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E & M Unit II – Worksheet 2 Key
Gravitational & Electrical Equipotentials
E
F
A contour map of a
large hill is shown to
the left. Assume all
heights given are in
meters.
1. What do the contour lines on the map represent? How do you know?
Lines of gravitational equipotential. The elevation is the same at all heights along a line.
2.
Rank the points A,B,C, D, E and F in order of increasing gravitational potential.
D < A = C < B<E<F
3. Calculate the gravitational potential difference between:
a) point A and the base of the hill GP = gh = (10N/kg)-100m = -1000J/kg
b) point B and the base of the hill GP = gh = (10N/kg)-200m = -2000J/kg
c) point C and point A GP = gh = (10N/kg) 0m = 0
d) point C and point E GP = gh = (10N/kg)200m = 2000J/kg
4. Would a gravitational field gain more energy if an object was moved from D to A or from E to F?
Why?
The same. Same change in elevation
5. Describe the motion of a ball placed at point C and released.
The ball would roll down perpendicular to the equipotential line.
6. How would a trip for a ball from B to A differ from F to E? How would they be alike? (Assume
the ball was released from rest.)
The ball would accelerate more rapidly from F to E than B to A. Both would have the same
change in velocity and the same kinetic energy.
7. What information does the spacing of the contour lines convey? Describe the behavior of a test
mass when it is released in a region where the lines are:
a) closely spaced corresponds to a steeper slope. The acceleration of the mass is great.
b) widely separated corresponds to a gentle slope. The acceleration of the mass is lower.
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E2 Potential - ws 2 v4.0
Consider the diagram above in which charges were packed closely together to form parallel plates.
Suppose also that the potential difference between the plates of charge is 7.0 volts and that the
equipotential lines are at 1.0 volt intervals. Based on your experience in the activity, answer
questions 6 - 10.
8. What do the lines in the diagram above represent? Explain.
They are lines where every point on the line is at the same electric potential. They are called
equipotential lines.
9. Rank the points A,B,C, D, E and F in order of increasing electric potential, relative to the ground.
B<C=E<D<F<A
10. Calculate the electrical potential difference between
A. point A and the negative plate 0V – 6V = -6V
B. point B and the negative plate 0V – 0V = 0V
C. point C and point A 6V – 2V = 4V
D. point C and point B 0V – 2V = -2V
11. a. Describe the motion of a positive charge placed at point E and released. How do you know?
It will accelerate perpendicular to the equipotential line and toward lower potential ( ).
The field lines are perpendicular to the equipotential lines and field lines indicate the
direction of the force on a positive charge. Field lines point away from positive charge and
toward negative.
b. Describe the motion of a negative charge placed at point E. How do you know?
It will accelerate perpendicular to the equipotential line and toward higher potential (
The negative charge has a force in the opposite direction of a positive.
).
c. In general, can we say that if a charged object is free to move in an electrical field, it will
move from an area of high potential to an area of low potential? Explain.
Only if it is a positive charge. A negative charge will move from low to high.
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12. What is the approximate magnitude of the strength of the electric field at the location of the * in
the center?
The two equipotential lines are about .8cm apart. The scale is 1 cm = .1cm so in the actual
1V
V
distance is .08cm or .0008m. E  Vx  .0008
m  1250 m The field is not quite uniform so this is not
exact.
13. What information does the spacing of the equipotential lines convey?
It indicates the strength of the electric field. The closer the lines the stronger the field, the
farther apart, the weaker.
Describe the behavior of a positive test charge when it is released in a region where the lines are:
a. closely spaced it will accelerate rapidly.
b. widely separated the acceleration will be less.
14. Consider the diagram at left in which the field produced by two point charges is shown.
a. Draw five equipotential lines on the diagram.
b. Draw the vector representing the force acting
on a + particle placed at A.
A
B
15. Would this particle follow the field line to B? Why or why not?
No it would not. It will begin by accelerating (and moving) in the direction of the force
vector. After it has moved a bit it will have a force in a new direction and a v in that
v v f  vo
direction. Since a =
, v is the same direction as the acceleration. v= vf-vo,

t
t
so we can get the direction of the new velocity vf by adding vo and v. As can be seen, the new
velocity is not in the direction of the net force.
B
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vo
v
a
vf
vo
E2 Potential - ws 2 v4.0
Use the diagram of the electric field produced by a dipole below to answer questions 16 – 20. The
difference in potential for each equipotental line is 2 volts.
B
A
Scale: 1.0 cm = .01 cm
*
+
-
D
C
16. What is the change in potential for a charged particle as it moves from A to D? Briefly explain.
There is no change, both are on the same equipotential line.
17. Describe the “change in elevation” of a proton if it could “walk” along the field line from A to B.
The first part of the trip would be “downhill” with medium steepness, then it would be downhill
but less steep for a while, then get back to medium steepness downhill.
18. What would be the change in the electrical potential energy stored in the electric field (in J and in
eV) if a proton was moved from A to B?
Ee  qV , 1e×-8V=-8eV 1.6 1019C(-8V)=-1.3 1018 J
19. What would be the change in the electrical potential energy stored in the electric field (in J and
in eV) if an electron was moved from C to B?
Ee  qV , -1q×-2V=2eV -1.6 1019C(-2V)=3.2 1019 J
20. Estimate the magnitude of the electric field strength at the location of the * in the center? Is this
estimate likely to be as accurate as the value in question 12? Why or why not?
The two equipotential lines are about .3cm apart. The scale is 1 cm = .01cm so in the actual
2V
=67,000 mV . The field is not uniform between the
distance is .003cm or .00003m. E  Vx  .00003m
two charges therefore there is less confidence in this answer.
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E2 Potential - ws 2 v4.0