Transcript SPH4UI

SPH4UI
Electric Potential
Mr. Burns
Electricity has Energy
To separate negative and positive charges from each other, work must be done
against the force of attraction. Therefore seperated charges are in a higherenergy state. When the charges are brought back together again, energy must
be released. It may be in the form of a spark.
Consider a small test charge in the electric field around a positive charged
object. The field is pushing it in the direction of the field lines. If some external
force pushes it the other way, against the direction of the field, work is being
done on it.
Therefore, a positive charge has its greatest electric potential energy at the
upper end (closest to charged object) of the field. Since the field pushes a
negative charge the other way, its potential energy is greatest at the lower end.
F

Review: Gravitational Force is conservative
Consider a comet in a highly elliptical orbit
U(r)
pt 2
pt 1
0
U(r1)
U (r )  

At point 1, particle has a lot of potential
energy, but little kinetic energy

At point 2, particle has little potential
energy, but a lot of kinetic energy
GMm
r
More
potential
energy
U(r2)
Less
potential
energy
Total energy = K + U
is constant!
Electric Potential Energy
We see that potential energy increases as opposite charges recede,
and that potential energy increases as like charges approach.
Electric Potential Energy
Electric Potential Energy:
kq1q2
U
r
Doesn’t it
kinda look
familiar?
Electric Potential Energy



The electrostatic force is a conservative (=“path
independent”) force
It is possible to define an electrical potential
energy function with this force
Work done by a conservative force is equal to
the negative of the change in potential energy.
Think of the work done by gravity to raise an object to a
higher location, since gravity points down and the
displacement is up, therefore the work is negative. Yet,
the potential energy increases.
Electric Force is conservative

Consider a charged particle traveling through a
region of static electric field:

--

++



A negative charge is attracted to
the fixed positive charge
negative charge has more potential
energy and less kinetic energy far
from the fixed positive charge,
and…
more kinetic energy and less
potential energy near the fixed
positive charge.
But, the total energy is conserved
We will now discuss electric
potential energy and the
electrostatic potential….
Electric Potential
Electric Potential
Consider the electric potential energy not just of any charge q2, but of a unit
positive test charge when in the field of any other charge q1.
Recall: Electric Potential Energy: U 
kq1q2
r
We call this value of potential energy per unit positive charge the electric
potential.
The units of electric potential are joules per coulomb, or volts
U Electric Potential Energy
V

q2
charge
kq1q2 1

r q2
kq1

r
Electric Potential
describes how much
work is done per unit
charge.
Electric Potential Difference
High Potential
Low Potential
+
Electric Potential Difference
The amount of work required per unit charge to move a positive
charge from one point to another in the presence of an electric
field.
The units are also volts
U
V 
q2
1 1
 kq1   
 rb ra 
Energy and Charge Movements


A positive charge gains electrical potential energy
when it is moved in a direction opposite the electric
field
If a charge is released in the electric field, it
experiences a force and accelerates, gaining kinetic
energy


As it gains kinetic energy, it loses an equal amount of
electrical potential energy
A negative charge loses electrical potential energy
when it moves in the direction opposite the electric
field
Energy and Charge Movements



When the electric field is
directed downward, point B is
at a lower potential than point
A
A positive test charge that
moves from A to B loses
electric potential energy
It will gain the same amount of
kinetic energy as it loses
potential energy
F=
Summary of Positive Charge Movements and Energy

When a positive charge is placed in an
electric field




It moves in the direction of the field
It moves from a point of higher potential to a point
of lower potential
Its electrical potential energy decreases
Its kinetic energy increases
Summary of Negative Charge Movements and Energy

When a negative charge is placed in an
electric field





It moves opposite to the direction of the field
It moves from a point of lower potential to a point
of higher potential
It moves from a point where it has higher potential
energy to a point where it has lower potential
energy.
Its electrical potential energy decreases
Its kinetic energy increases
Review of Potential and Potential Energy
Force:
+
+
- Moves from High Potential (V) to Low Potential (V)
- Potential energy (U) decreases
- Electric Field (E) does positive work
Force: 
+
-
- Moves from Low Potential (V) to High Potential (V)
- Potential energy (U) decreases
- Electric Field (E) does positive work
Force: 
-
+
- Moves from High Potential (V) to Low Potential (V)
- Potential energy (U) decreases
- Electric Field (E) does positive work
Force: 
-
-
- Moves from Low Potential (V) to High Potential (V)
- Potential energy (U) decreases
- Electric Field (E) does positive work
Electric potential energy

Imagine two positive charges, one with charge Q1, the other
with charge Q2:
Q2
Q1

Initially the charges are very far apart, so we say that the
initial energy Ui of interaction is zero
(we are free to define the energy zero somewhere).

If we want to push the particles together, this will require
work (since they want to repel).
Q1
Q2
 the final energy Uf of the system will increase by the
same amount: U = Uf – Ui = W
Electric potential Energy
R


W   F
 dr
you
Pretend q1 is fixed at the origin.

What is the work requiredto move q2, initially at infinity, to
 FE  drQ1
a distance r away?
Q2

R
r
  kq q 
Remember – work is force
distance:
1times
2
   2   dr
R r r

 kq1q2
W   F  dl  

 1 r
 kq1q2   
 What if q2 were negative (but q
r positive)?
R
1 still
 Then the work “required” by us would be negative
 tokq
 kq1q2 
1q2 together.
 the charges would like
come
 



 In this case the final energy
is
negative!

R

 

kq1q2

Particles will move to minimize
the final potential energy.
R
Electric Potential Energy
•
In addition to discussing the energy of a “test” charge in a Coulomb
field, we can speak of the electric potential energy of the field itself!
•
Reasons?
– Work had to be done to assemble the charges (from infinity)
into their final positions.
– This work is the potential energy of the field.
– The potential energy of a system of N charges is defined to be
the algebraic sum of the potential energy for every pair of
charges.
•
This theme continues in the course with E in capacitors and B in
inductors – these devices store electric and magnetic energy.
•
We will start with a couple of example calculations…
Electric Potential Energy
• Example 1: What is the potential energy of
this collection of charges?
+2q
-q
2d
d
d
-q
Step 1: Bring in +2q from infinity. This costs nothing.
Step 2: Bring in one -q charge. The force is
k (2q)(q)
U
attractive! The work required is negative:
d
Step 3: Bring in 2nd -q charge. It is attracted to the +2q, but repelled
from the other -q charge. The total work (all 3 charges) is
kq 2 
1 
 k (2q)(q) k (2q)(q) k (q)(q) 
U 




4




d
d
d 
2d 
2

A negative amount of work was required to bring these charges
from infinity to where they are now (i.e., the attractive forces
between the charges are larger than the repulsive ones).
ACT 1
• Consider the 3 collections of point charges shown
below.
– Which collection has the smallest potential energy?
d
-Q
-Q
d
d
d
-Q
+Q
-Q
d
d
d
-Q
+Q
(a)
(b)
+Q
d
+Q
(c)
ACT 1
• Consider the 3 collections of point charges shown
below.
• Which collection has the smallest potential energy?
d
-Q
+Q
-Q
d
d
d
d
d
d
-Q
-Q
-Q
+Q
(a)
(b)
+Q
d
+Q
(c)
•
We have to do positive work to assemble the charges in (a) since they all have the
same charge and will naturally repel each other. In (b) and (c), it’s not clear whether
we have to do positive or negative work since there are 2 attractive pairs and one
repulsive pair.
(a)
kQ 2
U  3
d
(b) (c)
(b)
kQ 2
U 
d
(c)
kQ 2
U 
2d
U
0
(a)
Electric Potential
A
Two charges which are equal in
magnitude, but opposite in sign
are placed at equal distances
from point A.
If a third charge is added to the system and placed at point A,
how does the electric potential energy of the charge
collection change ?
a) increases
b) decreases
c) doesn’t change
Electric potential
•
Consider that we have three charges fixed in space.

The potential energy of an added test charge q0
at point P is just
U of
q0 a t P
 Q1
Q3 
Q2
 q0  k
 k
k

 r
r2 p
r3 p 
 1p
Q1
Q2
r1p
r2p
Q3
r3p
q0

Note that this factors: q0 x (the effects of all other charges)

Just as we previously defined the electric field as the ratio of force/charge, we now
define the electric potential as the potential energy/charge:
V
U
q0
or
U = q0V
•
U depends on what qo is, but V is independent of qo (can be + or -)
•
Units of electric potential are volts: 1 V = 1 J/C
•
V is a scalar field, defined everywhere in space.
Work and Potential Energy




There is a uniform field
between the two plates
As the positive charge
moves from A to B, work
is done
WAB=F d=(q E) d
ΔU =-W AB=-q E d

only for a uniform field
Work and Potential Energy
A proton moves from rest in an electric field of 8.0104 V/m along the +x
axis for 50 cm. Find:
a) the change in in the electric potential
b) the change in the electrical potential energy
c) the speed after it has moved 50 cm.
V   Ed
V

V    8.0 104   0.5m 
m

 4.0 104 V
Work and Potential Energy
b) the change in the electrical potential energy
U E  qV
U E  1.6 1019 C  4.0 104 V 
 6.4 1015
c) the speed after it has moved 50 cm.
Ki  U i  K f  U f
Ui  K f
K f  Ui
1 2
mv  6.4  10 15 J
2
2(6.4 1015 J )
v
1.67 1027 kg
 2.8 106 m / s
Potential Difference (=“Voltage Drop”)

The potential difference (V) between points A
and B is defined as the change in the
potential energy (U) (final value minus initial
value) of a charge q moved from A to B
divided by the size of the charge
U B  U A W U
V  VB  VA 
 
q
q
q

Potential Difference is not the same as
Potential Energy
Electric potential difference, in terms of E

Suppose charge q0 is moved from pt
A to pt B through a region of space
described by electric field E.
Fwe supply = -Felec
Felec
q0
A
E
B
• Force on the charge due to E  work WAB≡WAB will have to be
done to accomplish this task:
U WAB

VB  VA 

q0
q0
•
To get a positive test charge from lower potential to higher potential
you need to invest energy - you need to do work.
•
The overall sign of this: A positive charge would “fall” from a higher
potential to a lower one
•
If a positive charge moves from high to low potential, it can do work
on you; you do “negative work” on the charge.
Potential Difference and a Battery
Electric circuits are all about the movement of charge between varying
locations and the corresponding loss and gain of energy which accompanies
this movement. The concept of electric potential can be applied to a simple
battery-powered electric circuit. Work must be done on a positive test charge
to move it through the battery from the negative terminal to the positive
terminal. This work would increase the potential energy of the charge and
thus increase its electric potential. As the positive test charge moves through
the external circuit from the + terminal to the negative terminal, it decreases
its electric potential energy and thus is at low potential by the time it returns
to the negative terminal. If a 12 volt battery is used in the circuit, then every
coulomb of charge is gaining 12 joules of potential energy as it moves
through the battery. And similarly, every coulomb of charge loses 12 joules of
electric potential energy as it passes through the external circuit.
Potential Difference and a Battery
With a clear understanding of electric potential difference, the role of a
battery in a simple circuit can be correctly understood. The battery simply
supplies the energy to do work upon the charge to move it from the
negative terminal to the positive terminal. By providing energy to the
charge, the battery is capable of maintaining an electric potential
difference across the two ends of the external circuit. Once the charge has
reached the high potential terminal, it will naturally flow through the wires
to the low potential
Potential difference and Energy Question
What is the potential difference across an air conditioner if 72C of charge
transfer 8.5 x 103 J of energy to the fan and compressor?
V
U
q
8.5  103 J

72C
 1.2  10 2 V
Potential difference and Energy Question
A static electric shock delivered to a student from a friend transfers
15 J of electric energy through a potential difference of 500V. What
is the quantity of charge transferred in the spark?
U
q
U
q
V
15 J

500V
 0.03C
V
0.03C  6.25 1018
e
C
1.9 1017 e
Electric potential
A
E
C
B
Points A, B, and C lie in a uniform electric field. What is the
potential difference between points A and B?
ΔVAB = VB - VA
a) ΔVAB > 0
b) ΔVAB = 0
c) ΔVAB < 0
Point C is at a higher potential than point A?
True
False
Electric potential Question
A positive charge is released from rest in a region of
electric field. The charge moves:
a) towards a region of smaller electric potential
b) along a path of constant electric potential
c) towards a region of greater electric potential
Electric potential Understanding Question
If you want to move in a region of electric field without
changing your electric potential energy. How would you
choose the direction ?
You would have to move
perpendicular to the field
if you wish to move without changing
electric potential.
Electric Potential Difference
A single charge ( Q = -1mC) is fixed at the origin. Define point A
at x = + 5m and point B at x = +2m.
What is the sign of the potential difference between A and B?
(Δ VAB  VB - VA )
(a) ΔVAB < 0 (b) Δ VAB = 0
B

-1mC
(c) Δ VAB > 0
A

x
•The simplest way to get the sign of the potential difference is to imagine
placing a positive charge at point A and determining which way it would move.
Remember that a positive charge will always “fall” to lower potential.
•A positive charge at A would be attracted to the -1mC charge;
therefore NEGATIVE work would be done to move the charge from A to B.
Δ VAB is Independent of Path
WAB
VB  VA 
q0

Δ VAB is the same for any path
chosen to move from A to B
(because electric forces are
conservative).
q0
A
E
B
Electric potential
Equipotential lines are connected lines of the same potential. These
often appear on field line diagrams. Equipotential lines are always
perpendicular to field lines, and therefore perpendicular to the force
experienced by a charge in the field. If a charge moves along an
equipotential line, no work is done; if a charge moves between
equipotential lines, work is done.
Equipotentials and Electric Fields Lines (Positive Charge):


The equipotentials for a
point charge are a
family of spheres
centered on the point
charge
The field lines are
perpendicular to the
electric potential at all
points
Electric potential Question
A positive charge Q is moved from A to B
along the path shown by the arrow. What
is the sign of the work done to move the
charge from A to B?
(a) WAB < 0

(b) WAB = 0
A
B
(c) WAB > 0
A direct calculation of the work done to move a positive charge from point A to
point B is not easy.
•
Neither the magnitude nor the direction of the field is constant along the
straight line from A to B.
But, you DO NOT have to do the direct calculation.

•
•
Remember: potential difference is INDEPENDENT OF THE PATH!!
Therefore we can take any path we wish.
Choose a path along the arc of a circle centered at the charge.
Electric Potential: Where is it zero?
•
Define the electric potential of a point
in space as the potential difference
between that point and a reference
point.
•
a good reference point is infinity ... we
often set V = 0
•
Thus, at a distance r from a spherical
point charge q, the electric potential is
given by:
kq
V 
r
Potential from N charges
The potential from a collection of N charges is just the
algebraic sum of the potential due to each charge separately
(this is much easier to calculate than the net electric field,
which would be a vector sum).
Q1
Q2
r1p
r2p
Q3
r3p
P

Vat P
N
N
qn
V (r )  Vn (r )  k 
n1
n1 rn
kQ1 kQ2 kQ3



r1 p r2 p r3 p
Voltage
+5 μC
-3 μC
A
Two charges q1 = + 5 μC, q2 = -3μC are placed at equal distances
from the point A. What is the electric potential at point A?
a) VA < 0
b) VA = 0
c) VA > 0
Vat A
kQ1 kQ2


r1 p r2 p
k  5m C  k  3mC 


R
R
k  2m C 

R
The Electron Volt

The electron volt (eV) is defined as the energy
that an electron (or proton) gains when
accelerated through a potential difference of 1 V




Electrons in normal atoms have energies of 10’s of eV
Excited electrons have energies of 1000’s of eV
High energy gamma rays have energies of millions of
eV
1 V=1 J/C  1 eV = 1.6 x 10-19 J
Summary
Electric potential energy
Is the energy of an electrically charged particle In an
electric field.
Defined as work that must be done to move it from an
infinite distance away to its present location.
Can be positive or negative (same signs then positive,
opposite signs then negative)
kQq0
U W 
r
Summary
Electric Potential
Is the value of the potential energy per unit of positive
charge in Volts.
V
kQ
r
Summary
Electric Potential Difference
Is the amount of work required per unit charge to move
a positive charge from one point to another in the
presence of an electric field.
The electrical potential energy for each coulomb of
charge in a circuit is called the electric potential
difference
V 
U
q0
1 1
 kQ   
 rb ra 
Summary Acts
How much work does it take to move a charge of +30 nanocoulombs through
a potential difference of 6.0 volts
Solution:
We are given a
potential
difference
We want work or
change in
potential energy
U
Therefore: V  q
U  qV
J

  30 109 C   6.0 
C

 1.8 107 J
Summary Acts
Find the speed of the mass of 2.5 x 10-6 kg if it moves through a potential
difference of 30.0 V and has a charge of 4.1 x 10-6 C.
Solution:
We require velocity,
therefore think about energy
(½ mv2).
Work =change kinetic energy
W  qV
1 2
mv  qV
2
2qV
v
m

2  4.1106 C   30.0V 
 9.9
2.5  106 kg
m
s
Summary Equations
kQq0
1 Qq0 (magnitude only)
F 2 
r
4 0 r 2
F kQ
E  2
q0 r
kQq0
Ue 
r
Electric Field is a Vector quantity (N/C)
Potential energy
Q

A
U
kQ
V e 
 rE
q0
r
Electric potential: Where q0 is the test charge and Q is
the charge creating the field
W  U  qV
Work done
Charge density for parallel plates.
Summary Equations
q  ne
n is the number of electrons
q
 V
E
 
o A o
d
For parallel plates
 E  EA  E  4 r 2 
Total flux closed circle
E 
E
Qenclosed
0
Any enclosed surface

2 0 r
Electric field of cylinder at a distance r from centre, Is
charge density.
Q 0 A
C

V
d
C  C1  C2 ,
1 1
1
 
C C1 C2
Capacitance of capacitor
In series, in parallel
Flash – Electric Potential