →
+
=0
r1
r2
r1 r2
|x| |x−a|
 x<0
2
1
2
1

→
−
=0 →
=
3 cases : 0 < x < a
|x| |x−a|
|x| |x−a|
 x>a

Extraneous root:
x<0 amd x=2a are
not compatible
x < 0 → | x |= − x, | x − a |= a − x → 2( a − x ) = − x → 2a − 2 x = − x → x = 2a
0 < x < a → | x |= x, | x − a |= a − x → 2( a − x ) = x → 2a − 2 x = x → 3x = 2a → x =
2
a
3
x > a → | x |= x, | x − a |= x − a → 2( x − a ) = x → 2 x − 2a = x → x = 2a
9
19.5 Capacitors and Dielectrics
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
Charge q (-q) is proportional to the
potential difference between the plates.
q = CV
SI Unit of Capacitance:
coulomb/volt = farad (F)
1 F=1 C/V=1 C2/J
Memory Aid
q
C=
V
coulomb
farad =
volt
France
Britain :
Italy
+q
−q
10
19.5 Capacitors and Dielectrics
A
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
Potential at distance y from negative plate;
V ( y ) = Ey
+q
V
d
0
Potential at positive plate is then
qd
V = Ed = E0 d =
ε0 A
−q
A
Here the subscript 0 indicates the special case of vacuum/air gap
Electric field in vacuum/air (downward) :
σ
q
=
E = E0 =
ε0 ε0 A
C = C0 =
qd ε 0 A
q
=q÷
=
ε0 A
d
V
Parallel plate capacitor
filled with vacuum/air
ε0 A
C = C0 =
d
11
19.5 Capacitors and Dielectrics
A
Example: A parallel plate capacitor have the following attributes:
Area of plates: A=4.50m2
The gap is filled with air
Plate separation: d=2.00cm
+
q
Find its capacitance C
d
V
-q
A
12
19.5 Capacitors and Dielectrics
A
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
d
V
Parallel plate capacitor
filled with vacuum/air
ε0 A
C = C0 =
d
[8.85 × 10 −12 C 2 /( N ⋅ m 2 )]( 4.50 m 2 )
C = C0 =
(0.0200 m)
=
1.99 × 10 −9 F = 1.99 nF
-q
A
Note that a farad (F) is a very
large capacitance
This 4.50 m2 plates at 2 .00 cm
separation gives only ~2 nF
13
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).
+V
+
+
+
+
+q +
+
+
+
∆q
0
−
−
−
−
− −q
−
−
−
V=q/C
V
∆q
∆W
= (∆q)V
q
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:
∆EPE = WEXT = −W− + (done by the electric field)
= ( ∆q) Ed = ( ∆q)V = area of the shaded stripe in V vs. q graph
14
Capacitors as an energy storage device:
(continued)
+V
+
+
+
+
+q +
+
+
+
∆q
0
−
−
−
−
− −q
−
−
−
V=q/C
V
Energy =
∆W
U = area of triangle
∆q
= (∆q)V
2
=½ qV = ½ q /C
q
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.
q2
Energy = U =
2C
but q = CV
Energy = U = 12 CV 2
15
Example: Parallel plate capacitor
Area of plates: A=4.50m2
Plate charge: q=22.5µC
Plate separation: d=2.00cm
The gap is filled with air
A
+q
(a) Find the voltage (potential difference) of the capacitor.
(b) Find the energy stored in this capacitor
d
-q
A
16
Example: Parallel plate capacitor
A
A=4.50m2
Area of plates:
Plate separation: d=2.00cm
q2 1
Plate charge: q=22.5µC
Energy = U =
= 2 CV 2
The gap is filled with air
2C
+q
d
(a) Find the voltage (potential difference) of the capacitor.
(b) Find the energy stored in this capacitor
22.5 × 10 −6 C
q 22.5 × 10 −9 C
(a) V = =
=
= 1.13 × 10 4 V
2
−9
−9
C 1.99 × 10 F 1.99 × 10 C / J
(b) Energy = U = 12 CV 2 = 12 (1.99 × 10 −9 C 2 / J )(1.13 × 10 4 V) 2
-q
A
= 12 q 2 / C = 12 ( 22.5 × 10 −9 C) 2 /(1.99 × 10 −9 F) = 0.127 J
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
Resident
Evil?
So how do we make capacitance
much bigger (approx. ×105)?
17
19.5 Capacitors and Dielectrics
Dielectrics are materials that can be electrically
polarized by the application of an external electric field.
Eo =
q
ε0 A
E=
Eo
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
Dielectric constant
Eo
κ=
E
Field in absence
of dielectric
Field in presence
of dielectric
κ
18
19.5 Capacitors and Dielectrics
The Effect of a Dielectric When a
Capacitor Has a Constant Charge
Eo =
q
ε0 A
E=
Eo
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?
Electric field in dielectric (κ > 1) :
q is held constant → E is reduced
E=
E0
κ
=
σ
q
=
κε 0 κε 0 A
Potential difference between plates (dividing out q0 ) :
V = ( ∆V ) − + = E ( ∆y ) − + = Ed =
V is reduced by a factor of κ
q κε A
C= = 0
V
d
C increases κ - fold
qd
κε 0 A
Parallel plate capacitor
filled with dielectric
κε 0 A
C = κC0 =
d
κ
19
Example: Parallel plate capacitor in a defibrillator
A
(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
+q
d
Find the required plate separation d and the capacitance C
2
κε
κε
A
AV


2
0
0
U = 12 CV = 
V =
2d
 d 
κε 0 AV 2
→d =
=
2U
(5.00 × 10 3 )[8.85 × 10 −12 C 2 /( N ⋅ m 2 )](0.0100 m 2 )(1.00 × 10 3 J/C) 2
2( 200 J)
2
1
2
-q
A
d = 1.11 × 10 −6 m = 1.11 μm
C=
κε 0 A
d
(5.00 × 10 3 )[8.85 × 10 −12 C 2 /( N ⋅ m 2 )](0.0100 m 2 )
=
1.11x10 −6
C = 3.98 × 10 −4 F = 0.398 mF
20
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
kq
V=
r
The net electric force does no work on a charge as
it moves on an equipotential surface.
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)
21
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)
kq
V=
r
22