Lecture 4 Electric Potential/Energy Chp. 25

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Lecture 4 Electric Potential/Energy Chp. 25

• Cartoon - There is an electric energy associated
  with the position of a charge.
• Opening Demo -
• Warm-up problem
• Physlet
• Topics
• Electric potential energy and electric potential
• Equipotential Surface
• 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 filed from the potential
• Electric potential energy from a system of point
  charge
• Potential of a charged isolated conductor
• Demos
• teflon and silk
• Charge Tester, non-spherical conductor, compare
  charge density at
• Radii

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Electric potential

• The electric force is mathematically the same as
  gravity so it too must be a conservative force.
  We will find it useful to define a potential
  energy as is the case for gravity. Recall that
  the change in the potential energy in moving
  from one point a to point b is the negative of
  the work done by the electric force.
• ?U Ub -Ua - Work done by the electric force
• Since FqE, ? U and
• Electric Potential difference Potential energy
  change/ unit charge
• ? V ? U/q
• ?V Vb -Va - ? E.ds (independent of path ds)

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• ?U Ub -Ua - Work done by the electric force

? V ? U/q

• ?V Vf -Vi - ? E.ds (independent of path ds)

Therefore, electric force is a conservative force
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?V Vf -Vi - ? E.ds (independent of path ds)

• The SI units of V are Joules/ Coulomb or Volt. E
  is
• N/C or V/m.
• 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 point a to point b.
• We are free to choose V to be 0 at any location.
  Normally
• V is chosen to be 0 at infinity for a point
  charge.

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Examples

• What is the electric potential difference for a
  positive charge moving in an uniform electric
  field?

E
E
d
x direction
a
b
d
?V -Ed
?U q ?V
?U -qEd
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Examples

• 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
-9V x 10-6 C
?U -9V x 10-6 Joules
?U -9 microJoules
Potential energy is lower by 9 microJoules
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Examples

• A proton is placed in an electric field of E105
  V/m and released. After going 10 cm, what is its
  speed?

Use conservation of energy.
E 105 V/m d 10 cm
a b

?V Vb Va -Ed
?U q ?V
?U ?K 0
?K (1/2)mv2
?K -?U
(1/2)mv2 -q ?V qEd
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Electric potential due to point charges

• Electric field for a point charge is E (kq/r2).
  The change in electric potential a distance dr
  away is dV - E.dr.
• Integrate to get V - ? kq/r2 dr kq/r
  constant.
• We choose the constant so that V0 at r ?.
• Then we have V kq/r for point charge.
• V is a scalar, not a vector
• 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 sum V ? kqi/ri.

y 2 1 p r3 3
r1 r2
x
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Replace R with r
eqn 25-26
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Electric potential for a positive point charge
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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?
• V kq/r 8.991010 N m2//C2 1.610-19
  C/0.52910-10m
• V 27. 2 J/C 27. 2 Volts

-
r 0.529 A
r

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Ways of Finding V

• 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.

Point charge V kq/r. For an element of charge
dq, dV kdq/r. r is a constant as we integrate.
V ? kdq/r ? kdq/(z2R2)1/2
k/(z2R2)1/2 ? dq k/(z2R2)1/2 Q
This is simpler than finding E because V is not a
vector.
V kq/r
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More Ways of finding V and E

• Use Gauss Law to find E, then use V - ? E.ds to
  get V
• Suppose Ey 1000 V/m. What is V? V -1000 y
• More generally, If we know V, how do we find E?
  dV - E. ds
• ds i dx j dyk dz and dV - Exdx -
  Eydy - Ezdz
• Ex - dV/dx,
• Ey - dV/dy,
• Ez - dV/dz.
• So the x component of E is the derivative of V
  with respect to x, etc.
• If Ex 0, then V constant in that direction.
  Then lines or surfaces
• on which V remains constant are called
  equipotential lines or surfaces.
• See examples

(Uniform field) V Ed
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Equipotential Surfaces

• Example For a point charge, find the
  equipotential surfaces. First draw the field
  lines, then find a surface perpendicular to these
  lines. See slide.
• What is the equipotential surface for a uniform
  field in the y direction? See slide.
• What is the obvious equipotential surface and
  equipotential volume for an arbitrary shaped
  charged conductor?
• See physlet 9.3.2 Which equipotential surfaces
  fit the field lines?

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a) Uniform E field E Ex , Ey 0 ,
Ez 0 Ex dv/dx V Ex d V constant in y
and z directions

• Point charge
• (concentric shells)

• Two point charges
• (ellipsoidal concentric shells)

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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. E0 inside conductor 2. Electric field lines
are perpendicular to the surface. Suppose they
werent? 3. Does E s/e0 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?

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Dielectric Breakdown Application of Gausss 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 E 3106 V/m. To
  examine this we model the shape of a conductor
  with two different spheres at each end

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The surface is at the same potential everywhere,
but charge density and electric fields are
different. For a sphere, V q/(4? ?0 r) and q
4?r2?, then V (?/ ?0 )r. Since E ?/ ?0
near the surface of the conductor, we get VEr.
Since V is a constant, E must vary as 1/r and ?
as 1/r. Hence, for surfaces where the radius is
smaller, the electric field and charge will be
larger. This explains why
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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.

Cloud

Van de Graaff
Show lightning rod demo with Van de
Graaff
- - - -
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A metal slab is put in a uniform electric field
of 106 N/C with the field perpendicular to both
surfaces.

• Draw the appropriate model for the problem.
• Show how the charges are distributed on the
  conductor.
• Draw the appropriate pill boxes.
• What is the charge density on each face of the
  slab?
• Apply Gausss Law. ? E.dA qin/?0

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• Slab of metal In a uniform Electric field

E 106 N/C
- - - - -

Gaussian Pill Box
Slab of metal In a uniform Electric field
Evaluate E.dA qin/?0

• Left side of slab - EA 0A A?/?0,
• E - ? /?0, ? - 106 N/C 10-11 C2/Nm2
  -10-5 N/m2

.Right side of slab 0A EA A?/?0, E ?
/?0, ? 106 N/C 10-11 C2/Nm2 10-5 N/m2
(note that charges arranges themselves so that
field inside is 0)
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What is the electric potential of a uniformly
charged circular disk?
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Warm-up set 4
1. HRW6 25.TB.37. 119743 The equipotential
surfaces associated with an isolated point charge
are concentric cylinders with the charge on
the axis vertical planes horizontal
planes radially outward from the charge
concentric spheres centered at the charge 2.
HRW6 25.TB.17. 119723 Two large parallel
conducting plates are separated by a distance d,
placed in a vacuum, and connected to a source of
potential difference V. An oxygen ion, with
charge 2e, starts from rest on the surface of
one plate and accelerates to the other. If e
denotes the magnitude of the electron charge,
the final kinetic energy of this ion is 2eV
eVd Vd / e eV / d eV / 2 3. HRW6
25.TB.10. 119716 During a lightning discharge,
30 C of charge move through a potential
difference of 1.0 x 108 V in 2.0 x 10-2 s. The
energy released by this lightning bolt is
1.5 x 1011 J 1500 J 3.0 x 109 J 3.3 x
106 J 6.0 x 107 J