Download Electric Potential

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Electric Potential
Chapter 24
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What is Physics?
How can we account for the force? The Electric Field (E)
Where does the energy come from when released? Potential Energy (U)
How do we account for U? Electric Potential (V)
Our goals in this chapter are to:
1. define electric potential
2. discuss how to calculate it for various arrangements of charged particles and objects
3. discuss how electric potential V is related to electric potential energy U.
Electric Potential
We are going to define the electric potential (or potential for
short) in terms of electric potential energy, so our first job is
to figure out how to measure that potential energy.
𝑈 = −𝑊
Where work W is done by the field and 𝑈 = 0 at ∞ .
The electric potential V at a point P in the electric field
of a charged object is
V
W U

q0
q0
where W∞ is the work that would be done by the electric
force on a positive test charge q0 were it brought from an
infinite distance to P, and U is the electric potential energy
that would then be stored in the test charge–object system.
(a)
(b)
A test charge has been
brought in from infinity to point
P in the electric field of the
rod.
We define an electric potential
V at P based on the potential
energy of the configuration in
(a).
Electric Potential
If a particle with charge q is placed at a point where
the electric potential of a charged object is V, the
electric potential energy U of the particle–object
system is
U  q0V
(a) A test charge has been brought
in from infinity to point P in the
electric field of the rod.
(b) We define an electric potential V
at P based on the potential
energy of the configuration in (a).
Electric Potential
2 Cautions
• Potential Energy ≠ Potential
• Electric Potential is a scalar (and there was much rejoicing)
Electric Potential
Change in Electric Potential. If the particle moves through a potential
difference ΔV, the change in the electric potential energy is
U E  qV  q V f  Vi 
Work by the Field. The work W done by the electric force as the particle moves
from i to f:
W  U E   qV   q V f  Vi 
Electric Potential
Conservation of Energy. If a particle moves through a change ΔV in electric
potential without an applied force acting on it, applying the conservation of
mechanical energy gives the change in kinetic energy as
K   qV   q V f  Vi 
Work by an Applied Force. If some force in addition to the electric force acts
on the particle, we account for that work in this way,
K  U  Wapp   qV  Wapp
Electron-volts. A more convenient unit of energy for subatomic measures is
defined as the work required to move a single elementary charge e through a
potential difference of 1 volt.
1 𝑒𝑉 ≡ 𝑒(1𝑉)
Sample 24.01
Equipotential Surfaces and the Electric Field
Adjacent points that have the same electric potential form an equipotential
surface, which can be either an imaginary surface or a real, physical surface.
Figure shows a family of equipotential surfaces associated with the electric field due to
some distribution of charges. The work done by the electric field on a charged particle
as the particle moves from one end to the other of paths I and II is zero because each
of these paths begins and ends on the same equipotential surface and thus there is no
net change in potential.
Equipotential Surfaces and the Electric Field
The work done as the charged particle moves from one end to the other of paths III
and IV is not zero but has the same value for both these paths because the initial and
final potentials are identical for the two paths; that is, paths III and IV connect the
same pair of equipotential surfaces.
Equipotential and Electric Field Lines
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Calculating the Potential from the Field
Link to APlusPhysics.com
The differential work 𝑑𝑊 done on a particle by a force 𝐹
during a displacement 𝑑 𝑠 is
𝑑𝑊 = 𝐹 ∙ 𝑑 𝑠
or
𝑑𝑊 = 𝑞0 𝐸 ∙ 𝑑 𝑠
To find the total work 𝑊 done on the particle by the field
from point 𝑖 to 𝑓, sum or integrate the differential works.
𝑓
𝑊 = 𝑞0
𝐸 ∙ 𝑑𝑠
𝑖
Since 𝑊 = −𝑞∆𝑉, the electric potential difference
between two points 𝑖 and 𝑓 is
𝑓
𝑉𝑓 − 𝑉𝑖 = −
𝐸 ∙ 𝑑𝑠
𝑖
A test charge q0 moves from
point i to point f along the path
shown in a non-uniform electric
field. During a displacement
ds, an electric force q0 E acts
on the test charge. This force
points in the direction of the
field line at the location of the
test charge.
Calculating the Potential from the Field
The electric potential difference between two points 𝑖 and 𝑓 is
𝑓
𝑉𝑓 − 𝑉𝑖 = −
𝐸 ∙ 𝑑𝑠
𝑖
where the integral is taken over any path connecting the points. If the integration is
difficult along any particular path, we can choose a different path along which the
integration might be easier. If we choose Vi = 0, we have, for the potential at a
particular point,
𝑓
𝑉=−
𝐸 ∙ 𝑑𝑠
𝑖
Calculating the Potential from the Field
In a uniform field of magnitude 𝐸, the change in potential from a higher
equipotential surface to a lower one, separated by distance Δx, is
V   E x
Calculating the Potential from the Field
Applying this to a uniform field,
𝑉𝑓 − 𝑉𝑖 = −
𝑓
𝐸 ∙ 𝑑𝑠
𝑖
If we move along a path parallel to the field lines,
the angle between 𝐸 and 𝑑𝑠 is zero,
𝐸 ∙ 𝑑 𝑠 = 𝐸 𝑑𝑠 cos 0 = 𝐸 𝑑𝑠
So,
𝑓
𝑉𝑓 − 𝑉𝑖 = −𝐸
𝑑𝑠
𝑖
In a uniform field of magnitude 𝐸, the change in
potential from a higher equipotential surface to a
lower one, separated by distance Δx, is
V   E x
Sample 24.02
Potential due to a Charged Particle
We find the potential by moving a test charge q0 from P to infinity. The test charge is
shown at distance r from the particle, during differential displacement ds.
We know that the electric potential
difference between two points i and f is

V f  Vi    E dr
For radial path
R
1
q
E
4 0 r 2
The magnitude of the electric field at
the site of the test charge
We set Vf =0 (at ∞)
and Vi =V (at R)
0 V  
q
4 0
Solving for V and switching R to r,
we get


R
1
q
dr

r2
4 0
1 q
V
4 0 r

1
 r 
R
Potential due to a Group of Charged Particles
The potential due to a collection of charged particles is
n
V = åVi =
i =1
n
qi
å
4pe 0 i =1 ri
1
[n charged particles]
Thus, the potential is the scalar sum of the individual potentials, with no
consideration of directions.
Answer: Same net potential (a)=(b)=(c)
Sample Problem 24.03
Sample Problem 24.04
Potential due to a Electric Dipole
The net potential at P is given by
2
V   Vi  V(  )  V(  )
i 1
1  q q 




4 0  r(  ) r(  ) 
q  r(  )  r(  ) 
V


4 0  r(  ) r(  ) 
(a) Point P is a distance r from the
midpoint O of a dipole. The line OP
makes an angle θ with the dipole axis.
(b) If P is far from the dipole, the lines of
lengths r(+) and r(-) are approximately
parallel to the line of length r, and the
dashed black line is approximately
perpendicular to the line of length r(-).
Potential due to a Electric Dipole
We can approximate the two lines to P as being
parallel and their length difference as being the
leg of a right triangle with hypotenuse d (Fig.
b). Also, that difference is so small that the
product of the lengths is approximately r 2.
r(  )  r(  )  d cos 
r(  ) r(  )  r 2
We can approximate V to be
q d cos 
V
4 0 r 2
where θ is measured from the dipole axis
as shown in Fig. a. And since 𝑝 = 𝑞𝑑, we
have
V
p cos 
4 0 r 2
1
[electric dipole]
(a) Point P is a distance r from the
midpoint O of a dipole. The line OP
makes an angle θ with the dipole axis.
(b) If P is far from the dipole, the lines of
lengths r(+) and r(-) are approximately
parallel to the line of length r, and the
dashed black line is approximately
perpendicular to the line of length r(-).
Potential due to a Continuous Charge Distribution
For a continuous distribution of charge (over an extended object), the potential
is found by
(1) dividing the distribution into charge elements dq that can be treated as
particles and then
(2) summing the potential due to each element by integrating over the full
distribution:
1
dq
V   dV 
4 0  r
We now examine two continuous charge distributions, a line and a disk.
Potential due to a Line of Charge
Fig. (a) has a thin conducting rod of length L. As shown in Fig. (b), the element
of the rod has a differential charge of dq   dx
This element produces an electric potential dV at point P (fig c) given by
1 q
V
4 0 r
dV 
dq
1
 dx

1
4 0 r
4 0  x 2  d 2  2
1
Potential due to a Line Charge
We now find the total potential V produced by the rod at point P by integrating dV
along the length of the rod, from x = 0 to x = L (Figs.d and e)
V   dV  
L
0
1
Where
𝑥2
So,
𝑉=
±
𝑎2
𝑑𝑉 =
1

4 0  x 2  d 2 
𝑑𝑥 = ln 𝑥 + 𝑥 2 ± 𝑎2
1
2

dx 
4 0
0
x  d
2
2
(Integral 17 in Appendix E)
𝜆
𝐿
ln 𝑥 + 𝑥 2 + 𝑑 2
4𝜋𝜖0
0
𝜆
𝑉=
ln 𝐿 + 𝐿2 + 𝑑 2 − ln 𝑑
4𝜋𝜖0
Which simplifies to,

dx
L
 L   L2  d 2 

V
ln 
4 0 
d

1
2





1
2
Potential due to a Charged Disk
In figure, consider a differential element consisting of a
flat ring of radius R’ and radial width dR’. Its charge has
magnitude
dq   A    2 R  dR 
in which (2𝜋𝑅’)(𝑑𝑅’) is the upper surface area of the
ring. The contribution of this ring to the electric potential
at P is
dq
1   2 R  dR 
dV 

1
4 0 r
4 0  z 2  R2  2
1
A plastic disk of radius R, charged
on its top surface to a uniform
surface charge density σ. We wish
to find the potential V at point P on
the central axis of the disk.
Potential due to a Charged Disk
We find the net potential at P by adding (via integration)
the contributions of all the rings from R’=0 to R’=R:
V   dV  
R
0

V
2 0

1   2 R  dR 
z  R
2
 z  R
4 0
2
2

1
2
z
2

1
2


2 0

R
0
R  dR 
 z  R
2
2

1
2

Note that the variable in the second integral is R’
and not z
A plastic disk of radius R, charged
on its top surface to a uniform
surface charge density σ. We wish
to find the potential V at point P on
the central axis of the disk.
Calculating the Field from the Potential
Suppose that a positive test charge q0 moves
through a displacement ds from one equipotential
surface to the adjacent surface. The work the
electric field does on the test charge during the
move is – 𝑞0𝑑𝑉. On the other hand the work done
by the electric field may also be written as the
scalar product (𝑞0𝑬) ∙ 𝑑𝒔. Equating these two
expressions for the work yields
or,
Since 𝐸cos𝜃 is the component of E in the
direction of ds, we get,
A test charge q0 moves a distance ds
from one equipotential surface to
another. (The separation between the
surfaces has been exaggerated for
clarity.) The displacement ds makes
an angle θ with the direction of the
electric field E.
Electric Potential Energy of a System of Charged Particles
The electric potential energy of a system of
charged particles is equal to the work needed to
assemble the system with the particles initially
at rest and infinitely distant from each other. For
two particles at separation r,
1 q1q2
U
4 0 r
[2-particle system]
Two charges held a fixed
distance r apart.
Sample Problem 24.06
U
1 q1q2
4 0 r
[2-particle system]
Three charges are fixed at the
vertices of an equilateral triangle.
What is the electric potential
energy of the system?
Potential of a Charged Isolated Conductor
(a) A plot of V(r) both
inside and outside a
charged spherical
shell of radius 1.0 m.
(a) A plot of E(r) for the
same shell.
Potential of a Charged Isolated Conductor
It is wise to enclose yourself in a cavity
inside a conducting shell, where the
electric field is guaranteed to be zero. A car
(unless it is a convertible or made with a
plastic body) is almost ideal.
Summary
Mechanical Energy
Electric Potential
• The electric potential V at point P in
the electric field of a charged
object:
• Applying the conservation of
mechanical energy gives the
change in kinetic energy:
Eq. 24-2
Electric Potential Energy
• Electric potential energy U of the
particle-object system:
Eq. 24-3
• If the particle moves through
potential ΔV:
Eq. 24-4
Eq. 24-9
• In case of an applied force in a
particle
Eq. 24-11
• In a special case when ΔK=0:
Eq. 24-12
Finding V from E
• The electric potential difference
between two point I and f is:
Eq. 24-18
Summary
Potential due to a Charged
Particle
• due to a single charged particle at a
distance r from that particle :
Potential due to a Continuous
Charge Distribution
• For a continuous distribution of
charge:
Eq. 24-26
• due to a collection of charged
particles
Eq. 24-27
Potential due to an Electric
Dipole
• The electric potential of the dipole is
Eq. 24-30
Eq. 24-32
Calculating E from V
• The component of E in any
direction is:
Eq. 24-40
Electric Potential Energy of a
System of Charged Particle
• For two particles at separation r:
Eq. 24-46
Concept Check – Electric Potential
Two test charges are brought separately into the vicinity of a charge
+Q. First, test charge +q is brought to point A a distance r from +Q.
Next, +q is removed and a test charge +2q is brought to point B a
distance 2r from +Q. Compared with the electrostatic potential of the
charge at A, that of the charge at B is
1. greater.
2. smaller.
3. the same.
Concept Check – Electric Potential
Two test charges are brought separately into the vicinity of a charge
+Q. First, test charge +q is brought to point A a distance r from +Q.
Next, +q is removed and a test charge +2q is brought to point B a
distance 2r from +Q. Compared with the electrostatic potential of the
charge at A, that of the charge at B is
1. greater.
2. smaller.
3. the same.
The electrostatic potential at A is VA=kQ/r, whereas at point B it is VB=kQ/2r.
The magnitudes of the charges q and 2q at A and B do not enter into the
expression for the electrostatic potential
Concept Check – Electrostatic Potential Energy
Two test charges are brought separately into the vicinity of a charge
+Q. First, test charge +q is brought to a point a distance r from +Q.
Then this charge is removed and test charge –q is brought to the same
point. The electrostatic potential energy of which test charge is greater:
1. +q
2. –q
3. It is the same for both.
Concept Check – Electrostatic Potential Energy
Two test charges are brought separately into the vicinity of a charge
+Q. First, test charge +q is brought to a point a distance r from +Q.
Then this charge is removed and test charge –q is brought to the same
point. The electrostatic potential energy of which test charge is greater:
1. +q
2. –q
3. It is the same for both.
The electrostatic potential energy of a test charge depends on both its position
and its charge. For the +q charge, the electrostatic potential energy equals
Qq/r, whereas for the –q charge it equals –Qq/r
Concept Check – Electric Potential
An electron is pushed into an electric field where it acquires a 1-V
electrical potential. Suppose instead that two electrons are pushed the
same distance into the same electric field. The electrical potential of the
two electrons is
1. 0.25 V.
2. 0.5 V.
3. 1 V.
4. 2 V.
5. 4 V.
Concept Check – Electric Potential
An electron is pushed into an electric field where it acquires a 1-V
electrical potential. Suppose instead that two electrons are pushed the
same distance into the same electric field. The electrical potential of the
two electrons is
1. 0.25 V.
2. 0.5 V.
3. 1 V.
4. 2 V.
5. 4 V.
The electrostatic potential of a charged particle in an electric
field depends only on the position of the particle, not on its charge.
Concept Check – Electric Potential
What is the electric potential at point B?
1. V > 0
2. V = 0
3. V < 0
A
B
Concept Check – Electric Potential
What is the electric potential at point B?
1. V > 0
2. V = 0
3. V < 0
A
B
Since Q2 and Q1 are equidistant from point B, and since they
have equal and opposite charges, then the total potential is zero.
Concept Check – Electric Potential (2)
What is the electric potential at point A?
1. V > 0
2. V = 0
3. V < 0
A
B
Concept Check – Electric Potential (2)
What is the electric potential at point A?
1. V > 0
2. V = 0
3. V < 0
A
B
Since Q2 (which is positive) is closer to point A than Q1 (which is
negative), and since the total potential is equal to V1 + V2, then
the total potential is positive.
Electric Potential
F
PEf
m
B
PEi
A
Fg
F
PEi
E
per unit charge 
B
A
mg
Fe
Gravity Field
 PE
qo
PEf
qoE
Electric Field
Electric Potential 
UE
qo
The Electron Volt
U E  qe V
1 eV=1.6×10-19 J
Electric Potential – Uniform Field
F
VB
+ B
VA
A
E
W  A  B   qo Ed
U E  qo Ed
VB  VA 
U E
qo
VB  VA   Ed
Electric Potential – Uniform Field
d
+
A
B
C
E
Topographical Maps
Topographical Maps
Topographical Maps
Equipotential Lines
Potential of Point Charge (+)
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9
8
7
6
5
4
3
2
1
0
Potential of Dipole
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8
6
4
2
0
-2
-4
-6
-8
-10
Potential of 2 Positive Charges
12
10
8
6
4
2
0
Concept Check – Electric Potential Energy
A proton and an electron are in a uniform electric field created by
oppositely charged plates. You release the proton from the positive
side and the electron from the negative side. Which feels the larger
electric force?
1.
2.
3.
4.
5.
proton
electron
both feel the same force
neither–there is no force
they feel the same magnitude force
but in opposite directions
electron
electron
-
+

E

E
proton
proton
Concept Check – Electric Potential Energy
A proton and an electron are in a uniform electric field created by
oppositely charged plates. You release the proton from the positive
side and the electron from the negative side. Which feels the larger
electric force?
1.
2.
3.
4.
5.
proton
electron
both feel the same force
neither–there is no force
they feel the same magnitude force
but in opposite directions
electron
electron
-
+

E

E
proton
proton
Since F = qE and the proton and electron have the same charge in
magnitude, they both experience the same force. However, the
forces point in opposite directions because the proton and
electron are oppositely charged.
Concept Check – Electric Potential Energy (2)
A proton and an electron are in a constant electric field created by
oppositely charged plates. You release the proton from the positive
side and the electron from the negative side. When it strikes the
opposite plate, which one has more KE?
1.
2.
3.
4.
5.
proton
electron
both acquire the same KE
neither – there is no change of KE
they both acquire the same KE but
with opposite signs
electron
electron
-
+

E

E
proton
proton
Concept Check – Electric Potential Energy (2)
A proton and an electron are in a constant electric field created by
oppositely charged plates. You release the proton from the positive
side and the electron from the negative side. When it strikes the
opposite plate, which one has more KE?
1.
2.
3.
4.
5.
proton
electron
both acquire the same KE
neither – there is no change of KE
they both acquire the same KE but
with opposite signs
electron
electron
-
+

E

E
proton
proton
Since PE = qV and the proton and electron have the same charge
in magnitude, they both have the same electric potential energy
initially. Because energy is conserved, they both must have the
same kinetic energy after they reach the opposite plate.
Concept Check – Equipotential of a Point Charge
Which two points have the same potential?
1.
2.
3.
4.
5.
A and C
B and E
B and D
C and E
no pair
A
C
B
E
Q
D
Concept Check – Equipotential of a Point Charge
Which two points have the same potential?
1.
2.
3.
4.
5.
A and C
B and E
B and D
C and E
no pair
A
C
B
E
Q
D
Since the potential of a point charge is: V  ko
q
r
only points that are at the same distance from charge Q are at the
same potential. This is true for points C and E. They lie on an
Equipotential Surface.
Concept Check – Electric Potential
Four point charges are arranged at the corners of a square. Find the
electric field E and the potential V at the center of the square.
1.
2.
3.
4.
5.
E=0 V=0
E=0 V≠0
E≠0 V≠0
E≠0 V=0
E = V regardless of the value
-Q
+Q
-Q
+Q
Concept Check – Electric Potential
Four point charges are arranged at the corners of a square. Find the
electric field E and the potential V at the center of the square.
1.
2.
3.
4.
5.
E=0 V=0
E=0 V≠0
E≠0 V≠0
E≠0 V=0
E = V regardless of the value
-Q
+Q
-Q
+Q
The potential is zero: the scalar contributions from the two
positive charges cancel the two minus charges. However, the
contributions from the electric field add up as vectors, and they
do not cancel (so it is non-zero).
Concept Check – Equipotential Surfaces
Which of these configurations gives V = 0 at all points on the x-axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
1)
+2mC
-2mC
x
-2mC
-1mC
x
+1mC
-1mC
2)
3)
4) all of the above
5) none of the above
Concept Check – Equipotential Surfaces
Which of these configurations gives V = 0 at all points on the x-axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
1)
+2mC
-2mC
x
-2mC
-1mC
x
+1mC
-1mC
2)
3)
4) all of the above
5) none of the above
Only in case (1), where opposite charges lie directly across the xaxis from each other, do the potentials from the two charges
above the x-axis cancel the ones below the x-axis.
Concept Check – Equipotential Surfaces
Which of these configurations gives V = 0 at all points on the y-axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
1)
+2mC
-2mC
x
-2mC
-1mC
x
+1mC
-1mC
2)
3)
4) all of the above
5) none of the above
Concept Check – Equipotential Surfaces
Which of these configurations gives V = 0 at all points on the y-axis?
+2mC
+1mC
+2mC
+1mC
x
-1mC
-2mC
1)
+2mC
-2mC
x
-2mC
-1mC
x
+1mC
-1mC
2)
3)
4) all of the above
5) none of the above
Only in case (3), where opposite charges lie directly across the yaxis from each other, do the potentials from the two charges
above the y-axis cancel the ones below the y-axis.
Concept Check – Work & Electric Potential
Which requires the most work, to move a positive charge from P to
points 1, 2, 3 or 4 ? All points are the same distance from P.
1.
2.
3.
4.
5.
P  1
P  2
P  3
P  4
all require the same amount of work
3
2
1
P

E
4
Concept Check – Work & Electric Potential
Which requires the most work, to move a positive charge from P to
points 1, 2, 3 or 4 ? All points are the same distance from P.
1.
2.
3.
4.
5.
P  1
P  2
P  3
P  4
all require the same amount of work
3
2
1
P
4

E
For path 1, you have to push the positive charge against the E
field, which is hard to do. By contrast, path #4 is the easiest, since
the field does all the work.
Concept Check – Work & Electric Potential (2)
Which requires zero work, to move a positive charge from P to points 1,
2, 3 or 4 ? All points are the same distance from P.
1.
2.
3.
4.
5.
P  1
P  2
P  3
P  4
all require the same amount of work
3
2
1
P

E
4
Concept Check – Work & Electric Potential (2)
Which requires zero work, to move a positive charge from P to points 1,
2, 3 or 4 ? All points are the same distance from P.
1.
2.
3.
4.
5.
P  1
P  2
P  3
P  4
all require the same amount of work
3
2
1
P
4

E
For path 3, you are moving in a direction perpendicular to the field
lines. This means you are moving along an equipotential, which
requires no work (by definition).