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Ch – 29 Electric Potential

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					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.   Chapter 29 begins with a discussion of:
a.   Electric potential
b.   Energy and Work
c.   Displacement in a uniform field
d.   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 mv 2
• 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.
Work is defined as:                 f
                                 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 10 6
    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

v F = 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

v F = 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 Parallel-
           Plate 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
a. 8 cm: vmax occurs at
   Kmax
b. 5 μJ
c. turning point at x = 3                             U as a function of position



   cm (why not 1 cm                  3


                                     2


   as well?)                         1




d. 1.0 x 10-4 N to the
                                     0
                            Energy        0   2   4       6          8         10   12   14
                                                                                              U μJ
                                     -1
                                                                                              TE μJ


   right                             -2


                                     -3


                                     -4


                                     -5
                                                          position (cm)
                  Answers
a.   8 cm: v max occurs at Kmax
b.   5 μJ
c.   turning point at x = 3 cm
d.   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 non-
  constant?
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.   Va = 900V                 Vb = 300V
2.   |∆V| = 600V
3.   Ua = 14.4 x10-17 J   Ub = 4.8 x10-17 J
4.   5.24 x 105 m/s
        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)
                    q1       q2        q3        1  q1 q2 q3 
VA  V1  V2  V3                                      
                    4 0 r1 4 0 r2 4 0 r3 4 0  r1 r2 r3 
                          5  10 9 C 5  10 19 C 10  10 9 C 
                        
    9.0  109 N m 2 /C2                                         3140 V
                          0.02 m        0.04 m       0.0447 m 

				
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