Version 120 – Exam 1 – hoffmann – (57505)
This print-out should have 20 questions.
Multiple-choice questions may continue on
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before answering.
001 (part 1 of 3) 10.0 points
Consider a solid insulating sphere of radius
b with nonuniform charge density ρ = a r,
where a is a constant.
b
dr
b
r
O
Find the charge contained within the radius
r < b as in the figure. The volume element dV
for a spherical shell of radius r and thickness
dr is equal to 4 π r 2 dr.
1. Qr =
aπ
r2
2. Qr = π a r 2
a r4
π
a r2
4. Qr =
π
3. Qr =
5. Qr = 0
r4
πa
r3
7. Qr =
πa
a r3
8. Qr =
π
6. Qr =
9. Qr = π a r 3
10. Qr = π a r 4 correct
Explanation:
A charge element is given by
dq = ρ dV = (a r)(4 π r 2 dr) = 4 π a r 3 dr .
For r < b, we integrate to find the total
1
charge within radius r:
Z
Z r
Q = dq = 4 π a
r ′3 dr ′
0
r4
= π a r4 .
= 4πa
4
002 (part 2 of 3) 10.0 points
If a = 5 × 10−6 C/m4 and b = 1 m, find E at
r = 0.7 m. The permittivity of a vacuum is
8.8542 × 10−12 C2 /N · m2 .
1. 35294.0
2. 18070.5
3. 126494.0
4. 60988.0
5. 90352.6
6. 21176.4
7. 72282.1
8. 30494.0
9. 69176.2
10. 83011.5
Correct answer: 69176.2 N/C.
Explanation:
Let :
a = 5 × 10−6 C/m4 ,
r = 0.7 m , and
ǫ0 = 8.8542 × 10−12 C2 /N · m2 .
The problem exhibits spherical symmetry;
pick a spherical Gaussian surface of radius
r, concentric with the first. Over this surface, the flux is constant due to symmetry, so
Gauss’ Law gives
Qencl
π a r4
=
ǫ0
ǫ0
2
ar
E=
4 ǫ0
(5 × 10−6 C/m4 )(0.7 m)2
=
4 (8.8542 × 10−12 C2 /N · m2 )
4 π r2 E =
= 69176.2 N/C ,
directed outward.
003 (part 3 of 3) 10.0 points
Version 120 – Exam 1 – hoffmann – (57505)
Find the charge Qb contained within the radius r, when r > b.
1. Qb = 0
a b3
π
πa
3. Qb = 4
b
a b4
4. Qb =
π
2. Qb =
2
3. -20.0
4. -156.0
5. -1295.0
6. -468.0
7. -100.0
8. -63.75
9. -18.75
10. -80.0
Correct answer: −156 kV.
Explanation:
4
5. Qb = π a b correct
Let : Ex = (1 kN/C) x3 ,
x1 = 1 m , and
x2 = 5 m .
a b2
6. Qb =
π
7. Qb = π a b2
8. Qb =
π b2
a
Ex = −
9. Qb = π a b3
π b3
a
Explanation:
10. Qb =
Let :
b = 1 m.
For r > b,
Q=
Z
dq = 4π a
Z
0
b
b4
r dr = 4π a = π a b4 .
4
3
The charge enclosed within radius r does
not depend on r when we go outside the
sphere; it is just the charge of the entire
sphere. Therefore, we could have used the
result Q = π a r 4 from Part 1 with r = b
instead of carrying out the integration again.
While the charge does not change any more
as we go outside the sphere, the electric field
1
falls off as 2 due to Gauss’ Law.
r
004 10.0 points
An electric field is given by Ex = (1 kN/C) x3 .
Find the potential difference between the
points on the x-axis at x = 1 m and x = 5 m.
1. -11.25
2. -255.0
dV
,
dx
so
dV = −Ex dx
Z x2
Ex dx
V2 − V1 = −
x1
Z x2
x3 dx
= −(1 kN/C)
x1 1
= −(1 kN/C)
x42 − x41
4
1
= −(1 kN/C)
4
4
× (5 m) − (1 m)4
= −156 kV .
005 (part 1 of 6) 10.0 points
A positive test charge is placed at the center
of a spherical Gaussian surface.
What happens to the net flux through the
Gaussian surface when the surface is replaced
by a cube of the same volume whose center is
at the same point?
1. The net flux increases.
2. The net flux decreases but is nonvanishing.
3. No change correct
Version 120 – Exam 1 – hoffmann – (57505)
4. The net flux is zero.
Explanation:
By Gauss’ Law,
I
~ · dA
~ = Q.
ΦS =
E
ǫ0
S
As long as the charge remains the same, and
it is enclosed by any Gaussian surface, the net
flux will remain the same.
006 (part 2 of 6) 10.0 points
What happens to the net flux through the
Gaussian surface when the sphere is replaced
by a cube of one-third the volume centered at
the same point?
1. The net flux is zero.
2. No change correct
3. The net flux increases.
4. The net flux decreases but is nonvanishing.
3
1. The net flux increases.
2. The net flux decreases but is nonvanishing.
3. No change
4. The net flux is zero. correct
Explanation:
If the charge is moved outside the Gaussian
surface, then the amount of enclosed charge is
zero and the net flux is zero. One can imagine
the field lines going out from the charge. Any
line entering the closed Gaussian surface will
also leave, so the net flux is zero.
009 (part 5 of 6) 10.0 points
The positive test charge is again placed at
the center of a spherical Gaussian surface.
What happens to the net flux through
the Gaussian surface when a second positive
charge is placed near, but outside, the original
sphere?
1. The net flux increases.
Explanation:
2. The net flux is zero.
007 (part 3 of 6) 10.0 points
What happens to the net flux through the
Gaussian surface when the charge is moved
off center in the original sphere, but remains
within the sphere?
1. The net flux is zero.
2. The net flux decreases but is nonvanishing.
3. The net flux increases.
4. No change correct
Explanation:
008 (part 4 of 6) 10.0 points
What happens to the net flux through the
Gaussian surface when the charge is moved
just outside the original sphere?
3. The net flux decreases but is nonvanishing.
4. No change correct
Explanation:
From Part 4 it is clear that an exterior
charge does not contribute to the flux, so the
flux remains the same.
010 (part 6 of 6) 10.0 points
What happens to the net flux through the
Gaussian surface when a second positive
charge is placed inside the original spherical
Gaussian surface?
1. No change
2. The net flux decreases but is nonvanishing.
Version 120 – Exam 1 – hoffmann – (57505)
4
field from charge q3 points along the positive
x-axis, since q3 and q1 are of the same sign
(repulsive force).
3. The net flux is zero.
4. The net flux increases. correct
Explanation:
Since the net enclosed charge increases, the
flux must increase.
F12
II
−10 µC
b
1.15 µC
b
I
1.49 m
b
x
O −1.29 µC
2.98 m
III
IV
Find the direction of the resultant electric
force on the −1.29 µC charge at the origin
due to the other two charges. The value of the
Coulomb constant is 8.9875 × 109 N · m2 /C2 .
1. Along the positive y-axis
2. In quadrant II
3. Along the negative x-axis
q2 > 0
F1
q3 < 0
b
011 (part 1 of 3) 10.0 points
Consider 3 charges arranged as shown.
y
b
b
q1 < 0
F13
By inspection, the resultant points into
quadrant I.
012 (part 2 of 3) 10.0 points
Determine the magnitude of the resultant
electric force on the −1.29 µC charge at the
origin due to the other two charges.
1. 0.0653261
2. 0.4737
3. 0.179137
4. 0.118415
5. 0.0930883
6. 0.0143706
7. 0.0667232
8. 0.152478
9. 0.0504534
10. 0.0767198
Correct answer: 0.0143706 N.
Explanation:
4. Along the negative y-axis
5. Along the positive x-axis
6. In quadrant I correct
7. In quadrant III
Let : q1 = −1.29 µC ,
q2 = 1.15 µC ,
q3 = −10 µC ,
a = 1.49 m , and
b = 2.98 m ,
8. In quadrant IV
Explanation:
From Coulomb’s law, the magnitude of the
force from point charge q1 on point charge q2
at distance x is
q1 q2
F = ke 2 .
x
The electric field from charge q2 points along
the positive y-axis, since q2 and q1 are of
opposite sign (attractive force). The electric
q1 q2
a2
= 8.9875 × 109 N · m2 /C2
(−1.29 µC) (1.15 µC)
×
(1.49 m)2
= −0.00600557 N and
F12 = ke
F13 = ke
q1 q3
b2
Version 120 – Exam 1 – hoffmann – (57505)
5
Therefore, the electric field points into the
= 8.9875 × 109 N · m2 /C2
third quadrant.
(−1.29 µC) (−10 µC)
×
(2.98 m)2
014 10.0 points
= 0.0130556 N , so
A charge of 8 pC is uniformly distributed
throughout the volume between concentric
spherical surfaces having radii of 1.3 cm and
q
3.3 cm.
2 + F2
F1 = F12
13
What is the magnitude of the electric
q
field 2.2 cm from the center of the sur= (−0.00600557 N)2 + (0.0130556 N)2
faces? The value of Coulomb’s constant is
8.98755 × 109 N · m2 /C2 .
= 0.0143706 N .
1. 13.6987
2. 26.3778
013 (part 3 of 3) 10.0 points
3. 18.2372
Determine the direction of the resultant elec4. 19.5869
tric field on the −1.29 µC charge at the origin
5. 19.789
due to the other two charges.
6. 23.1868
7. 5.7174
1. In quadrant II
8. 37.2091
9. 20.329
2. Along the positive x-axis
10. 36.483
3. Along the negative x-axis
Correct answer: 37.2091 N/C.
Explanation:
4. Along the negative y-axis
5. In quadrant I
Let : qtot
r1
r2
r
6. In quadrant III correct
7. In quadrant IV
= 8 pC = 8 × 10−12 C ,
= 1.3 cm ,
= 3.3 cm ,
= 2.2 cm = 0.022 m , and
ke = 8.98755 × 109 N · m2 /C2 .
8. Along the positive y-axis
Explanation:
The electric field at the origin is opposite in
direction to the electric force we just found,
since the electric field is defined to be in the
direction in which a positive test charge would
feel a force.
b
q3 < 0
b
E2
E
F = qE
q2 > 0
E3
and
q<0
By Gauss’ law,
Φc =
I
~ · dA
~ = qin
E
ǫ0
The tricky part of this question is to determine the charge enclosed by our Gaussian
surface, which by symmetry considerations is
chosen to be a concentric sphere with radius
r. Since the charge q is distributed uniformly
within the solid,
qin
Vin
=
qtot
Vtot
where qin and Vin are the charge and volume
enclosed by the Gaussian surface, so
Version 120 – Exam 1 – hoffmann – (57505)
6
q1 = n qe and q2 = −n qe ,
r 3 − r1
= qtot
r23 − r13
(2.2 cm)3 − (1.3 cm)3
−12
= (8 × 10
C) ×
(3.3 cm)3 − (1.3 cm)3
= 2.00379 × 10−12 C
qin = qtot
Vin
Vtot
3
and by Gauss’s Law,
qencl
E = ke 2
r
= (8.98755 × 109 N · m2 /C2 )
2.00379 × 10−12 C
×
(0.022 m)2
= 37.2091 N/C .
015 10.0 points
In a charging process, 4 × 1013 electrons
are removed from one small metal sphere and
placed on a second identical sphere. Initially
both metal spheres were neutral. After the
charging process the electrical potential energy associated with the two spheres is found
to be −0.068 J .
What is the distance between the two
spheres? The value of the Coulomb constant
is 8.98755 × 109 N · m2 /C2 and the elemental
charge is 1.6 × 10−19 C .
1. 4.97473
2. 0.31518
3. 5.41368
4. 1.43801
5. 5.49448
6. 0.371099
7. 5.259
8. 7.77302
9. 0.348608
10. 1.27823
Ue = k e
so
q1 q2
r
−(n qe )2
q1 q2
= ke
Ue
Ue
9
= −(8.98755 × 10 N · m2 /C2 )
(4 × 1013 )2 (1.6 × 10−19 C)2
×
−0.068 J
r = ke
= 5.41368 m .
016 (part 1 of 4) 10.0 points
Two charges are located in the (x, y) plane as
shown. The fields produced by these charges
are observed at a point p with coordinates
(0, 0).
p
1.7 m
1.7 m
−9.1 C
8.3 C
2.2 m
3m
Find the x-component of the electric
field at p.
The Coulomb constant is
9
8.98755 × 10 N · m2 /C2 .
1. 1721310000.0
2. -1591370000.0
3. -241919000.0
4. 13620700000.0
5. 899503000.0
6. 1568060000.0
7. -1210860000.0
8. 11524100000.0
9. -3914000000.0
10. 2122420000.0
Correct answer: 5.41368 m.
Correct answer: 1.36207 × 1010 N/C.
Explanation:
Explanation:
Let :
n = 4 × 1013 ,
qe = −1.6 × 10−19 C,
Ue = −0.068 J , and
ke = 8.98755 × 109 N · m2 /C2 .
Let : ke = 8.98755 × 109 N · m2 /C2 ,
(xp , yp) = (0, 0) ,
(x1 , y1) = (3 m, −1.7 m) ,
Q1 = −9.1 C ,
Version 120 – Exam 1 – hoffmann – (57505)
(x2 , y2) = (−2.2 m, −1.7 m) ,
Q2 = 8.3 C .
and
Ex = Ex1 + Ex2
= 5.98455 × 109 N/C
+ 7.63612 × 109 N/C
p
y2
y1
q2
q1
x2
x1
Consider the electric field vectors
E2
θ
E1
−9.1 C
8.3 C
r1 =
q
(3 m)2 + (−1.7 m)2
= 3.44819 m and
r2 =
q
= 1.36207 × 1010 N/C .
017 (part 2 of 4) 10.0 points
Find the electric potential at p with coordinates (0, 0) . Let V = 0 at infinity.
1. -54527300000.0
2. 45607700000.0
3. 941991000.0
4. 3111810000.0
5. 58492500000.0
6. 2069910000.0
7. -53102800000.0
8. 1493610000.0
9. -49715900000.0
10. 8545370000.0
Correct answer: 3.11181 × 109 V.
Explanation:
The potential for a point charge Q is
(−2.2 m)2 + (−1.7 m)2
V = ke
= 2.78029 m .
Horizontally, the contributions from the
two charges are
Ex1
Q1 −x1
Q1
= ke 2 cos θ1 = ke 2
r1
r1 r1
= −(8.98755 × 109 N · m2 /C2 )
3m
−9.1 C
×
2
(3.44819 m) 3.44819 m
= 5.98455 × 109 N/C , and
E x 2 = ke
7
Q2
Q2 −x2
cos θ2 = ke 2
2
r2
r2 r2
= −(8.98755 × 109 N · m2 /C2 )
8.3 C
−2.2 m
×
2
(2.78029 m) 2.78029 m
= 7.63612 × 109 N/C , so
Q
,
r
so
Q1
r1
= (8.98755 × 109 N · m2 /C2 )
−9.1 C
×
3.44819 m
= −2.37188 × 1010 V and
V1 = ke
Q2
r2
= (8.98755 × 109 N · m2 /C2 )
8.3 C
×
2.78029 m
= 2.68306 × 1010 V , and
V2 = ke
Version 120 – Exam 1 – hoffmann – (57505)
Vp = V1 + V2
Q2
sin(180◦ − θ2 )
2
r2
Q2 −y2
= ke 2
r2 r2
= −2.37188 × 1010 V
+ (2.68306 × 1010 V)
E y2 = ke
= 3.11181 × 109 V .
018 (part 3 of 4) 10.0 points
Find the y-component of the electric field at
p by taking the appropriate partial derivative
of the potential function from part 2.
1. -13815100000.0
2. 39790600.0
3. -2465090000.0
4. 2103230000.0
5. -12181900000.0
6. 2509400000.0
7. 6907540000.0
8. -2887060000.0
9. -1344010000.0
10. -6336990000.0
Correct answer: 2.5094 × 109 N/C.
Explanation:
p
r = y 2 + x2 ,
so
−1/2
1
∂r
= y 2 + x2
·2y
∂y
2
∂V
Q
= −k 2 , so
∂r
r
and
∂V
∂V ∂r
=−
∂y
∂r ∂y
Q 1 2
2 −1/2
=k 2
y +x
·2y
r 2
Qy
Q y
=k 2· =k 3 ,
r r
r
Ey = −
= −(8.98755 × 109 N · m2 /C2 )
−1.7 m
8.3 C
×
2
(2.78029 m) 2.78029 m
= 5.90064 × 109 N/C , and
E y = E y1 + E y2
= −3.39124 × 109 N/C
+ 5.90064 × 109 N/C
= 2.5094 × 109 N/C .
019 (part 4 of 4) 10.0 points
~ · d~s over a
Find Vs − Vp by integrating −E
straight line path between the two points.
Let Point s be at position (0, 1).
1. 9321390000.0
2. -1957170000.0
3. 12519400000.0
4. 11299300000.0
5. 10328000000.0
6. 10194800000.0
7. -1912560000.0
8. -335440000.0
9. 11149200000.0
10. 3277440000.0
Correct answer: −1.95717 × 109 V.
Explanation:
In general, the potential difference between
any two points A and B is
Q1
sin(180◦ − θ1 )
2
r1
Q1 −y1
= ke 2
r1 r1
E y1 = ke
9
8
2
2
= −(8.98755 × 10 N · m /C )
−9.1 C
−1.7 m
×
2
(3.44819 m) 3.44819 m
= −3.39124 × 109 N/C , and
VB − VA = −
Z
B
A
~ · d~s .
E
Let A correspond to the origin p and B to
the point s on the positive y-axis
Version 120 – Exam 1 – hoffmann – (57505)
Vs − Vp = −
Z
s
p
~ · d~s = −
E
Z
s
V2 ,s = ke
Ey dy
p
Vs − Vp = −k Q
s
p
Q2
r2 ,s
= (8.98755 × 109 N · m2 /C2 )
8.3 C
×
3.48281 m
= 2.14185 × 1010 V , and
and from Part 3
Z
9
y
dy
r3
where
d
dy
Vs = V1 ,s + V2 ,s
1 y
1
y
=− 2 · =− 3.
r
r r
r
Using the chain rule and
= −2.02639 × 1010 V
+ 2.14185 × 1010 V
= 1.15464 × 109 V .
∂r
from Part 3,
∂y
Q s k Q k Q
−
,
Vs − Vp = k =
r p
rs
rp
kQ
kQ
and Vp =
, so all we need
rs
rp
to do is evaluate the potential at point s and
at point p and subtract them.
From above, Vp = 3.11181 × 109 V. The
distances from each charge to point s are
The potential difference ∆V is
∆V = Vs − Vp
= 1.15464 × 109 V
− 3.11181 × 109 V
where Vs =
r1 ,s =
=
q
q
= −1.95717 × 109 V .
020 10.0 points
Pictured below is a distribution of 6 point
charges and their surrounding electric field.
x21 + (−y1 + 1)2
+Q
(3
m)2
+ (2.7
= 4.03609 m ,
Gaussian surface
m)2
and
-Q
+Q
-Q
+Q
-Q
r2 ,s
q
= x22 + (−y2 + 1)2
q
= (−2.2 m)2 + (2.7 m)2
= 3.48281 m ,
V1 ,s = ke
so
Q1
r1 ,s
= (8.98755 × 109 N · m2 /C2 )
−9.1 C
×
4.03609 m
= −2.02639 × 1010 V ,
What is the total electric flux through the
closed Gaussian surface shown?
−Q
ǫ0
2Q
2.
ǫ0
−2 Q
3.
correct
ǫ0
Q
4.
ǫ0
1.
5. 0
Version 120 – Exam 1 – hoffmann – (57505)
6Q
ǫ0
−6 Q
7.
ǫ0
Explanation:
The total charge within the Gaussian surface is −2 Q, so the total electric flux is
6.
φ=
−2 Q
.
ǫ0
10