Version 001 – Electrostatics I – tubman – (12125)
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AP EM 1993 MC 55
001 10.0 points
Two metal spheres that are initially uncharged are mounted on insulating stands, as
shown.
−
− −−
X
Y
Two charged particles of equal magnitude
(−Q and +Q) are fixed at opposite corners of
a square that lies in a plane (see figure below).
A test charge +q is placed at a third corner.
−Q
+q
+Q
What is the direction of the force on the
test charge due to the two other charges?
1.
A negatively charged rubber rod is brought
close to but does not make contact with sphere
X. Sphere Y is then brought close to X on the
side opposite to the rubber rod. Y is allowed
to touch X and then is removed some distance
away. The rubber rod is then moved far away
from X and Y.
What are the final charges on the spheres?
Sphere X
1
correct
2.
3.
4.
5.
Sphere Y
6.
1. Positive
Negative correct
2. Zero
Zero
3. Positive
Positive
4. Negative
Negative
5. Negative
Positive
Explanation:
When the negatively charged rod moves
close to the sphere X, the negatively charged
electrons will be pushed to sphere Y. If X and
Y are separated before the rod moves away,
those charges will remain on X and Y, so
X is positively charged and Y is negatively
charged.
AP EM 1998 MC 39 40
002 (part 1 of 2) 10.0 points
7.
8.
Explanation:
The force between charges of the same sign
is repulsive and between charges with opposite signs is attractive.
−Q
+Q
+q
The resultant force is the sum of the two
vectors in the figure.
Version 001 – Electrostatics I – tubman – (12125)
003 (part 2 of 2) 10.0 points
If F is the magnitude of the force on the test
charge due to only one of the other charges,
what is the magnitude of the net force acting on the test charge due to both of these
charges?
1. Fnet = 0
2F
2. Fnet = √
3
3. Fnet = 3 F
F
4. Fnet = √
3
√
5. Fnet = 2 F correct
Correct answer: 66.7322 N.
Explanation:
Let : q1
q2
q3
x1
x2
x3
= −4 µC = −4 × 10−6 C ,
= −2 µC = −2 × 10−6 C ,
= 9 µC = 9 × 10−6 C ,
= 6 cm = 0.06 m ,
= 8 cm = 0.08 m , and
= −2 cm = −0.02 m .
Coulomb’s law (in vector form) for the electric force exerted by a charge q1 on a second
charge q3 , is
! 13 = ke q1 q3 r̂13 ,
F
r2
6. Fnet = F
7. Fnet =
2
where r̂13 is a unit vector directed from q1 to
q3 ; i.e., !r13 = !r3 − !r1 .
3F
2
8. Fnet = 2 F
F
9. Fnet = √
2
2F
10. Fnet =
3
Explanation:
The individual forces form a right angle, so
the magnitude of the net force is
!
√
Fnet = F 2 + F 2 = 2 F .
−2 µC
−4 µC
9 µC
Charge on Third Particle
004 10.0 points
A particle with charge −4 µC is located on
the x-axis at the point 6 cm , and a second
particle with charge −2 µC is placed on the
x-axis at 8 cm .
2 4 6 8 10
x → (cm)
What is the magnitude of the total electrostatic force on a third particle with charge
9 µC placed on the x-axis at −2 cm ? The
Coulomb constant is 8.9875 × 109 N · m2 /C2 .
−10 −8 −6 −4 −2
x13 = x3 − x1
= (−2 cm) − (6 cm) = −0.08 m
x23 = x3 − x2
= (−2 cm) − (8 cm) = −0.1 m
x3 − x1
x̂13 = !
= −ı̂
(x3 − x1 )2
x3 − x2
x̂23 = !
= −ı̂
(x3 − x2 )2
Since the forces are collinear, the force on
the third particle is the algebraic sum of the
forces between the first and third and the
second and third particles:
! =F
! 13 + F
! 23
F
#
"
q2
q1
= ke 2 r̂13 + 2 r̂23 q3
r13
r23
= 8.9875 × 109 N · m2 /C2
"
(−4 × 10−6 C)
(−ı̂)
×
(−0.08 m)2
#
(−2 × 10−6 C)
+
(−ı̂)
(−0.1 m)2
×(9 × 10−6 C)
= 66.7322 N ,
Version 001 – Electrostatics I – tubman – (12125)
with a magnitude of 66.7322 N .
Charges in a Thundercloud
005 (part 1 of 2) 10.0 points
In a thundercloud there may be an electric
charge of 28 C near the top of the cloud and
−28 C near the bottom of the cloud.
If these charges are separated by about
7 km, what is the magnitude of the electric force between these two sets of charges?
The value of the electric force constant is
8.98755 × 109 N · m2 /C2 .
Correct answer: 1.43801 × 105 N.
Explanation:
Let :
q1 = 28 C ,
q2 = −28 C ,
d = 7 km = 7000 m , and
k = 8.98755 × 109 N · m2 /C2 .
Considering the two sets of electric charges
as point-charges, separated by a distance of
7 km, we can apply Coulomb’s law:
q1 q2
F =k 2
r
= 8.98755 × 109 N · m2 /C2
(28 C) (−28 C)
×
(7000 m)2
= −1.43801 × 105 N ,
with a magnitude of 1.43801 × 105 N .
006 (part 2 of 2) 10.0 points
The electrostatic force between the top and
the bottom of this thundercloud is
The negative sign indicates that the force
is attractive; i.e., the charges are being pulled
towards one another.
Compare Two Coulomb Forces
007 10.0 points
Two charges q1 and q2 are separated by a
distance d and exert a force F on each other.
What is the new force F " , if charge 1 is
increased to q1" = 5 q1 , charge 2 is decreased
q2
to q2" = , and the distance is decreased to
2
d
"
d = ?
2
5
1. F " = F
2
5
2. F " = F
4
3. F " = 100 F
4. F " = 50 F
5. F " = 25 F
6. F " = 10 F correct
7. F " =
25
F
2
8. F " = 20 F
9. F " = 5 F
25
F
4
Explanation:
10. F " =
k q1" q2"
=
F =
r" 2
"
1. repulsive.
= 10
2. attractive. correct
3. undetermined.
4. zero.
Explanation:
3
$
k (5 q1 )
$ %2
d
2
q2
2
%
k q1 q2
= 10 F .
d2
Conceptual 16 Q03
008 10.0 points
If you double the charge on one of two charged
objects, how does the force between them
change?
Version 001 – Electrostatics I – tubman – (12125)
1. Doubles correct
2. No change correct
2. Does not change
3. Triples
3. Quadruples
4. Quadruples
4. Halves
5. Doubles
4
Explanation:
The charge on each object doubles, as does
the distance:
5. Triples
Explanation:
q1 q2
∝ q1 q2
r2
The factor is (2 q1 ) q2 = 2 (q1 q2 ) .
The force doubles because the product of
the charge doubles.
Fe = ke
Conceptual 16 Q02
009 10.0 points
If you double the distance between two
charged objects, by what factor is the electric force affected?
F ∝
q1 q2
(2 q1 ) (2 q2 )
q1 q2
=
= 2 .
2
2
r
(2 r)
r
The electrical force between them does not
change.
Induction and Grounding
011 10.0 points
1) Two uncharged metal balls, Y and X, each
stand on a glass rod and are touching.
Correct answer: 0.25.
X
Y
Explanation:
Fe = ke
1
q1 q2
∝ 2
2
r
r
The factor is
1
1
1
=
= 0.25 2 .
2
2
(2 r)
4r
r
Conceptual 16 Q14
010 10.0 points
Object A and object B are initially uncharged
and are separated by a distance of 1 meter.
Suppose 10,000 electrons are removed from
object A and placed on object B, creating an
attractive force between A and B. An additional 10,000 electrons are removed from A
and placed on B and the objects are moved so
that the distance between them increases to 2
meters.
By what factor does the electric force between them change?
1. Halves
2) A third ball carrying a positive charge, is
brought near the first two.
+
X
Y
3) While the positions of these balls are fixed,
ball Y is connected to ground.
+
X
Y
4) Then the ground wire is disconnected.
+
X
Y
Version 001 – Electrostatics I – tubman – (12125)
5) While Y and X remain in touch, the ball
carring the positive charge is removed.
X
Y
6) Then ball Y and X are separated.
X
Y
After these procedures, what are the signs
of the charge qY on Y and qX on X?
1.
5
charge. The charge on X is redistributed in
the system Y Y; i.e., they share the negative
charge (equally if identical).
Finally we separate Y and X. The signs
of the charge on Y and that on X are both
negative.
Force Between Electrons
012 10.0 points
Two electrons in an atom are separated by
1.6 × 10−10 m, the typical size of an atom.
What is the force between them? The
Coulomb constant is 9 × 109 N · m2 /C2 .
Correct answer: 9.02251 × 10−9 N.
Explanation:
qY is positive and qX is neutral
2. qY is negative and qX is negative correct
3.
qY is positive and qX is negative
4.
qY is neutral and qX is positive
5.
qY is positive and qX is positive
6.
qY is negative and qX is positive
7.
qY is neutral and qX is negative
8.
qY is negative and qX is neutral
9.
qY is neutral and qX is neutral
Explanation:
When the ball with positive charge is
brought nearby, the free charges inside Y
and X rearrange themselves. The negative
charges are attracted and go to the right
(i.e., move to X), leaving positive charges on
the left hand side of the system Y Y, i.e., in
Y.
When we ground Y, electrons flow from the
ground to Y (making it neutral), whereas the
negative charges in X are still held enthralled
by the positive charge on the third ball. We
break the ground.
Now we remove the third ball with positive
Let :
d = 1.6 × 10−10 m and
ke = 9 × 109 N · m2 /C2 .
The force between the electrons is
ke q1 q2
ke qe2
=
d2
d2
9 × 109 N · m2 /C2
=
1.6 × 10−10 m2
× (1.602 × 10−19 C)2
F =
= 9.02251 × 10−9 N .
Hewitt CP9 22 E20
013 10.0 points
How does the magnitude of the electrical force
between a pair of charged particles change
when they are brought to half their original
distance of separation?
1. Reduces to one half of original value
2. quadruples correct
3. doubles
4. Reduces to one quarter of original value
5. Doesn’t change
Explanation:
Version 001 – Electrostatics I – tubman – (12125)
By the inverse square law, at half the distance the electric force field is four times as
strong:
1
4
F ∝ & '2 = 2
1
r
2r
6