Honors Physics
Mr. Payne
April 30, 2012
April 30, 2012
• Today: Uniform Fields
– Gravitational and Electrical
– Homework: Problems 5, 6, 11, 12, page 489
– (Problems 1-4, 9 and 10 were due today)
• Tomorrow: Non-Uniform Fields and Capacitance
– Gravitational and Electrical
– Homework: Problems 14, 19, 31, pages 489-490
• Wednesday: Potential and Work
• Friday: Finish Potential, Start Circuits
• Monday, May 7:
– Electrical and Gravitational Potential Test
• A field is an area in which the effects of a force
can be felt.
– Uniform fields are consistent throughout. The
earth’s gravitational field strength is about 9.8
N/kg any where on Earth’s surface, which means a
kilogram of mass on the surface of the earth
weighs 9.8 N. The force due to the gravitational
field on the surface of Mars is 3.71 N/kg, and
Jupiter produces a force of 24.92 N on each
kilogram of mass at its surface.
• The electric field strength near a charge is generally
consistent, and it is the total charge at a point that
produces the force in a similar manner to the mass of
an object gravitationally. Both obey the inverse
square law, and the consistency is based on objects
not changing position by enough to have a large
effect on the field strength.
Gravitation: Uniform Field
• How do we measure the gravitational effect on our mass on
Earth? (What equation do we use to find weight?)
• Fg = mg
• Using G = 6.67x10-11N-m2/kg2, me=5.98x1024kg, and
re=6.38x106m, determine the numeric answer with units to
• Gme/(re)2
• What is the significance of this number and the appropriate
• This gives us a value of 9.8 N/kg, which is “g”
• What altitude would result in a value of 9.7 N/kg?
• The answer is about 6.41*106 m, which is 30 km, or about
18 miles above the earth’s surface.
Gravitational and Electric Potential
• Mass is a fundamental property, and
corresponds to charge.
• Gravity as “g” is also a fundamental property,
and is more properly stated as gravitational
field strength (9.8 N/kg on the surface of the
Earth), and E is the corresponding electric
field strength. The Universal Gravitation
Constant “G” corresponds to the electric
proportionality constant “k”
Uniform Field Shape
• What is the shape of the gravitational field?
• What is the shape of the electric field?
Field Strength
• Gravitational field strength is in N/kg, and is
calculated as g = Fg / m
• Electric field strength is in N/C, and is
calculated as E = Fe / q
Potential Energy
• The standard symbol for potential energy is the letter
• For gravitational potential energy, the formula is
U=mgh, where “h” is a height above a
predetermined elevation, or radius.
• Change in gravitational potential energy is DU = mgDh
• For electric potential energy, the formula is U=qEd,
where “d” is a distance from a charge, equal to the
• Change in electric potential energy is DU = qEDd
• The potential is the effect of a field, generally
associated with the effect on a test mass,
usually 1 kg, or a test charge, usually a small
positive test charge. The symbol for potential
is “V”
• Gravitational potential is calculated using the
formula DV = g·Dh
• Electric potential is calculated by using the
formula DV = E ·Dx
Uniform Field Graphs
• How does Fg vary with distance in a uniform
• Since this is a uniform field, “g” is constant, so
Fg is constant.
• How is this relationship graphed?
Uniform Field Graphs
• How does Fe vary with distance in a uniform
• Since this is a uniform field, E is constant, so
Fe is constant.
• How is this relationship graphed?
Uniform Field Graphs
• What would graphs look like for the fields (“g”
and E ) as distance increases? Remember,
this is a uniform field.
• How would graphs be made for energy and
potential for gravitational fields and electrical
Work in a Field
• Work has been defined as the product of the
magnitude of the displacement times the
component of the force parallel to the
displacement (Giancoli, p. 137). Work in an
electric field follows the same principle:
W = Fd = qEd
Electric Potential and Potential
• Electric potential is the electric
potential per unit charge
Va =
• Potential Difference between
two points is the amount of
work necessary to move a charge between
two points: V = V -V = E - E = - W
Problem 1
• (I) How much work does the electric field do
in moving a -7.7μC charge from ground to a
point whose potential is +55 V higher?
Vba = -
® Wba = -qVba = - -7.7 ´10-6 C +55 V = 4.2 ´10-4 J
Problem 2
• 2. (I) How much work does the electric field
do in moving a proton from a point with a
potential of +125 V to a point where it is -55V
Express your answer both in joules and
electron volts.
• The work done by the electric field can be
found from Eq. 17-2b.
V =® W = -qV = - (1.60 ´10 C) ( -1.80 ´10 V ) = 2.88 ´10 J
( )(
= - 1 e -180 V = 1.80 ´102 eV
Problem 3
• (I) How much kinetic energy will an electron
gain (in joules and eV) if it accelerates through
a potential difference of 23,000 V in a TV
picture tube?
• The kinetic energy gained is equal to the work done on the
electron by the electric field. The potential difference must
be positive for the electron to gain potential energy. Use Eq.
17-2b. V = - W ® W = -qV = - -1.60 ´10 C 2.3´10 V = 3.7 ´10
( )(
= - -1 e 2.3´104 V = 2.3´104 eV
Other Problems
• 4. (I) An electron acquires 7.45 X 10-16 J of kinetic energy
when it is accelerated by an electric field from plate A to
plate B. What is the potential difference between the
plates, and which plate is at the higher potential?
• 9. (II) Two parallel plates, connected to a 200-V power
supply, are separated by an air gap. How small can the gap
be if the air is not to become conducting by exceeding its
breakdown value of 3 X 106 V/m?
• 10. (II) The work done by an external force to move a -8.5
μC charge from point a to point b is 15.0 x 10-4J. If the
charge was started from rest and had 4.82 X 10-4J of kinetic
energy when it reached point b, what must be the potential
difference between a and b?

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