Download Chapter 19 - Electric Potential Energy and Electric Potential

CHAPTER 19 - ELECTRIC
POTENTIAL ENERGY AND
ELECTRIC POTENTIAL
Sections 1 - 5
Objectives: After completing this unit, you
should be able to:
• Understand an apply the concepts of electric
potential energy, electric potential, and electric
potential difference.
• Calculate the work required to move a known
charge from one point to another in an electric
field created by point charges.
• Write and apply relationships between the electric
field, potential difference, and plate separation for
parallel plates of equal and opposite charge.
Review: Work and Energy
Work is defined as the product of displacement x
and a parallel applied force F.
Work = Fx; Units: 1 J = 1 N m
Potential Energy, U, is defined as the ability to do
work by virtue of position or condition. (Joules)
Kinetic Energy, K, is defined as the ability to do work
by virtue of motion (velocity). (Joules)
Signs for Work and Energy
Work (Fx) is positive if an applied force, F, is
in the same direction as the displacement, x.
The force, F, does positive work.
B
F
P
mg
A
x
The force, mg, does negative work.
The P.E. at B relative to A is positive
because the field can do positive
work if object P is released.
P.E. at A relative to B is negative;
outside force needed to move object P
from A to B.
Gravitational Work and Energy
Consider work against g to move object
P from A to B, a vertical height h.
B
F
P
Work = Fh = mgh
At level B, the potential energy U is:
mg
A
U = mgh (gravitational)
The external force does positive work;
the gravity, g, does negative work.
The external force, F, against the gravity field
increases the potential energy, U.
h
g
Electrical Work and Energy
An external force, Fe , moves a +q charge
from A to B against the electric field force,
qE, (from Fe = qE).
++++
B
Fe
Work = Fx = (qE)d
+q
At level B, the potential energy, U is:
qE
U = qEd (Electrical)
A
d (x)
E
- - - -
The Electric field does negative work. The
external force does positive work.
The external force, Fe , against the Electric field increases the
potential energy.
Electric Potential Difference
Since work depends on the magnitude of the charge, qe, we
can define this in terms of energy per charge by dividing both
sides of the equation, U = Fd, by the charge qe .
U
q

W
 Volt  J / C
q
The electrical potential energy is measured in Joules. It can
also be measured in another term called the electron Volt, eV.
1eV = 1.60 x 10-19 J
Electric Potential Difference
One electron volt is the magnitude of the amount by which the
potential energy of an electron changes when the electron
moves through a potential difference of one volt.
W or Potential Energy (U)  qV
As you can see, there must be a difference in potential for there
to be voltage. So we can define another name for voltage as
POTENTIAL DIFFERENCE.
The potential here however is caused by an electric force which
is related to electric energy.
This is important to understand! A FORCE drives the +q
DOWNWARD and it also covers a DISPLACEMENT.
Electric Potential Difference
Therefore, if we consider the definition for WORK, and use it
algebraically, a new relationship for the Voltage results.
U
Fx
and U  Fx, so V  e , and
q
q
F
E  e , solve for Fe and sub in above, you get
q
Eqx
Fe
V
 Ed, if d  x, and or sub in E for
,
q
q
we get the same final equation
V  Eavg d, or
V
Eavg
V
d
Parallel Plates
Consider Two parallel plates of equal and
opposite charge, a distance d apart.
VA
++++
d
+q
Constant E field: F = qE
Work = Fd = (qE)d
VB - - - -
Also, Work = q(VA – VB)
So that: qVAB = qEd, so:
VAB = Ed or E = V/d
The potential difference between two oppositely charged
parallel plates is the product of E and d.
E
Example
1. (a) Calculate the speed of a proton that is
accelerated from rest through a potential
difference of 120 V.
(b) Calculate the speed of an electron that is
accelerated through the same potential difference.
• Answer: a. 1.52 x 105 m/s; b. 6.49 x 106 m/s
Example
2. A deuteron (a nucleus which consists of one
proton and one neutron) is accelerated through
a 2.7 kV potential difference.
(a) How much energy does it gain?
(b) (b) How fast would it be going if it started from
rest?
Answer: a. 4.33 x 10-16 J
b. 5.09 x 105 m/s
Example
3. An ion accelerated through a potential
difference of 115 V experiences an increase in
kinetic energy of 7.37 x 10-17 J. Calculate the
charge on the ion.
Answer: 6.41 x 10-19 C
Example
4. A positron, when accelerated from rest between
two points at a fixed potential difference,
acquires a speed of 30% of the speed of light.
What speed will be achieved by a proton if
accelerated from rest between the same two
points?
Answer: 2.10 x 106 m/s
Example
5. The electric field between two charged parallel
plates separated by a distance of 1.8 cm has a
uniform value of 2.4 x 104 N/C. Find the
potential difference between the two plates.
How much kinetic energy would be gained by
a deuteron in accelerating from the positive to
the negative plate?
Answer: 6.92 x 10-17 J or 432 eV
Example
6. A proton moves in a region of uniform electric
field. The proton experiences an increase in
kinetic energy of 5 x 10-18 J after being
displaced 2 cm in a direction parallel to the
field. What is the magnitude of the electric
field?
Answer: 1.56 x 103 N/C
Electric Potential Difference Created by
Point Charges
Recall that work is force times a distance, substitute this in for
work.
Then substitute the
electrical force formula
in for force and you get
the formula for finding
the Potential of a point
charge at a given point
from a charge or set of
charges.
W Fe x

, x  r, and
q
q
kq1q 2
Fe 
, sub into formula above,
2
r
kq1q 2
r
2
kq1q 2r kq1
r
V


, or
2
q
r
r q
V
Total Voltage (V)  k

i
qi
r
Potential For Multiple Charges
When two or more charges are present, the potential due to
all the charges is obtained by adding together the individual
potentials.
Q1
-
r1
 A
r2
r3
Q3
-
+
Q2
kq1 kq2 kq3
VA 



r1
r2
r3
qi
V k
r
BE CAREFUL, since voltage is potential difference, that means you
add the signs. DO NOT USE THE ABSOLUTE VALUE of the equation!
You must use the sign or signs of the charge or charges!
Example
7. At what distance from a point charge of 8 μC would the
potential equal 3.6 x 104 V?
• Answer: 2.0 m
Example
8. At a distance r away from a point charge q, the
electrical potential is V = 400 V and the magnitude of
the electric field is E = 150 N/C. Determine the value of
q and r.
• Answer: r = 2.67 m,
q = 119 nC
Example
9. The three charges shown in the figure are at
the vertices of an isosceles triangle.
Calculate the electric potential at the
midpoint of the base, taking q = 7 μC.
•
Answer: 1.097 x 107 V
Example
10. Two point charges, Q1 = +5 nC and Q2 = -3 nC, are
separated by 35 cm.
(a) What is the potential energy of the pair? What is the
significance of the algebraic sign of your answer?
(b) What is the electric potential at a point midway between
the charges?
Answer: (a) -3.857 x 10-7 J, Energy needed to separate
them.
(b) 102.857 V
•
•
Assignment
Pages 594 - 596,
#1, 5, 8, 13, 18, 27, 35, 38