Transcript Lecture 4 Electric potential

Lecture 4 Work, Electric Potential
and Potential Energy Ch. 25
• Topics
• Work, electric potential energy and electric potential
• Calculation of potential from field
• Potential from a point charge
• Potential due to a group of point charges, electric dipole
• Potential due to continuous charged distributions
• Calculating the electric field from a potential
• Electric potential energy from a system of point charges
• Equipotential Surface
• Potential of a charged isolated conductor
• Demos
• Teflon and silk
• Charge Tester, non-spherical conductor, compare charge density at
different radii.
•Elmo
•Potential in the center of four charges
•Potential of a electric dipole
• Polling
1
f
r r f r r
W   F  ds   qE  ds
i
i
U  W
2
U
V 
q
3
Work, Potential Energy and Electric potential
• The electric force is mathematically the same as gravity so it too must
be a conservative force. We want to show that the work done is
independent of the path and only depends on the endpoints. Then
the force is said to be a conservative force.
• First start with work
Work done by the electric
forcef =
f
r r
r r
W   F  ds   qE  ds
i
i
• Then we will find it useful to define a potential energy as is the case
for gravity.
U  W
• And the electric potential
U
V 
q
4
Lets start with a uniform electric field and
find the work done
for a positive test charge.
f
r r
W   qE  ds  qE cos s
i
P2
a
b
P3
c
E
P1
5
Find work done along path W12 for a positive test
charge
f
r
r
W   qE  ds  qE cos s
i
b
P2
P3
F
a
c
E
P1
W12  qE cos 90 a
W12  0
6
Find Work along path W23
f
r r
W   qE  ds  qE cos s
i
P2
F
a
b
P3
c
E
P1
W23  qE cos 0 b
W23  qEb
7
W12 + W23 = 0 + qEb =qEb
Compare this work done along path W13
8
Work done along path W13
f
r r
W   qE  ds  qE cos s
i
b
P2

P3
c
a

F
P1
b
W13  qEcos c  qE c  qEb
c
9
Conclusion
• Work
done along path W12 + W23 = W13.
• Work is independent of the particular path.
• Although we proved it for a uniform field, it is
true for any field that is a only a function of r and
is along r.
• It only depends on the end points i and f.
•This means we can define a function at every
point in space and when we take the difference of
that function between any two points, it is equal
to the negative of the work done.
10
b
P2
P3
c
a

F
P1
When we go from P1 to P3 we evaluate the Work function at
P3 and subtract the value at P1 and then the a difference
equals the negative of the work done in going form P1 to P3.
This function is called the potential energy function
11
Example of finding the Potential Energy
Function U in a Uniform Field
What is the electric potential difference for a unit positive
charge moving in an uniform electric field from a to b?
E
E
d
a
U  Ub  Ua
b
b
b
r r
U  Ub  U a  W    qE  ds  Eq  dx  Eq(xb  xa )
a
a
If we set the origin at xb = 0, and measure from b to a, then
U  qEd
Ub=0 and the potential energy function is U=qEd
This is analogy with gravitation where we U =mgh.
12
Now define the Electric Potential Difference V
which does not depend on charge.
U
V 
q
U  W
W
V 
q0
• The potential difference is the negative of the work done
per unit charge by an electric field on a positive unit charge
when it moves from one point to another.
13
Find the potential difference V for a
uniform electric field
U
V 
q
U  qEd
qEd
V 
q
V  Ed
For a battery of potential difference of 9 volts you
would say that the positive terminal is 9 volts above
the negative terminal.
14
Note relationship between
potential and electric field
W
F
V 
    ds    E  ds
q0
q0
dV  Edx
E  dV / dx
• V is a scalar not a vector. Simplifies solving problems.
• We are free to choose V to be 0 at any location. Normally
V is chosen to be 0 at the negative terminal of a battery or
0 at infinity for a point charge.
15
Example for a battery in a circuit
• In a 9 volt battery, typically used in IC circuits, the
positive terminal has a potential 9 v higher than the
negative terminal. If one micro-Coulomb of positive
charge flows through an external circuit from the positive
to negative terminal, how much has its potential energy
been changed?
q
16
Generalize concept of electric
potential energy and potential
difference for any electric field
17
f
U  U f  Ui = - Work done by the electric force =
  F  ds
i
y
V 
U
q
x
V  Vf  Vi    E  ds (independent of path, ds)
Therefore, electric force is a conservative force.
18
Find the electric potential when moving from
one point to another in a field due to a point
charge?
V    E  dr
kq
E  2 r̂
r
f
Vf  Vi    E  dr
i
19
Potential of a point charge at a distance R
f
Vf  Vi    E  dr

i


1
1
1 1
V f  Vi    E  dr  kq cos 0  2 dr  kq
 kq(  )
r
rR
 R
R
R
q
V f  Vi  0  Vi  k
R
kq
V
R
1
k
4 0
Replace R with r
1
q
V
4 0 r
eqn 25-26
20
Electric potential for a positive point charge
kq
V (r) 
r
r  x 2  y2
• V is a scalar
• V is positive for positive charges, negative for negative charges.
• r is always positive.
• For many point charges, the potential at a point in space is the
simple algebraic sum (Not a vector sum)
21
Electric potential due to a positive point charge
Hydrogen atom.
• What is the electric potential at a distance of 0.529 A from
the proton? 1A= 10-10 m
kq 
V

2
Nm
8.99  10
R
V  27.2
9
 1.6  10
C 
2
19
C
r = 0.529 A
.529  10 10 m
J
 27.2Volts
C
•What is the electric potential energy of the electron
at that point?
U = qV= (-1.6 x 10-19 C) (27.2 V)
= - 43.52 x 10-19 J
or - 27.2 eV where eV stands for electron volt.
Note that the total energy E of the electron in the ground state
of hydrogen is - 13.6 eV
Also U= 2E = -27.2 eV. This agrees with above formula.
22
What is the electric potential due to several
point charges?
For many point charges, the potential at a point in space is
the simple algebraic sum (Not a vector sum)
kqi
V 
ri
i
y
q2
q1
r1
 q1 q 2 q 3 
V  k   
 r1 r 2 r 3 
r2
r3
q3
x
23
Four Point charges
24
Two point charges that are opposite and equal
What is the potential
due to a dipole?
25
Potential for a Continuous Distribution of
Charge
Point charge
V
kq
r
For an element of charge
Integrate
V

dq, dV 
kdq
r
kdq
r
26
Chaper 24 Problem 22. With V = 0 at infinity, what is the
electric potential at P, the center of curvature of the uniformly
charged nonconducting rod?
27
Chapter 24 Problem 26. What is the magnitude of the net electric
potential at the center?
1. A thin rod of charge -3.0 µC that forms a full circle of radius 6.0 cm
2. A second thin rod of charge 2.0 µC that forms a circular arc of radius
4.0 cm, subtending an angle of 90° about the center of the full circle
3. An electric dipole with a dipole moment that is perpendicular to a
radial line and that has magnitude 1.28 multiplied by 10-21 C·m
28
Figure 24-44 shows a thin plastic rod of length L and
uniform positive charge Q lying on an x axis. With V =
0 at infinity, find the electric potential at point P1 on the
axis, at distance d from one end of the rod.
29
Potential due to a ring of charge
• Direct integration. Since V is a scalar, it is easier to evaluate V
than E.
• Find V on the axis of a ring of total charge Q. Use the formula
for a point charge, but replace q with elemental charge dq and
integrate.
kq
Point charge V 
r
For an element of charge dq, dV 
kdq
r
r is a constant as we integrate.
V



kdq
r
kdq
(z 2  R 2 )
k
(z  R )
2
2
 dq
V
k
(z  R )
This is simpler than finding E because V
is not a vector.
2
2
Q
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Potential due to a line charge
We know that for an element of charge dq
dq
the potential is
dV  k
r
For the line charge let the charge density be .
Then dq=dx
So, dV  k
dx
r
Then , dV  k
But, r  x2  d2
dx
x 2  d2
Now, we can find the total potential V produced by the rod at point P by
integrating along the length of the rod from x=0 to x=L
L
L
V   dV   k
0
0
dx
L
x d
2
2
 k 
0
dx
x d
2
So, V  k(ln( L  L2  d2 )  ln d)
 V  k ln(x  x  d )
2
2
2
L
0
 L  L2  d 2 
or, V  k  ln 

d


31
A new method to find E if the potential is known.
If we know V, how do we find E?
V    E  ds
ˆ  ĵdy  k̂dz
ds  idx
E  Exiˆ  Ey ĵ  Ez k̂
dV  E  ds
dV  E x dx  E y dy  E z dz
dV
dx
dV
Ey  
dy
dV
Ez  
dz
Ex  
So the x component of E is the derivative of V with respect to x, etc.
–If V = a constant, then Ex = 0. The lines or surfaces on which V
remains constant are called equipotential lines or surfaces.
–See example on next slide
32
Now find the electric field at point P1 on the axis, at
distance d from one end of the rod. Find the x and y
components Ex and Ey.
33
Equipotential Surfaces
• Three examples
• What is the equipotential surface and equipotential
volume for an arbitrary shaped charged conductor?
• See physlet 9.3.2 Which equipotential surfaces fit the
field lines?
34
x
Blue lines are the electric field lines
Orange dotted lines represent the equipotential surfaces
c)
Electric Dipole
(ellipsoidal concentric shells)
b)
Point charge
(concentric shells)
a)
Uniform E field
E  Ex , Ey  0, Ez  0
dV
dx
V  Ex d
Ex  
V = constant in y and z
directions
35
Electric Potential Energy U of a system of
charges
How much work is required to set up the arrangement of
Figure 24-46 if q = 3.20 pC, a = 54.0 cm, and the particles are
initially infinitely far apart and at rest?
q1
q2
W=U
U  k(
q4
q3
q1q2 q1q3 q1q4 q2 q3 q2 q4 q3q4





)
r12
r13
r14
r23
r24
r34
U  k(
qq qq qq qq qq qq




 )
a
a
2a a
2a a
4q 2 2q 2
U  k(

)
a
2a
36
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
37
Dielectric Breakdown: Application of Gauss’s Law
If the electric field in a gas exceeds a certain value, the gas
breaks down and you get a spark or lightning bolt if the gas
is air. In dry air at STP, you get a spark when
4 V
E  3  10
cm
V = constant on surface of conductor
Radius r2
r1
1
2
38
This explains why:
•
Sharp points on conductors have the highest electric fields and cause
corona discharge or sparks.
•
Pick up the most charge with charge tester from the pointy regions of the
non-spherical conductor.
•
Use non-spherical metal conductor charged with teflon rod. Show
variation of charge across surface with charge tester.
Radius R
V = constant on surface of conductor
1
+
Cloud
+
+
+
-
-
-
2
Van de Graaff
-
39
How does a conductor shield the interior from an
exterior electric field?
• Start out with a uniform electric field
with no excess charge on conductor.
Electrons on surface of conductor adjust
so that:
1. E=0 inside conductor
2. Electric field lines are perpendicular
to the surface. Suppose they weren’t?
s
3. Does E =
 0 just outside the conductor
4. Is s uniform over the surface?
5. Is the surface an equipotential?
6. If the surface had an excess charge, how would your answers change?
40
What is the electric potential of a uniformly charged circular disk?
We can treat the disk as a set of ring charges.
The ring of radius R’ and thickness dR’ has an
area of 2R’dR’ and it’s charge is dq = sdA =
s2R’)dR’ where s=Q/(R2), the surface
charge density. The potential dV at point P due
to the charge dq on this ring given by
dV 

q
kdq
(z 2  (R')2 )
dV 
ks 2 R'dR'
(z 2 _(R')2 )
Integrating R’ from R’=0 to R’=R
R
V  ks 2 
0
R'dR'
(z 2  (R')2 )
 V  2ks ( z  R  z)
2
2
41
Chapter 24 Problem 19
The ammonia molecule NH3 has a permanent electric dipole
moment equal to 1.31 D, where 1 D = 1 debye unit = 3.34
multiplied by 10-30 C·m. Calculate the electric potential due to an
ammonia molecule at a point 44.0 nm away along the axis of the
dipole. (Set V = 0 at infinity.)
42
Chapter 24 Problem 55
Two metal spheres, each of radius 1.0 cm, have a center-to-center
separation of 2.2 m. Sphere 1 has charge +2.0 multiplied by 10-8 C.
Sphere 2 has charge of -3.8 multiplied by 10-8 C. Assume that the
separation is large enough for us to assume that the charge on each
sphere is uniformly distributed (the spheres do not affect each other).
Take V = 0 at infinity.
(a) Calculate the potential at the point halfway between the centers.
(b) Calculate the potential on the surface of sphere 1.
(c) Calculate the potential on the surface of sphere 2.
43
Chapter 24 Problem 57
A metal sphere of radius 11 cm has a net charge of 2.0 multiplied
by 10-8 C.
(a) What is the electric field at the sphere's surface?
(b) If V = 0 at infinity, what is the electric potential at the sphere's
surface?
(c) At what distance from the sphere's surface has the electric
potential decreased by 500 V?
44