Download PHY 1361 General Physics II Fall 2006 Practice Test #2

PHY 1361 General Physics II Fall 2006
Practice Test #2
1
I. Multiple-Choice Questions
1.
The velocity of a particle (m = 10 mg, q = –4.0 µC) at t = 0 is 20 m/s in the positive
x direction. If the particle moves in a uniform electric field of 20 N/C in the
positive x direction, what is the particle's speed at t = 5.0 s?
a.
b.
c.
d.
e.
2.
60 m/s
20 m/s
45 m/s
40 m/s
70 m/s
A positively charged particle is moving in the +y-direction when it enters a
region with a uniform electric field pointing in the +x-direction. Which of the
diagrams below shows its path while it is in the region where the electric field
exists. The region with the field is the region between the plates bounding each
figure. The field lines always point to the right. The x-direction is to the right; the
y-direction is up.
q
q
(a)
3.
(c)
q
(d)
q
(e)
Two concentric imaginary spherical surfaces of radius R and 2R respectively
surround a positive point charge Q located at the center of the surfaces. When
compared to the electric flux Φ 1 through the surface of radius R, the electric flux
Φ 2 through the surface of radius 2R is
a.
b.
c.
d.
e.
4.
(b)
q
1
Φ1 .
4
1
Φ 2 = Φ1.
2
Φ 2 = Φ1 .
Φ 2 = 2Φ 1 .
Φ 2 = 4Φ 1 .
Φ2 =
Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is
uniform and given by E = (4i + 3j) N/C. What is the potential difference
VA − VB ?
a.
b.
c.
d.
e.
33 V
27 V
30 V
24 V
11 V
2
5.
A particle (charge = +2.0 mC) moving in a region where only electric forces act
on it has a kinetic energy of 5.0 J at point A. The particle subsequently passes
through point B which has an electric potential of +1.5 kV relative to point A.
Determine the kinetic energy of the particle as it moves through point B.
a.
b.
c.
d.
e.
6.
3.0 J
2.0 J
5.0 J
8.0 J
10.0 J
Identical 2.0-µC charges are located on the vertices of a square with sides that are
2.0 m in length. Determine the electric potential (relative to zero at infinity) at the
center of the square.
a.
b.
c.
d.
e.
38 kV
51 kV
76 kV
64 kV
13 kV
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II. Problems
7. Three point charges are arranged as shown in the figure below. (a) Find the vector
electric field that the 6.00-nC and –3.00-nC charges together create at the origin. Express
the result using unit vectors. (b) Find the vector force on the 5.00-nC charge. Express the
result using unit vectors. (c) Find both the magnitude and direction of the electric field
and the force in (a) and (b).
8. A solid, insulating sphere of radius a has a uniform charge density ρ and a total charge
Q. Concentric with this sphere is an uncharged, conducting hollow sphere whose inner
and outer radii are b and c, as shown in the figure below. (a) Find the magnitude of the
electric field in the regions r < a, a < r < b, b < r < c, and r > c. (b) Determine the
induced charge per unit area on the inner and outer surfaces of the hollow sphere.
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III. Bonus Question
9. A thin rod of length ℓ and uniform charge per unit length λ lies along the x axis, as
shown in the figure below. (a) Show that the electric field at P, a distance y from the rod
along its perpendicular bisector, has no x component and is given by E = 2ke λ sin θ0/y. (b)
What If? Using your result to part (a), show that the field of a rod of infinite length is E
= 2ke λ /y.
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Solutions to Practice Test #2, PHY 1361, Fall 2006
I. Multiple-Choice Questions
1. b. 2. d. 3. c. 4. d. 5. b. 6. b.
II. Problems
7.
(a)
E1 =
E2 =
ke q1
( )
( 8.99 × 10 )( 3.00 × 10 )
( )
( 8.99 × 10 )(6.00 × 10 )
−ˆ
j=
r12
ke q2
r22
ˆ=
−i
−9
9
( 0.100) 2
−9
9
( 0.300)
(
2
( −ˆj) = − ( 2.70 × 10
3
( −iˆ) = − ( 5.99× 10
2
) (
)
N C ˆ
j
)
ˆ
N C i
)
ˆ− 2.70 × 103 N C ˆ
E = E 2 + E1 = − 5.99 × 102 N C i
j
(b)
)(
(
)
ˆ− 2 700ˆ
F = qE = 5.00 × 10−9 C −599i
j N C
(
)
( −3.00iˆ− 13.5ˆj) µN
ˆ− 13.5 × 10−6 ˆ
F = −3.00 × 10−6 i
j N =
(c) It is Your Turn to find the magnitude and direction of the field and force!
8.
(a)
(b)
∫ E ⋅ dA
(
)
= E 4π r2 =
qin
∈0
For r < a,
4

qin = ρ  π r3 
3

so
E=
For a < r < b and c < r ,
qin = Q .
So
E=
For b ≤ r ≤ c ,
E = 0 , since E = 0 inside a conductor.
ρr
3 ∈0
.
Q
.
4π r2 ∈0
Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0
inside the conductor, the total charge enclosed by a spherical surface of radius
b ≤ r ≤ c must be zero.
Therefore,
q1 + Q = 0
and
σ1 =
q1
4π b
2
=
−Q
.
4π b2
Let q2 = induced charge on the outside surface of the hollow sphere. Since the
hollow sphere is uncharged, we require
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q1 + q2 = 0
σ2 =
and
q1
4π c2
Q
.
4π c2
=
III. Bonus Question
9.
(a)
The electric field at point P due to each element of length dx, is
k dq
dE = 2 e 2 and is directed along the line joining the element to
x +y
point P. By symmetry,
Ex = ∫ dEx = 0
and since
dq = λ dx ,
E = Ey = ∫ dEy = ∫ dE cosθ
where
cosθ =
Therefore,
l2
E = 2keλ y ∫
0
(b)
(x
For a bar of infinite length,
2
dx
)
2 32
+y
=
θ 0 = 90°
y
2
x + y2
.
2keλ sin θ 0
.
y
and
Ey =
2keλ
.
y
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