Transcript Lect09

Electric Potential
Q V
4pe0 R
Q
4pe0 r
R
r
C
B
r
a
B
q
r
A
A
path independence
a
Today…
• Conservative Forces and Energy Conservation
– Total energy is constant and is sum of kinetic and potential
• Introduce Concept of Electric Potential
– A property of the space and sources as is the Electric Field
– Potential differences drive all biological & chemical reactions,
as well as all electric circuits
• Calculating Electric Potentials
put V(infinity)=0
– Charged Spherical Shell
– N point charges
– Example: electric potential of a charged sphere
• Electrical Breakdown
– Sparks
– Lightning!!
Conservation of Energy of a particle from phys 1301
• Kinetic Energy (K)
– non-relativistic
• Potential Energy (U)
1 2
K  mv
2
– determined by force law
U ( x, y, z )
• for Conservative Forces: K+U is constant
– total energy is always constant
• examples of conservative forces
– gravity; gravitational potential energy
– springs; coiled spring energy (Hooke’s Law): U(x)=kx2
– electric; electric potential energy (today!)
• examples of non-conservative forces (heat)
– friction
– viscous damping (terminal velocity)
– electrical resistance
Example: Gravitational Force is conservative
(and attractive)
• Consider a comet in a highly elliptical orbit
U(r)
pt 2
pt 1
0
U(r1)
GMm
U (r )  
r
• At point 1, particle has a lot of potential
energy, but little kinetic energy
• At point 2, particle has little potential
energy, but a lot of kinetic energy
More
potential
energy
U(r2)
Less
potential
energy
Total energy = K + U
is constant!
Electric forces are conservative, too
• Consider a charged particle traveling
through a region of static electric field:
+
• A negative charge is attracted to
the fixed positive charge
• negative charge has more
potential energy and less kinetic
energy far from the fixed positive
charge, and…
• more kinetic energy and less
potential energy near the fixed
positive charge.
• But, the total energy is conserved
• We will now discuss electric
potential energy and the
electrostatic potential….
Electric potential and potential energy
• Imagine a positive test charge, Qo, in an external
electric field,
E  x, y, z  (a vector field)
• What is the potential energy, U(x,y,z) of the charge
in this field?
– Must define where in space U(x,y,z) is zero,
perhaps at infinity (for charge distributions that
are finite)
– U(x,y,z) is equal to the work you have to do to
take Qo from where U is zero to point (x,y,z)
Electric potential and potential energy
• Define
V  x, y, z  
U  x, y, z 
Q0
(U = QV)
• U depends on Qo , but V is independent of Qo
(which can be + or -)
• V(x,y,z) is the electric potential in volts
associated with E  x, y, z  . (1V = 1 J/c)
– V(x,y,z) is a scalar field
Electric potential difference
• Suppose charge q0 is moved from
pt A to pt B through a region of
space described by electric field E.
Fwe supply = -Felec
Felec
q0
A
E
B
• To move a charge in an E-field, we must supply a force just
equal and opposite to that experienced by the charge due to
the E-field.
• Since there will be a force on the charge due to E, a
certain amount of work WAB≡WAB will have to be done to
accomplish this task.
Electric potential difference, cont.
• Remember: work is force times distance
\

• To get a positive test charge from the lower potential to the
higher potential you need to invest energy - you need to do
work
• The overall sign of this: A positive charge would “fall” from
a higher potential to a lower one
• If a positive charge moves from high to low potential, it can
do work on you; you do “negative work” on the charge
Question 1
A
E
C
B
2) Points A, B, and C lie in a uniform electric field. What is the
potential difference between points A and B and A and C?
ΔVAB = VB - VA and VAC = VC - VA
a) ΔVAB > 0 and VAC > 0
b) ΔVAB = 0 and VAC < 0
c) VAB = 0 and VAC > 0
d) ΔVAB < 0 and VAC < 0
Question 1
A
E
C
B
2) Points A, B, and C lie in a uniform electric field. What is the
potential difference between points A and B and A and C?
ΔVAB = VB - VA and VAC = VC - VA
a) ΔVAB > 0 and VAC > 0
b) ΔVAB = 0 and VAC < 0
c) VAB = 0 and VAC > 0
d) ΔVAB < 0 and VAC < 0
Motion in a region of electric field
•A positive charge is released from rest in a region of electric
field
•The charge moves towards a region of lower potential
energy, i.e lower electrical potential
–Away from a positive charge (+ve electric field), you
have to do work to push them together
–Towards a negative charge (-ve electric field), you have
to do work to pull them apart
•To move without changing the electrical potential energy
–Must move perpendicular to the field ( F dl  0 )
Question 2
• A single charge ( Q = -1mC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
– What is the sign of the potential difference
between A and B? (Δ VAB  VB - VA )
(a) ΔVAB < 0 (b) Δ VAB = 0
B

-1mC
(c) Δ VAB > 0
A

x
Question 2
• A single charge ( Q = -1mC) is fixed at
the origin. Define point A at x = + 5m
and point B at x = +2m.
B

– What is the sign of the potential difference
between A and B? (Δ VAB  VB - VA )
(a) ΔVAB < 0 (b) Δ VAB = 0
A

-1mC
(c) Δ VAB > 0
• Imagine placing a positive charge at point A and determining which
way it would move. Remember that it will always “fall” to lower
potential.
•A positive charge at A would be attracted to the -1mC charge;
therefore NEGATIVE work would be done to move the charge from A
to B.
•You can also determine the sign directly from the definition:
Since
, ΔVAB <0 !!
x
Δ VAB is Independent of Path
q0
E
A
B
• The integral is the sum of the tangential (to the path) component
of the electric field along a path from A to B.
• This integral does not depend upon the exact path chosen to
move from A to B.
• Δ VAB is the same for any path chosen to move from A to B
(because electric forces are conservative).
Does it really work?
• Consider case of constant field:
– Direct: A - B
B
h
A
q
C
r
dl
• Long way round: A - C – B
• So here we have at least one example of a case in which the
integral is the same for BOTH paths.
• In fact, it works for all paths (proof next time)
E
Question 3
A positive charge Q is moved from A to B
along the path shown. What is the sign of
the work done to move the charge from A
to B?
(a) WAB < 0
(b) WAB = 0
(c) WAB > 0
A
B
Question 3
A positive charge Q is moved from A to B
along the path shown. What is the sign of
the work done to move the charge from A
to B?
(a) WAB < 0
•
(b) WAB = 0
A
B
(c) WAB > 0
A direct calculation of the work done to move a positive
charge from point A to point B is not easy.
• Neither the magnitude nor the direction of the field is
constant along the straight line from A to B.
•
But, you DO NOT have to do the direct calculation.
•
•
Remember: potential difference is INDEPENDENT OF
THE PATH!!
Therefore we can take any path we wish.
Choose a path along the arc of a circle centered at
the charge. Along this path
at every point!!
Electric Potential: where is it zero?
• So far we have only considered potential differences.
• Define the electric potential of a point in space as the
potential difference between that point and a reference
point.
• a good reference point is infinity ... we often set V = 0
• the electric potential is then defined as:
• for a point charge at origin, integrate in from infinity along a line
r
• here “r” is distance to origin
Vr  V    dl  E

r
 q 
qdl
Vr  V   


2
4
pe
l
4
pe
l
0
0 


r
line
integral
Potential from charged spherical shell
• E-field (from Gauss' Law)
•
r < a: Er = 0
•
r >a:
• Potential
• r > a:
a
Radius = a
a
1
Q
Er =
4pe 0r 2
V
Q
4pe0 r
Q
4pe0
a
a
• r < a:
r
What does the result mean?
• This is the plot of the radial component of the electric field
of a charged spherical shell:
Er
Notice that inside the shell, the electric
field is zero. Outside the shell, the electric
field falls off as 1/r2.
The potential for r>a is given by the
integral of Er. This integral is simply the
area underneath the Er curve.
a R
r
V
Q
4pe0 a
Q
4pe0 r
a
a R
a
r