Chapter 15
„
Coulomb’s Law F12 = k
Q1Q2
r122
„
Electric Field E=F/q, (E=E1+E2+E3+…)
„
Conductors in Electric Field
Chapter 16
Electric Potential
Electric Energy
E=0 inside a good conductor
Any excess charge resides on the surface of the conductor
Electric
a
„
„
Work=force×distance
Potential energy (PE) is the energy
that depends on the position or
configuration
PE is a property of system of
interacting objects
-B
Earth
+A
Q
b
m
c
Earth
+A
PE=QEd (KE=0) W=ΔPE
d
F=mg
F=QE
-B
F=QE
PE=0
-B
W=QEd(charge
W=QEd(charge))
W=m
W=mgh(mass)
h(mass)
h
PE=mgh (KE=0)
F=mg
PE=0
Earth
+A
d
• Work done on Charge: W=ΔPE=QE·d-0 with
respect to B (PE=0 at B)
Work done on mass: W=ΔPE=mg·h-0 (PE=0
on the earth)
• There is no change in the energy of a charge
moved perpendicular to an electric field, as
there is no change in the energy of a mass
moved perpendicular to a gravitational field (
for instance, along the earth’s surface)
g
E
We have learned that
„
Gravitational
+A
PE=0
KE=QEd - B
PE=0
KE=mgh
Earth
Potential Difference
„
The potential difference between points
A and B is defined as the change in the
potential energy (final value minus
initial value) of a charge q moved from
A to B divided by the size of the charge
„
„
ΔV = VB – VA = ΔPE / q
Potential difference is not the same as
potential energy
Potential difference is the work done to move the charge
from one point to another. Only potential difference
between two points is physically meaningful
Potential Difference, cont.
„
„
„
Another way to relate the energy and the
potential difference: ΔPE = q ΔV
Both electric potential energy and potential
difference are scalar quantities
Units of potential difference
„
„
ΔV=Vab=Vb-Va=Wab/Q
Equipotential lines
E
V = J/C
b
A special case occurs when there is a uniform
electric field
„
Gives more information about units: N/C = V/m
Energy and Charge Movements, cont
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
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
d
Vab=-Ed (uniform electric field)
VB – VA= -Ed
„
a
„
„
„
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
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
Its electrical potential energy decreases
Its kinetic energy increases
P and Q are points within a uniform electric field that are separated
by a distance of 0.1 m as shown. The potential difference between P
and Q is 50 V.
A charge q = −4.0 µC is moved 0.25 m horizontally to
point P in a region where an electric field is 150 V/m
and directed vertically as shown. What is the change in
the electric potential energy of the charge?
(a) −2.4 × 10−3 J
(d) +1.5 × 10−4 J
(b) −1.5 × 10−4 J
(e) +2.4 × 10−3 J
1. Determine the magnitude of this electric field.
(a) 0.5 V/m
(c) 50 V/m
(b) 5.0 V/m
X(d) 500 V/m
2. How much work is required to move a +1000 μC point charge
from P to Q?
(a) 0.02 J
X (c) zero joules
X (b) 0.05 J
Electric Potential due to a Point Charge
V
„
r
+
r
r
„
r
„
„
„
V1 is the electric potential
due to q1 at some point P1
The work required to bring
q2 from infinity to P1 is
q2V1
This work is equal to the
potential energy of the two
particle system
PE = q2 V1 = k e
q1q2
r
The algebraic sum is used because
potentials are scalar quantities
Notes About Electric Potential Energy
of Two Charges
„
If the charges have the same sign, PE is
positive
„
„
„
P1
(e) 5000 J
(d) 1000 J
Superposition principle applies
The total electric potential at some
point P due to several point charges is
the algebraic sum of the electric
potentials due to the individual charges
„
Electrical Potential Energy of Two
Charges
(c) 200 J
Electric Potential of Multiple Point
Charges
V=kQ/r (V=0 at r=∞)
V
(e) 5000 V/m
Positive work must be done to force the two
charges near one another
The like charges would repel
If the charges have opposite signs, PE is
negative
„
„
The force would be attractive
Work must be done to hold back the unlike
charges from accelerating as they are brought
close together
Question
Is there a point along the line joining two
equal positive charges where the
electric field is zero? Where the electric
potential is zero?
Answer:
Q
Q
E=0
The sum of both potentials is nonzero at any finite distance
but becomes zero at infinite distance.
Summary
„
„
„
„
Electric potential: PE per unit charge, V=PE/q
Potential difference: work done to move the charge
from one point to another,
Vab=Vb-Va=ΔPE/q=Wab/q or Wab=qVab
For uniform electric field: Vab=-Ed
For electric field due to a point charge:
V=kQ/r, (V=0 at r=∞)
Three point charges –Q, –Q, and +3Q are arranged along a line
as shown in the sketch.
What is the electric potential at the point P?
(a) +kQ/R
(c) –1.6kQ/R
(b) –2kQ/R
(d) +1.6kQ/R X
(e) +4.4kQ/R