Download 02_E2_ws1_key

E&M Unit II: Worksheet 1 Key
Fields, Potential, and Energy
1. Two books, initially on the floor, are picked up and placed on a shelf 2.0 m off the floor. One,
Twenty Other Things I Like To Do With That Stuff Between My Toes, has a mass of 1.0 kg. The
other, Why Physics Rules The Universe, has a mass of 2.0 kg. The gravitational field strength
near the surface of the Earth is approximately 10 N/kg.
a. Calculate the gravitational force acting on each book as it rests on the shelf. What factors
determine the size of this force?
Fg=mg = (10N/kg)1.0kg = 10N (book 1)
Fg = (10N/kg)2.0kg = 20N (book 1)
The mass of the book is the main factor as close to the earth’s surface the gravitational
field strength is a constant 10 N/kg.
b. Calculate the change in the gravitational potential energy stored in the gravitational field as a
result of each book being picked up off the floor and placed on the shelf. What factors
determine the size of this change?
Eg = mgh = 10N(2.0m) = 20J
Eg = mgh = 20N(2.0m) = 40J
The mass of the book and the height the book is raised.
c. What is the difference in gravitational potential (potential energy per unit mass) between the
floor and the shelf (use data from each book to calculate this)? What factors determine the
size of this difference?
ΔE g
m

ΔE g
20J
 20 kgJ
1kg
m

40J
 20 kgJ
2kg
Both values are the same because the change in height is the same.
d. How does gravitational potential differ from gravitational potential energy?
Eg depends on mass, gravitational field strength and change in height.
Gravitational potential depends on gravitational field strength and change in height.
2. Below are two parallel conducting plates, each carrying an equal quantity of excess charge of
opposite type. The plates are separated by 2.0 m.
(+)
2.0 m
(-)
1C
2C
©Modeling Instruction - AMTA 2013
Between the pair of plates are two positively charged
objects; the object on the left carries 1.0 µC of excess
charge, the object on the right carries 2.0 µC. The
electric field strength between the plates is uniform, and
approximately 10 N/C. You move each charge from the
negative plate to the positive plate. [for this example,
neglect the effects of the gravitational field]
1
E2 Potential - ws 1 v4.0
a. Calculate the electrical force acting on each object when it is between the plates. What
factors determine the size of this force?
Fe=qE = (10N/C)1.010-6C = 1.010-5 N Fe=(10N/C)2.010-6C = 2.010-5 N
The amount of charge is the main factor as while it also depends on the electric field
strength, that is a constant 10 N/C.
b. Calculate the change in the electrical potential energy stored in the electric field as a result of
each charge being moved from the negative plate to the positive plate. What factors
determine the size of this change?
Ee = Fx = qEd = 1.010-5 N(2.0m) = 2.010-5J (object 1)
2.010-5 N(2.0m) = 4.010-5J (object 2)
The factors are the charge on the object, the electric field strength and the distance moved.
c. What is the difference in electric potential (potential energy per unit charge) between the
plates for each of the two charges? What factors determine the size of this difference?
ΔE e 2.0  10 -5 J

 20 CJ  20V
-6
q
1.0  10 C
ΔE e 4.0  10 -5 J

 20 CJ  20V
-6
q
2.0  10 C
Both values are the same because the change in position is the same for both.
d. What is the difference in electric potential between the negative plate and a point midway
between the plates?
ΔV 
Fx qEd

 Ed  10 NC (1.0m)  10V
q
q
e. How does electrical potential differ from electric potential energy?
Electric potential energy depends on charge, electric field strength and change in position.
Electric potential depends on electric field strength and change in position.
3. In unit 1, the units for electric field were given as
can also be given in
N
. It turns out that the electric field strength
C
V
. Show how these units, which appear very different, actually describe the
m
same quantity.
J
V
J
J
so
can be written as C which is the same as
. A joule is a Newton meter
m
C
Cm
m
Nm
N
so we can write the last step as
and the meters cancel out to yield .
Cm
C
A volt is a
©Modeling Instruction - AMTA 2013
2
E2 Potential - ws 1 v4.0