Physics 4B
FINAL EXAM
Chapters 23  34
Fall 1995
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Solve the following problems in the space provided.
Use the back of the page if needed.
Each problem is worth
10 points.
You
must
show your work in a logical fashion starting with the correctly applied physical principles
which are on the last page.
Your score will be maximized if your work is easy to follow because partial credit will
be awarded.
1.
A small plastic sphere of mass 5.00x10
16
kg is held motionless between two
charged parallel plates 4.00mm apart.
Assume that the plastic sphere has two
excess electrons on it.
Find (a)the electric force on the sphere, (b)the electric field
between the plates and (c)the potential difference between the plates.
(d)In the
diagram at the right, show the polarity of the plates, sketch the electric field and
show all the forces that act on the sphere.
(a)According to Newton's Second Law,
Σ
F
=
ma
⇒
F
e

F
g
=
0
⇒
F
e
=
F
g
.
Putting in the numbers, F
e
=
mg
=
(5.00x10

16
)(9.80)
=
4.90x10

15
N
.
(b)Using the definition of electric field,
E
≡
F
q
=
4.90x10

15
2(1.60x10

19
)
=
1.53x10
4
V
m
.
(c)The electric potential is
∆
V
= 
r
E
•
d
r
s
∫
.
Since the field between the plates is constant, V
=
Ed
=
(1.53x10
4
)(4.00x10

3
)
=
61.3V
.
2. Two long concentric cylinders of radii a and b have a potential difference V
o
established between them.
Find the electric field between the cylinders as a
function of the distance from the center. (Hint: You may want to start by
assuming a linear charge density
λ
exists on the cylinders, but for full credit you
must express
λ
in terms of the potential difference).
Applying Gauss's Law to an imaginary cylinder with a radius r between a and b.
Since the cylinders are long, symmetry requires the field to be constant over this
Gaussian cylinder as it must point radially outward.
Therefor,
r
E
•
d
r
A
∫
=
q
ε
o
⇒
E2
π
r
l
=
λ
l
ε
o
⇒
E
=
λ
2
πε
o
r
The electric potential is
∆
V
= 
r
E
•
d
r
s
∫
.
Integrating from the inner to the outer cylinder,
0

V
o
= 
λ
2
πε
o
r
a
b
∫
dr
=
λ
2
πε
o
ln
b
a
⇒ λ =
2
πε
o
V
o
ln
b
a
( 29
.
Finally, E
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 Fall '09
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 Work, Magnetic Field, Faraday

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