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Q21-1. Two uniformly charge spheres are attached to frictionless pucks on an air table. The
charge on sphere 2 is three times the charge on sphere 1. Which force diagram correctly shows
the magnitude and direction of the electrostatic forces on the two spheres?
A)
B)
C)
D)
E)
Answer: D (By Newton's 3rd Law and by the form of Coulomb's Law)
Q21-2. Two protons are near each other. Each feels an electrostatic repulsion of magnitude Felec
and a gravitational attraction of magnitude Fgrav, due to the other charge.
F
As the charges are moved apart, the ratio elec …
Fgrav
A) increases B) decreases C) remains constant
+
+
Answer: remains constant
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Q21-3. An electric dipole consists of two equal and opposite charges
(+Q and –Q) separated by some fixed distance d . A charge +q is brought near the dipole and is
positioned so that the distances to the +Q and the –Q charges are identical as shown below:
A
+q
D
+Q
B
–Q
d
C
E) Some other direction
What is the direction of the net electrostatic force on the +q charge?
Answer: to the right (Be sure you can draw the vector diagram, showing the forces on +q)
Q21-4. What is the correct expression for the y-component of the force on +q due to the other
charges +Q and –Q ?
y
+Q
A)
B)
r
C)
+q
r
Answer:
k
Qq
r2
–Q
x
D)
Qq
r2
Qq
k 2
r
k
Qq
r2
Qq
k 2 2
r
 2 k
E) None of these!
(Notice that the y-component of the force is negative)
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Q21-5. Vector Review: How many of these equation make no sense?

A  ˆj

B  3iˆ  ˆj  5jˆ


A
3
C  A  ˆi
C

C  4  ˆi
A) 0
(All make sense)
B) 1
C) 2
D) 3
E) 4
Answer: The last one C  4  ˆi makes no sense. You can't add a number and a vector. All the
others makes sense.
Q21-6. Trig Review: In a tilted xy coordinate system, the weight vector mg is straight down.
The coordinates are tilted at an angle  as shown. What is Wy , the y-component of the weight
mg?
A) +mg sin
y
B) –mg cos
C) –mg sin
D) +mg
E) 0

mg
x
Answer: –mg cos (careful with the signs!)
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Q21-7. An electric dipole (+Q and –Q separated by a distance d ) is placed along the x-axis as
shown. A positive test charge +q is placed at position A to the right of the dipole.
–Q
+Q
+q
x
A
The positive test charge feels a force that is
A) zero.
B) to the right.
C) to the left.
If the test charge +q is removed, electric field at position A is
A) zero.
B) to the right
C) to the left
D) There is no electric field at point A, when the test charge is removed.
If a negative test charge –q is used to measure the Electric field at A, the electric field at position
A is A) zero.
B) to the right
C) to the left
Answers: The force on +q is to the left. The E-field at A points to the left, regardless of whether
the test charge is positive (+q) or negative (–q).
Q21-8. Two charges +Q and – Q are located on the x-axis as shown, what is the magnitude of
the electric field at point A?
–Q
+Q
R
A)
kQ  1 
 1  
R 2  9
C) Zero
Answer:
A
x
R
R
B) kQ2  1  1 
R
D)

4
kQ  1 
   1
R2  9 
E) None of these.
kQ  1 
 1  
R 2  9
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Q21-9. Two positive charges, each of size +Q, are equal distances from the origin as shown.
What is the direction of the electric field at the point in empty space which forms a square with
the two charges and the origin?
y
Field
here?
+Q
x
+Q
A
B
C
D
E) No Field there
Answer: B (upper right)
Q21-10. Two vectors A and B are shown. Consider the vector sum C  A  B . What is Cy,
the y-component of C ?
A: 3
y
B: 2
x
A
B
C: –2
D: –4
E: None of these/don't know.
Answer: Cy = Ay + By = +1 – 3 = –2
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Q21-11. A point in empty space is near 3 charges as shown. The distances from the point to
each of the three charges are identical.
+Q
y
C
+Q
+2Q
A
x
B
The direction of the electric field at that point is..
A: Some angle less than 45o below the +x-direction.
B: 45o below the +x-direction.
C: along the +y directions
D: Some other angle.
E: The electric field at that point is zero.
Answer: 45o below the +x-direction. (Be sure you can draw the E-field diagram.)
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Q21-12. Three charges of equal magnitude are arranged as shown. What is the direction of the
electric field at point x?
BA
C
B
DA
C
x
x
E) None of these.
An electron is fired into the region of the three charges from the lower right as shown. What is
the direction of the acceleration of the electron when it is at point x?
B
C
A
D
x
E) None of these.
Answers: The direction of the E-field at point x is straight up (choice A). The direction of the
acceleration of an election is the direction of a = F/m = qE/m = –eE/m (vector quantities are in
bold here), which is the direction opposite of the E-field. So the direction of the acceleration is
straight down (None of these).
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Ch26-13. A charge Q is on the y-axis at y = d. What is the magnitude of the E-field at position
x, on the x-axis?
A)
B)
C)
kQ
d  x 
y
Q
2
kQ
x  d2
2
x
kQ
 d2 
2
2
d
kQ
D)

x2  d2
x
E) None of these
E
What is Ex the x-component of E at position x on the x-axis?
d
A) E
B) E
x 2  d2
Answers: E =
x
x 2  d2
C) E
x2
kQ
x
Ex = +E cos = E
2 ,
2
x d
x  d2
2
Ch26-14. What is the magnitude of the electric field
at point x on the x-axis due to the two +Q charges
shown on the y-axis?
A)
B)
C)
x
x
x
D) None of these
x 2  d2
y
+Q
2kQx
2
 d2 
3/ 2
d
E=?
2kQd
2
 d2 
1/ 2
x
d
2kQx
2
d

2 1/ 2
D) None of these
Answer:
x
+Q
2kQx
2
 d2 
3/ 2
(see the previous problem to see why)
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Q21-15. Two socks are observed to attract each other. Which, if any, of the first 3 statements
MUST be true? (emphasis on MUST)
A)
The socks both have a non-zero net charge of the same sign.
B)
The socks both have a charge, of opposite signs.
C)
Only one sock is charged; the other is neutral.
D)
None of the preceding statements must be true.
Answer: D. The socks might be attracting because they have opposite sign charges. Or it could
be that only one of them is charged and the other is neutral, in which case the attraction is due to
polarization of the neutral sock.
Q21-16. A dipole is placed in an external field as shown. In which situation(s) is the net force
on the dipole zero?
(1)
(3)
A:
B:
C:
D:
E:
(2)
(4)
(1)
(2)
(1) and (2)
(3) and (4)
(2) and (4)
Answer: The net force is zero in cases (3) and (4). In case 4, there is a torque, so the particle
will rotate, but there is not net force.
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Q21-17.

z


The integral E  dE means E
z
 x dE x
z
 y dE y
z
 z dE x .
Often, from symmetry, one can see that one or more of the three component integrals vanishes.
An infinite line of charge with linear charge density  is along the x-axis and extends to  . At
the point A shown, what can you can about the x- and y- components of
y

E?
A
++++++++++++++++++++++++++
x
A: Ex = 0, Ey <0
C: Ex = 0, Ey =+
E: Ex > 0, Ey > 0
B: Ex < 0, Ey = 0
D: Ex = 0, Ey >0
Answer: Ex = 0, Ey >0
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z
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Q21-18. A circular ring of radius R, uniformly charged
with tot al charge +Q, is in the xy plane centered on the
origin. The electric field dE at position z=h on the zaxis, due to a small piece of the ring with charge dQ, is
shown. What is the magnitude of the field dE?
A:
C:
kQ
h2
B:
dE

k dQ
h2
h
R
k dQ
k dQ
R2  h2
D:
x
R2  h2
y
dQ
E: None of these.
What is dEz, the z-component of dE ?
A: dE sin
B: dE cos
C: dE tan
D: None of these.
What is cos?
A:
h
B:
2
R  h2
h
C:
h
R
R2  h2

h
1 
cos


D:
E: None of these!
2
2
 R h 
Answers: dE =
k dQ
,
R2  h2
dEz = dE cos , cos =
2/6/2016 PHYS1120 Dubson Fa09
h
R  h2
2
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z
Q21-19. A circular ring uniformly charged with charge Q is in the xy

plane centered on the origin as shown. On the z-axis, E  E z
graph accurately represents the electric field Ez on the z-axis?
z . Which
y
Ez
Ez
x
z
A
z
B
Ez
Ez
z
z
D
C
E: None of these is an accurate representation of Ez
Answer: B
Q21-20. A charge +Q is fixed in space. A second charge +q is brought a distance r away. Then
+q charge is removed and another charge +2q is brought a distance 2r away. Then the +2q
charge is removed and a charge +5q is brought a distance 2r away.
+Q
Which charge feels the largest force?
A) +q
B) +2q
C) +5q
D) Two of the charges feel the same size
force.
+q
r
+Q
+2q
2r
+5q
+Q
Answer: +5q
2r
2/6/2016 PHYS1120 Dubson Fa09
©University of Colorado