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PHYS 1444
Lecture #3
Tuesday, June 12, 2012
Ryan Hall
for Dr. Andrew Brandt
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•
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Chapter 21: E-field examples
Chapter 22: Gauss’ Law Examples
Chapter 23: Potential
- Electric Potential Energy
- Electric Potential
Homework on Ch 21-22 due today at midnight
Homework #3 (Ch 23) Due next Tuesday June 19th at midnight
Tuesday June 12, 2012
PHYS 1444 Ryan Hall
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Example 21-7
Two point charges are separated by a distance of 10.0 cm. One has a
charge of -25 μC and the other +50 μC. (a) Determine the direction
and magnitude of the electric field at a point P between the two
charges that is 2.0 cm from the negative charge. (b) If an electron
(mass = 9.11 x 10-31 kg) is placed at rest at P and then released, what
will be its initial acceleration (direction and magnitude)?
Solution: a. The electric fields add in magnitude, as both are directed towards the negative charge. E = 6.3 x
108 N/C. b. We don’t know the relative lengths of E1 and E2 until we do the calculation. The acceleration is
the force (charge times field) divided by the mass, and will be opposite to the direction of the field (due to the
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negative charge of the electron). Substitution gives a = 1.1 x 1020 m/s2.
21-7 Electric Field Calculations for
Continuous Charge Distributions
A continuous distribution of charge may be treated as a
succession of infinitesimal (point) charges. The total field is
then the integral of the infinitesimal fields due to each bit of
charge:
Remember that the electric field is a vector; you will need a
separate integral for each component.
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21-7 Electric Field for Continuous Charge Distributions
Example 21-9: A ring of charge.
A thin, ring-shaped object of radius a holds a total charge +Q
distributed uniformly around it. Determine the electric field at a
point P on its axis, a distance x from the center. Let λ be the
charge per unit length (C/m).
Solution: Because P is on the axis, the transverse components of E must
add to zero, by symmetry. The longitudinal component of dE is
dE cos θ, where cos θ = x/(x2 + a2)1/2. Write dQ = λdl, and integrate4dl
from 0 to 2πa. Answer: E = (1/4πε0)(Qx/[x2 + a2]3/2)
Example 21-8
Calculate the total
electric field (a) at
point A and (b) at point
B in the figure due to
both charges, Q1 and
Q2.
Solution: The geometry is shown in the figure. For each point, the process is:
calculate the magnitude of the electric field due to each charge; calculate the x and
y components of each field; add the components; recombine to give the total field.
a. E = 4.5 x 106 N/C, 76° above the x axis.
b. E = 3.6 x 106 N/C, along the x axis.
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21-7 Electric Field Calculations for
Continuous Charge Distributions
Conceptual Example 21-10: Charge at the center of a ring.
Imagine a small positive charge placed at the center of a
nonconducting ring carrying a uniformly distributed
negative charge. Is the positive charge in equilibrium if it is
displaced slightly from the center along the axis of the ring,
and if so is it stable? What if the small charge is negative?
Neglect gravity, as it is much smaller than the electrostatic
forces.
Solution: The positive charge is in stable equilibrium, as it is
attracted uniformly by every part of the ring. The negative charge is
also in equilibrium, but it is unstable; once it is displaced from its
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equilibrium position, it will accelerate away from the ring.
21-7 Electric Field Calculations for
Continuous Charge Distributions
Example 21-12: Uniformly charged disk.
Charge is distributed uniformly over a thin circular disk of
radius R. The charge per unit area (C/m2) is σ. Calculate the
electric field at a point P on the axis of the disk, a distance z
above its center.
Solution: The disk is a set of
concentric rings, and we know
(from example 21-9) what the field
due to a ring of charge is. Write dQ
= σ 2πr dr. Integrate r from 0 to R.
See text for answer.
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Example 22 – 2
Flux from Gauss’ Law: Consider the two
Gaussian surfaces, A1 and A2, shown in the figure.
The only charge present is the charge +Q at the
center of surface A1. What is the net flux through
each surface A1 and A2?
• The surface A1 encloses the
charge +Q, so from Gauss’ law
we obtain the total net flux
• For A2 the charge, +Q, is
outside the surface, so the total
net flux is 0.
r r Q
E  dA 
Ñ

0
r r 0
E  dA   0
Ñ

0
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Example 22 – 5
Long uniform line of charge: A very long straight
wire possesses a uniform positive charge per unit
length, l. Calculate the electric field at points
near but outside the wire, far from the ends.
• Which direction do you think the field due to the charge on the wire is?
– Radially outward from the wire, the direction of radial vector r.
• Due to cylindrical symmetry, the field is constant anywhere on the
Gaussian surface of a cylinder that surrounds the wire.
– The end surfaces do not contribute to the flux at all. Why?
• Because the field vector E is perpendicular to the surface vector dA.
• From Gauss’ law
Solving for E
r r
Qencl ll

E  dA  E dA  E  2 rl  
Ñ

l
E
2 0 r
Ñ

0
0
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Example 22-4: Solid sphere of charge.
An electric charge Q is distributed
uniformly throughout a nonconducting
sphere of radius r0. Determine the
electric field (a) outside the sphere
(r > r0) and (b) inside the sphere (r < r0).
Solution: a. Outside the sphere, a gaussian surface encloses the
total charge Q. Therefore, E = Q/(4πε0r2).
b. Within the sphere, a spherical gaussian surface encloses a
fraction of the charge Qr3/r03 (the ratio of the volumes, as the
charge density is constant). Integrating and solving for the field
gives E = Qr/(4πε0r03).
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Electric Potential Energy
• Concept of energy is very useful in solving mechanical problems
• Conservation of energy makes solving complex problems easier.
• When can the potential energy be defined?
– Only for a conservative force.
– The work done by a conservative force is independent of the path. What
does it only depend on??
• The difference between the initial and final positions
– Can you give me an example of a conservative force?
• Gravitational force
• Is the electrostatic force between two charges a conservative
force?
– Yes. Why?
– The dependence of the force on distance is identical to that of the
gravitational force.
• The only thing matters is the direct linear distance between the object not the
path.
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Electric Potential Energy
• What is the definition of change in electric potential energy Ub –Ua?
– The gain (or loss) of potential energy as the charge moves from point a to
point b.
– The negative work done on the charge by the electric force to move it from a
to b.
• Let’s consider an electric field between two parallel
plates w/ equal but opposite charges
– The field between the plates is uniform since the gap is
small and the plates are infinitely long…
• What happens when we place a small charge, +q,
on a point at the positive plate and let go?
– The electric force will accelerate the charge toward
negative plate.
– What kind of energy does the charged particle gain?
• Kinetic energy
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Electric Potential Energy
• What does this mean in terms of energies?
– The electric force is a conservative force.
– Thus, the mechanical energy (K+U) is conserved
under this force.
– The charged object has only electric potential
energy at the positive plate.
– The electric potential energy decreases and
– Turns into kinetic energy as the electric force works
on the charged object and the charged object
gains speed.
• Point of greatest potential energy for
– Positively charged object
– Negatively charged object
PE= U
0
KE= 0
K
ME= U
K
U+K=const 13
Electric Potential
• How is the electric field defined?
– Electric force per unit charge: F/q
• We can define electric potential (potential) as
– The electric potential energy per unit charge
– This is like the voltage of a battery…
• Electric potential is written with a symbol V
– If a positive test charge q has potential energy Ua at
a point a, the electric potential of the charge at that
point is
Ua
Va 
q
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Electric Potential
• Since only the difference in potential energy is meaningful,
only the potential difference between two points is
measurable
• What happens when the electric force does “positive work”?
– The charge gains kinetic energy
– Electric potential energy of the charge decreases
• Thus the difference in potential energy is the same as the
negative of the work, Wba, done on the charge by the electric
field to move the charge from point a to b.
• The potential difference Vba is
U b  U a Wba
Vba  Vb  Va 

q
q
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A Few Things about Electric Potential
• What does the electric potential depend on?
– Other charges that create an electric field
– What about the test charge?
• No, the electric potential is independent of the test charge
• Test charge gains potential energy by existing in the potential created by other charges
• Which plate is at a higher potential?
– Positive plate. Why?
• Since positive charge has the greatest potential energy.
– What happens to the positive charge if it is let go?
• It moves from higher potential to lower potential
– How about a negative charge?
• Its potential energy is higher on the negative plate. Thus, it moves from negative plate to
positive. Potential difference is the same for
a negative charge at the negative plate as a positive Zero point of electric potential
can be chosen arbitrarily.
charge at the positive plate.
• The unit of the electric potential is Volt (V).
• From the definition, 1V = 1J/C.
Often the ground, a conductor
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connected to Earth is zero.
Example 23 – 1
A negative charge: Suppose a negative charge, such
as an electron, is placed at point b in the figure. If the
electron is free to move, will its electric potential energy
increase or decrease? How will the electric potential
change?
e-
• An electron placed at point b will move toward the positive plate
since it was released at its highest potential energy point.
• It will gain kinetic energy as it moves toward left, decreasing its
potential energy.
• Note the electron is moving from point b at a lower potential to
point a at a higher potential. DV=Va-Vb>0.
• This is because the potential is generated by the charges on the
plates not by the electron.
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Electric Potential and Potential Energy
• What is the definition of the electric potential?
– The potential energy difference per unit charge
• How would you express the potential energy that a charge q
would obtain when it is moved between point a and b with
the potential difference Vba?
U b  U a  q Vb  Va   qVba
– In other words, if an object with charge q moves through a
potential difference Vba, its potential energy changes by qVba.
• So electric potential is
– A measure of how much energy an electric charge can acquire in a
given situation
– related to how much work a given charge can do.
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Comparisons of Potential Energies
• Let’s compare gravitational and electric potential energies
m
•
2m
What is the potential energy of each rock? •
– mgh and 2mgh
•
– +QVba and +2QVba
Which rock has a bigger potential energy? •
– The rock with a larger mass
•
Why?
– It’s got a bigger mass.
What is the potential energy of each charge?
Which object has a bigger potential energy?
– The object with a larger charge.
•
Why?
– It’s got a bigger charge.
The “potential” is the same but the heavier rock or larger charge can do a greater19work.
Some Typical Voltages
Sources
Thundercloud to ground
Approximate Voltage
108 V
High-Voltage Power Lines
Power supply for TV tube
Automobile ignition
106 V
104 V
104 V
Household outlet
Automobile battery
Flashlight battery
Resting potential across nerve membrane
102 V
12 V
1.5 V
10-1 V
Potential changes on skin (EKG and EEG)
10-4 V
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Electric Potential and Potential Energy
• The electric potential difference gives potential energy (or
the possibility to do work) based on the charge of the object.
• So what is happening in batteries or generators?
– They maintain a potential difference.
– The actual amount of energy used or transformed depends on how
much charge flows.
– How much is the potential difference maintained by a car’s
battery?
• 12Volts
– If for a given period, 5C charge flows through the headlight lamp,
what is the total energy transformed?
C*J/C=J (Joules)
• Etot=5C*12V=60 What is the unit?
– If it is left on twice as long? Etot=10C*12V=120J.
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Example 23 – 2
Electrons in a TV tube: Suppose an electron in the picture tube of a
television set is accelerated from rest through a potential difference
Vba=+5000V. (a) What is the change in potential energy of the
electron? (b) What is the speed of the electron (m=9.1x10-31kg) as a
result of this acceleration? (c) Repeat for a proton (m=1.67x10-27kg)
that accelerates through a potential difference of Vba=-5000V.
• (a) What is the charge of an electron?
–
e  1.6  1019 C
• So what is the change of its potential energy?


DU  qVba  eVba  1.6 1019 C  5000V   8.0  1016 J
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Example 23 – 2
• (b) Speed of the electron?
– The entire potential energy of the electron is transformed into
kinetic energy. Thus the equation is
1
DK  me ve2  0  W  DU 
2

(DU  eVba )

  8.0  1016 J  8.0  1016 J
ve 
2  DK

me
2  8.0 1016
7

4.2

10
m/ s
31
9.1 10
• (c) Speed of a proton that accelerates through V=-5000V?
1
DK  m p v 2p  0  W  DU   e   Vba   eVba  8.0 1016 J
2
2  8.0 1016
2  eVba
5
vp 

9.8

10
m/ s

27
mp
1.67 10
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Electric Potential and Electric Field
• The effect of a charge distribution can be
described in terms of electric field or electric
potential.
– What kind of quantities are the electric field and the
electric potential?
• Electric Field: Vector
• Electric Potential: Scalar
– Since electric potential is a scalar quantity, it often
can make problem solving easier.
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Electric Potential and Electric Field
• The potential energy is an (independent of path) function
expressed in terms of a (conservative) force.
r
b r
U b  U a   F  dl
a
• The potential difference is the potential energy difference
per unit charge

r
r
b F
Ub  U a

Vba  Vb  Va 
 dl  
a q
q


b
a
r r
E  dl
– This formula can be used to determine Vba when the electric field is
given.
• When the field is uniform
Vb  Va  

b
a
r r
E  dl   E

b
a
dl   Ed
Unit of the electric field in terms of potential?
V/m
or
Vba   Ed
Can you derive this from N/C?
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Example 23 – 3
Uniform electric field obtained from voltage:
Two parallel plates are charged to a voltage of
50 V. If the separation between the plates is
5.0 cm, calculate the magnitude of the electric
field between them, ignoring any fringe effects.
5cm
50V
What is the relationship between electric field and the
potential for a uniform field?
V   Ed
Solving for E
50V
V
50V
 1000V / m


E
2
d
5.0cm 5  10 m
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Electric Potential due to Point Charges
• What is the electric field due to a point charge Q at a
distance r?
1 Q
Q
k 2
E
2
4 0 r
r
• Electric potential due to the field E for moving from point ra
to rb away from the charge Q is
r
rb r
rb rˆ
Q
Vb  Va   E  dl  
ˆ

rdr

2
ra
4 0 ra r


Q
4 0


rb
ra
1
Q 1 1
dr 
  
2
4 0  rb ra 
r
Notice how the integral is carried out in the radial direction.
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Electric Potential due to Point Charges
• Since only the differences in potential have physical
meaning, we can choose Vb  0 at rb   .
• The electrical potential V at a distance r from a single
point charge is
1 Q
V
4 0 r
• So the absolute potential from a single point charge
depends only on the magnitude of the point charge
and the distance from it
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Properties of the Electric Potential
• What are the differences between the electric potential and
the electric field?
1 Q
– Electric potential
V
4 0 r
• Electric potential energy per unit charge
• Inversely proportional to the distance
• Simply add the potential from each of the charges to obtain the total potential
from multiple charges, since potential is a scalar quantity
r
1 Q
– Electric field
E 
2
4

r
0
• Electric force per unit charge
• Inversely proportional to the square of the distance
• Need vector sums to obtain the total field from multiple charges
• Potential for a positive charge is large near the charge and
decreases to 0 at large distances.
• Potential for the negative charge is small (large magnitude but
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negative) near the charge and increases with distance to 0
Shape of the Electric Potential
• So, how does the electric potential look like as a function of
distance?
– What is the formula for the potential by a single charge?
1 Q
V
4 0 r
Positive Charge
Negative Charge
A uniformly charged sphere would have the same potential as a single point charge.
What does this mean?
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Uniformly charged sphere behaves like all the charge is on the single point in the center.
Example 23 – 6
Work to bring two positive charges close together: What
minimum work is required by an external force to bring a
charge q=3.00 μC from a great distance away ( r   ) to a
point 0.500 m from a charge Q=20.0 μC?
What is the work done by the electric field in terms of potential
energy and potential?
q Q Q
W  qVba  
  
4 0  rb ra 
Since rb  0.500m, ra  

q Q
q Q
W

0




4 0  rb
4 0 rb

8.99  10


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we obtain


N  m 2 C 2  3.00  106 C 20.00  106 C
0.500m
  1.08J
In other words, the external force must input work of +1.08J to bring the charge
3.00C from infinity to 0.500m from the 20.0C charge.
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