Download 29 Electric Potential

Ch – 29 Electric Potential
Reading Quiz Ch. 29
1. The units of the electric potential are:
a. V/C.
b. Ω/m
c.
V/m
d. N/C
e. J/C.
2.
a.
b.
c.
d.
Chapter 29 begins with a discussion of:
Electric potential
Energy and Work
Displacement in a uniform field
The voltage inside a capacitor
Learning Objectives – Ch 29
• To introduce electric potential energy and use it
in conservation of energy problems.
• To define the electric potential.
• To find and use the electric potential of point
charges, charged spheres, and parallel-plate
capacitors.
• To find the electric potential of a continuous
distribution of charge.
• To introduce and use potential graphs and
equipotential surfaces.
Energy Review
The mechanical energy of a system of
particles is made up of its kinetic energy
(K) and any potential energy (U):
Emech = K + U
• K = ½ mv2
• U is a relative value dependent on an
arbitrary zero value, U0 (analogy to
Celsius vs. Kelvin for temperature).
Energy Review
Emech is conserved for a system of particles that
interact with conservative forces:
ΔEmech = ΔK + ΔU = 0
A conservative force is one for which the work
done is path-independent.
f
Work is defined as:
W =  Fs (ds)

i
ΔU = - W for any work due to a conservative force.
This is not an arbitrary number and has the same
value, regardless of the U0 chosen.
The electric force is conservative.
Use the energy bar chart to help determine
whether ΔU is negative, positive or zero, as a
positive charge moves from i to f. At the initial
position i, it has v0 and U0 = 0.
Answers
• The particle will speed up
going from i to f.
• The work done is
positive.
• ΔU is negative.
• The particle will slow
down going from i to f.
• Work done is negative.
• ΔU is positive.
Use the energy bar chart to help determine
whether ΔU is negative, positive or zero, as a
negative charge moves from i to f
Answers
• ΔU is 0. The particle will
neither speed up nor
slow. No work is done.
• ΔU is negative. The
negative particle will
speed up; displacement
is in the same direction
as the force. The work
done is positive.
Conceptual Problem
A small, positively-charged
particle is shot toward a
larger fixed charge. For
both cases shown:
Does the particle speed up
or slow down?
Is the acceleration
constant?
Is ΔU positive, negative or
zero?
Answers
Slows down
Non-constant
acceleration
ΔU positive
Speeds up
Non-constant
acceleration
ΔU negative
Electric potential energy of a
capacitor/charged particle system
• Electric field and therefore
force (qE field) is constant
• Work done by conservative
electric force is positive:
W = |(qE field)| |Δr| cos 0
• ΔU is therefore negative as a
positive charge travels
towards the negative plate.
Electric Potential Energy of a
capacitor/charged particle system
• Anywhere inside the
capacitor:
U = U0 + qEs and
ΔK + ΔU = 0
U0 is potential energy value at
negative plate, usually
chosen to be zero.
U0 and U are not “real” values,
but ΔU is.
Capacitor Problem
The electric field strength is 2.0 x
104 N/C inside the capacitor
with a spacing of 1.0 mm. An
electron is released from rest at
the negative plate, where U0 =
0.
a. What is the value of the final
potential energy of the
electron?
b. What is the speed of the
electron when it reaches the
positive plate?
Answer
a. U = -3.2 x 10-18 J. Negative in this case
refers to less than the starting potential
energy of 0 J and is not related to the
direction of E field or direction of travel
b. v = 2.65 x 106 m/s. The electron gained
speed and lost potential energy as it
moved toward the negative plate.
Electric Potential Energy of a
System of 2 Point Charges
• This expression refers to
the energy of the system,
not to the energy of either
particle.
• The potential energy of 2
like charges is positive.
• The potential energy of 2
unlike charges is
negative.
• This equation is also valid
for the potential energy of
two charged spheres.
The Zero of Electric Potential Energy
• For both like and unlike particles, U approaches zero as r
approaches infinity.
• However, the zero of potential energy has no physical significance.
Only the change in potential energy matters.
• Like charges will always have a greater potential energy when they
are a finite value of r apart than when they are separated by infinity.
• Unlike charges will always have less potential energy when they are
a finite value of r apart than when they are separated by infinity.
• 2 unlike charges separated by a finite distance r have less potential
energy than 2 like charges of the same magnitude separated by the
same distance.
Negative energy
• A negative value for Emech means that the
system has less total energy (K + U) than
a system which is infinitely far apart (U=0)
and at rest (K=0).
• A system with a negative Emech is a bound
system. The particles cannot escape each
other (e.g. hydrogen atom)
Numerical Problem
A 2.0 mm-diameter bead is charged to
-1.0 x 10-9 C (changed from 10-12C)
a. A proton is fired at the bead from
far away with a speed of 1.00 x 106
m/s. It collides head-on. What is
the impact speed?
b. An electron is fired at the bead from
far away. It reflects, with a turning
point 0.10 mm from the surface of
the bead. What was the electron’s
initial speed?
Answer part a
Ki + Ui = KF + UF
where
Ui = U0 (far away) = 0
UF = - (kq1q2)/r at r = 1.0 mm
vF = 1.02 x 106 m/s
Answer part b
Ki + Ui = KF+ UF
where
Ui = U0 (far away) = 0
KF = 0 (turning point)
UF = (kq1q2)/r at r = 1.1 mm
vF = 2.40 x 108 m/s
Electric Potential
Uc = q(Es)
Uq,Q = q (KQ/r)
For both cases:
U = [charge q] X [potential for interaction with
source charges, caused by the alteration of
space by the source charges]
Uq,sources = qV
Electric Potential Inside a ParallelPlate Capacitor
At any point inside the
capacitor with electric
field strength E:
Uelec = Uq,sources = q(Es)
(U0 = 0 at negative plate)
V = U/q = qEs/q = Es
V_ = 0 (negative plate)
Voltage (∆V) across a Parallel-Plate
Capacitor
For a capacitor with
spacing of d
∆Vc = V+ - V- = Ed
Rearranging:
E = ∆Vc /d
This shows that E can be
expressed in units of
Volts per meter
Stop to think
Rank, in order, from
largest to smallest,
the potentials Va to Ve
Answer
V = Es, where s is the
distance from the
negative plate
Va = Vb > Vc > Vd = Ve
Graphical representations of electric
potential inside a capacitor
• Contour map –
– the green dashed contour
lines represent
equipotential surfaces.
– Any point on one of these
surfaces is at the same
potential.
– These surfaces are
perpendicular to the
direction of E.
– E points in the direction of
decreasing potential
Graphical representations of electric
potential inside a capacitor
Graph of potential vs x
(negative plate is at x
= 3 mm).
Electric field points in
the “downhill”
direction of the graph.
The Zero of Potential
These three arrangements represent the same
physical situation. The assigned zero is
arbitrary. ∆V is the same.
Energy Diagrams – potential energy needed
as a function of position
• The PE curve (blue) is the
potential energy necessary to
get to that position. It is a
function of the source charges
• The TE line (brown) shows the
total energy of the system.
• The distance from the x-axis to
the PE curve is U, the potential
energy of the system.
• The distance from the curve to
the TE line is K of the system.
Energy Diagrams
• The TE line crosses the
PE curve at a turning
point.
– total energy of the system
is potential
– v = 0 (why?)
• F = -dU/dx
– the negative value of the
tangent at any point
– force is zero at minima and
maxima (note that energy
is not necessarily 0 when
force is!
– Minima in the PE curve are
points of stable equilibrium.
Maxima are points of
unstable equilibrium.
Graphical Problem
The graph shows the electric potential
along the x-axis. Draw the
potential-energy diagram for a –20
nC charged particle that moves in
this potential. Suppose this
charged particle is shot in from the
right (at x > 12 cm) with a kinetic
energy of 1 μJ.
a. Where is the point of maximum
speed?
b. What is the particle’s kinetic
energy at that point?
c. Where is the turning point?
d. What is the force on the
particle at the turning point?
Answers to Graphical Problem
U as a function of position
3
2
1
0
Energy
a. 8 cm: vmax occurs at
Kmax
b. 5 μJ
c. turning point at x = 3
cm (why not 1 cm
as well?)
d. 1.0 x 10-4 N to the
right
0
2
4
6
8
10
12
14
U μJ
-1
TE μJ
-2
-3
-4
-5
position (cm)
Answers
a.
b.
c.
d.
8 cm: vmax occurs at Kmax
5 μJ
turning point at x = 3 cm
1 x 10-4 N to the right
Equipotential Lab - Capacitor
• What orientation do E
field vectors have relative
to equipotential lines?
• Do E field vectors point in
the direction of
increasing or decreasing
potential?
• Constant E field will has
what kind of equipotential
lines spacing?
Equipotential Lab – positively
charged sphere
• What did we do to set
“infinity” to zero?
• What orientation do E
field vectors have relative
to equipotential lines?
• Do E field vectors point in
the direction of
increasing or decreasing
potential?
• For the point charge, the
equipotential line spacing
not constant. Is the E
field constant or nonconstant?
The Electric Potential of a Point
Charge
V = (Utestq, sourceq)/ qtest = (1/4πε0)(qsource /r)
Ranking task - potential
Rank in order, from
largest to smallest,
the potentials at the
points indicated by
dots in each of the
figures below
(e.g V1>V2=V3)
Answer to ranking task
b. V3 > V1 = V2
c. V3 > V2 > V1
e. V1 > V2 > V2
Conceptual problem – electric
potential
determine whether ΔV is
negative, positive or zero
for the three cases shown
at right. The charges
shown are the source
charges. In the last case,
the E field due to the
source charges are
shown.
Answers to conceptual problem
ΔV is negative.
ΔV is positive.
ΔV is positive. E field
points in direction
of decreasing
potential.
Numerical Problem
For the situation shown in the
figure, find:
1. The potential at points a and
b.
2. The magnitude of the
potential difference between
a and b.
3. The potential energy of a
proton at a and b.
4. A proton has a velocity of
4.0 x 105 m/s to the right at
point a. What is its speed at
point b?
mp = 1.67 x 10-27 kg
e = 1.6 x 10-19 C
Answers 1-4
1.
2.
3.
4.
Va = 900V
|∆V| = 600V
Ua = 14.4 x10-17 J
5.24 x 105 m/s
Vb = 300V
Ub = 4.8 x10-17 J
Numerical Problem (5-6)
5. A proton has a velocity of
4.0 x 105 m/s to the left at
point b. What is its speed
at point a?
6. A proton moves along a
trajectory that passes
through points c and d.
The proton’s speed at point
c is 4.0 x 105 m/s. What is
its speed at point d?
Answer (5)
v = 2.12 x 105 m/s
Answer (6)
It’s the same as 5!
The electric force is a
conservative force so
the path taken doesn’t
matter, only the initial
and final distance.
Potential of a sphere
• Insulator - applies for a uniformly charged solid
sphere or shell.
• Conductor - Recall that all charge will be
uniformly distributed on the outside surface of
the sphere or shell
• At the surface of a sphere of radius R:
V = (1/4πε0)(qsphere /R)
• At a distance r > R:
V = (1/4πε0)(qsphere /R) or if you know (V0), the potential
at the surface:
V = R/r V0
The Electric potential of more than one
point charge
V = Σ (1/4πε0)(qi /ri)
What is the electric
potential at Point A?
Answer (HW #29)
1  q1 q2 q3 
 
 
4 0 r1 4 0 r2 4 0 r3 4 0  r1 r2 r3 
9
19
9

5

10
C

5

10
C
10

10
C
9
2
2
 9.0  10 N m /C 


  3140 V
0.04 m
0.0447 m 
 0.02 m
VA  V1  V2  V3 

q1


q2

q3
