Transcript Chapter 23 – Electric Potential

Chapter 23 – Electric Potential
What is the value of employing the concept of energy when
solving physics problems?
Potential energy can be defined for conservative forces. Is
the electrostatic force conservative?
For conservative forces, the work done in moving an object
between two points is independent of the path taken.
b
W  F d
a
a
b
E
Work and Potential Energy for gravity
dW  dU
Work and Potential Energy
dW  F d
Electric Field Definition:
F
E  lim  F  qE
q 0 q
dW  qE d
dW  dU  qE d
Work Energy Theorem
a
b
b
a
a
 dU  q E d
b
E
Electric Potential Difference
b
b
a
a
 dU  q E d
b
Ub  Ua  q  E d
a
a
b
E
U b  Ua
  E d
q
a
b
U b  Ua
Vba  Vb  Va 
  E d
q
a
b
Definition:
What the heck is dl
dl
What it means:
• Potential Difference, Vb-Va is the work per
unit charge an external agent must perform
to move a test charge from ab without a
change in kinetic energy.
Ub  Ua
Wba
Vba  Vb  Va 

q
q
a
b
E
An example
Units of Potential Difference
Ub  Ua
Wba
Vba  Vb  Va 

q
q
 Joules   J 
 Coulomb    C   Volt  V
Because of this, potential difference is often referred to as “voltage”
In addition, 1 N/C = 1 V/m - we can interpret the electric field
as a measure of the rate of change with position of the electric
potential.
So what is an electron Volt (eV)?
Electron-Volts
• Another unit of energy that is commonly used
in atomic and nuclear physics is the electronvolt
• One electron-volt is defined as the energy a
charge-field system gains or loses when a
charge of magnitude e (an electron or a
proton) is moved through a potential
difference of 1 volt
– 1 eV = 1.60 x 10-19 J
Example
Through what potential difference would one need to
accelerate an electron in order for it to achieve a velocity
of 10% of the velocity of light, starting from rest?
(c = 3 x 108 m/s)
Conventions for the potential “zero point”
Ub  Ua
Wba
Vba  Vb  Va 

q
q
0
Choice 1: Va=0
Ub  Ua
Vb  Va 
q
“Potential”
0
Ub
Vb 
q
0
0
Choice 2: V  0
U b  U
Vb  Vb  V 
  E d
q

b
b
Ub
Vb 
  E d
q

Potential difference for a uniform
electric field
b
Vba  Vb  Va   E d
+Q
a
b
a
E  Eoˆj
d
d  dxiˆ  dyjˆ
-Q
d


d
Vba  Vb  Va   Eo ˆj dxiˆ  dyjˆ   Eody  Eod
0
Ub  Ua  qEod
0
Potential difference for a point charge
b
Vba  Vb  Va   E d
a
kq
E  2 rˆ
r
d  drrˆ  rdˆ  r sin dˆ
+Q
rb
rb
rb
kq
kq
dr
Vba  Vb  Va    2 rˆ drrˆ    2 dr  kq  2
r
r
r
ra
ra
ra
1 1
 1
Vba  Vb  Va  kq     kq   
 r  ra
 rb ra 
rb
dl for a point charge
Recall the convention for the potential
“zero point”
1 1
Vba  Vb  Va  kq   
 rb ra 
V  0
1 1
Vb  Vb  V  kq   
 rb  
kq
V r 
r
Equipotential surfaces are concentric spheres
Electric Potential of a Point
Charge
• The electric
potential in the
plane around a
single point charge
is shown
• The red line shows
the 1/r nature of the
potential
E and V for a Point Charge
• The equipotential lines
are the dashed blue
lines
• The electric field lines
are the brown lines
• The equipotential lines
are everywhere
perpendicular to the
field lines
Potential of a charged conductor
Given:
Find:
Spherical conductor
Charge=Q
Radius=R
V(r)
R
The plots for a metal sphere
Determining the Electric Field
from the Potential
dV  E ds  E ds
V
Ex  
x
V
Ey  
y
V
Ez  
z
dV
E 
ds
V ˆ V ˆ V ˆ
E
i
j
k
x
y
z
E  V
Superposition of potentials
V0  V1  V2  V3  ...
+Q1
+Q2
+Q3
r10
r20
0
r30
kQ1 kQ2 kQ3
V0 


 ...
r10
r20
r30
N
kQi
V0  
i 1 ri0
Electric potential due to
continuous charge distributions
kdq
dV 
r
kQ
V r 
r
Single charge
+Q1
+Q2
+Q3
Single piece of a charge distribution
+
+
r10
r20
0
r30
+
+
dq
dV
0
kQi
V0  
i 1 ri0
dq
Vk 
r
all charge
Discrete charges
Continuous charge distribution
N
Electric Potential for a
Continuous Charge Distribution
• Consider a small
charge element dq
– Treat it as a point
charge
• The potential at
some point due to
this charge element
is
dq
dV  ke
r
Electric field due to continuous
charge distributions
+Q1
+Q2
+Q3
kQ
E 0  2 rˆ
r
kdq
dE 0  2 rˆ
r
Single charge
Single piece of a charge distribution
r10
E 03
E 02
r20
0
r30
E 01
+
+
+
+
dq
dE0
0
N
Qi
E0  k  2 rˆi0
i 1 ri0
dq
E0  k 
rˆ
2
r
all charge
Discrete charges
Continuous charge distribution
Example: A ring of charge
dq  ds  Rd
d
+ 
a
r2  x2  R 2
+
dV 
kRd

+
+
kdq
dV 
r
x
x a
2
2
dV
+
+
+
V
2
ka
x a
2
2
 d 
0
k2a
x a
2
2

kQ
x a
2
2
Electric field from a ring of charge
dq  ds  Rd
d
+ 
a
r2  x2  R 2
+

+
+
x
+
+
+
dV
kQ
V
x2  a2
dV ˆ
E  V  
i
dx
E
x
kQx
2
a

2 3/ 2
ˆi
Example: Electric field of a charged
ring directly
+ 
a
r  x R
2
+
2
y-components cancel by symmetry
2

+
+
kdq
dE  2 rˆ
r
dq  ds  Rd
d
dEx
x
+
+
dE x 
k  ad
x
dE  2 2
x a
x2  a2
dE
dEy
kdq
cos 
2
r
+
E
2
kxa
x
2
a
 d 
3
2 2 0

kxa
x
2
a
3
2 2

 2 
kQx
x
2
a
3
2 2

Potential due to a charged disk
a
V
r
x
dV
kQ
x r
2
2
dV 
kdq
x2  r2
dq  dA  rdrd  2rdr
a
V
0
k2rdr
x r
2
2
a
 k2
0
rdr
x2  r2
V  k2  x 2  a 2  x 


Uniformly Charged Disk
E
a
kQx
x
r
2
3
2 2

r
x
dE 
dE
kxdq
x
2
r
3
2 2

dq  dA  rdrd  2rdr
dE 
kx2rdr
x
2
r
a
E
3
2 2

0
kx2rdr
x
2
r
3
2 2

a
 kx
0
x2 a2
2rdr
x
2
r
3
2 2

 kx

du
x
u
2
3
2
x2 a2
x2 a 2
 kx

x2
3

2
u du  kx
u


1
2
1
2



1
1 
x
 2kx 

  k2 1 

2
2
2
2
2
x

a
x
x

a




x2
Electric Dipole
p  Q2a
kQ k  Q
1 
r
1
V

 kQ  
  kQ
r
r  r
r  r  r 
 r r  r 
r  r
r  2a cos 
kQ2a cos  kp cos 
V

2
r
r2
Electric Potential of a Dipole
• The graph shows
the potential (y-axis)
of an electric dipole
• The steep slope
between the
charges represents
the strong electric
field in this region
E and V for a Dipole
• The equipotential lines
are the dashed blue
lines
• The electric field lines
are the brown lines
• The equipotential lines
are everywhere
perpendicular to the
field lines
Potential energy due to multiple
point charges
+Q2
+Q1
r21
+Q1
r23
r13
+Q3
kq1
V
r12
kq1q 2
U  q2V 
r12
r21
+Q2
kq
V r 
r
kq1 kq 2
V

r13
r23
kq1q 2 kq1q3 kq 2q3
U


r12
r13
r23
Irregularly Shaped Objects
• The charge density is high
where the radius of
curvature is small
– And low where the radius of
curvature is large
• The electric field is large
near the convex points
having small radii of
curvature and reaches
very high values at sharp
points
Problem P25.23
Show that the amount of
work required to assemble
four identical point charges
of magnitude Q at the
corners of a square of side
s is 5.41keQ2/s.
U  U1U 2 U 3 U 4
U  0  U 12  U 13  U 23   U 14  U 24  U 34 
U  0
U 
keQ 2 keQ 2  1
1
 keQ 2 



1

1
 1



s
s  2 
s 
2 
keQ 2 
kQ2
2
4
 5.41 e


s 
s
2
Example P25.33
An electron starts from rest 3.00 cm
from the center of a uniformly charged
insulating sphere of radius 2.00 cm
and total charge 1.00 nC. What is the
speed of the electron when it reaches
the surface of the sphere?
keeQ keqQ 1 2

 mv
r1
r2
2
v
 2  8.99  109

v


2keeQ  1 1 

m  r1 r2 
C 2 1.60  1019 C 109 C 

1
1



31
9.11 10 kg
 0.030 0 m 0.020 0 m 
N m
2
v  7.26  106 m s
Example P25.37
The potential in a region
between x = 0 and x = 6.00
At
m is V = a + bx, where a = 10.0 V and b = –7.00
V/m. Determine
(a) the potential at x = 0, 3.00 m, and 6.00 m, and
(b) the magnitude and direction of the electric field
at x = 0, 3.00 m, and 6.00 m.
x 0
V  10.0 V
x 3.00 m V  11.0 V
x 6.00 m V  32.0 V
E
dV
 b    7.00 V m   7.00 N C in the  x direction
dx
Example 25.43
A rod of length L (Fig. P25.43) lies along the x axis with its
left end at the origin. It has a nonuniform charge density
λ = αx, where α is a positive constant.
(a) What are the units of α?
(b) Calculate the electric potential at A.

C
1
C
          2
m
 x m m
V  ke
Figure P25.43
L
dq
 dx
xdx
L 


 ke
 ke 
 ke L  dln  1  

r
r
d x
d 

0