Version 001 – Electostatics I – tubman – (2014B2)
This print-out should have 10 questions.
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before answering.
1
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
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
−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
correct
2.
3.
4.
5.
Sphere Y
1. Positive
Negative correct
6.
2. Negative
Positive
7.
3. Positive
Positive
8.
4. Zero
Zero
5. Negative
Negative
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
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 – Electostatics I – tubman – (2014B2)
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?
2F
3
F
=√
2
1. Fnet =
2. Fnet
3. Fnet = 0
5. Fnet = 2 F
√
6. Fnet = 2 F correct
F
8. Fnet = √
3
3F
9. Fnet =
2
2F
10. Fnet = √
3
Explanation:
The individual forces form a right angle, so
the magnitude of the net force is
p
√
Fnet = F 2 + F 2 = 2 F .
4 µC
Charge on Third Particle
004 10.0 points
A particle with charge 4 µC is located on
the x-axis at the point 10 cm , and a second
particle with charge 2 µC is placed on the
x-axis at −6 cm .
2 µC
Explanation:
Let : q1
q2
q3
x1
x2
x3
= 4 µC = 4 × 10−6 C ,
= 2 µC = 2 × 10−6 C ,
= 2 µC = 2 × 10−6 C ,
= 10 cm = 0.1 m ,
= −6 cm = −0.06 m , and
= −2 cm = −0.02 m .
~ 13 = ke q1 q3 r̂13 ,
F
r2
7. Fnet = 3 F
2 µC
Correct answer: 17.4757 N.
Coulomb’s law (in vector form) for the electric force exerted by a charge q1 on a second
charge q3 , is
4. Fnet = F
2 4 6 8 10
x → (cm)
What is the magnitude of the total electrostatic force on a third particle with charge
2 µC placed on the x-axis at −2 cm ? The
Coulomb constant is 8.9875 × 109 N · m2 /C2 .
−10 −8 −6 −4 −2
2
where r̂13 is a unit vector directed from q1 to
q3 ; i.e., ~r13 = ~r3 − ~r1 .
x13 = x3 − x1
= (−2 cm) − (10 cm) = −0.12 m
x23 = x3 − x2
= (−2 cm) − (−6 cm) = 0.04 m
x3 − x1
x̂13 = p
= −ı̂
(x3 − x1 )2
x3 − x2
x̂23 = p
= +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)
×
(−0.12 m)2
(2 × 10−6 C)
+
(+ı̂)
(0.04 m)2
×(2 × 10−6 C)
= 17.4757 N ,
Version 001 – Electostatics I – tubman – (2014B2)
with a magnitude of 17.4757 N .
Charges in a Thundercloud
005 (part 1 of 2) 10.0 points
In a thundercloud there may be an electric
charge of 59 C near the top of the cloud and
−59 C near the bottom of the cloud.
If these charges are separated by about
2 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 .
3
The negative sign indicates that the force
is attractive; i.e., the charges are being pulled
towards one another.
Conceptual 16 Q03
007 10.0 points
If you double the charge on one of two charged
objects, how does the force between them
change?
1. Triples
2. Quadruples
6
Correct answer: 7.82142 × 10 N.
Explanation:
3. Halves
4. Doubles correct
Let :
q1 = 59 C ,
q2 = −59 C ,
d = 2 km = 2000 m , and
k = 8.98755 × 109 N · m2 /C2 .
Considering the two sets of electric charges
as point-charges, separated by a distance of
2 km, we can apply Coulomb’s law:
q1 q2
r2
= 8.98755 × 109 N · m2 /C2
(59 C) (−59 C)
×
(2000 m)2
= −7.82142 × 106 N ,
F =k
with a magnitude of 7.82142 × 106 N .
006 (part 2 of 2) 10.0 points
The electrostatic force between the top and
the bottom of this thundercloud is
5. Does not change
Explanation:
Fe = ke
q1 q2
∝ q1 q2
r2
The factor is (2 q1 ) q2 = 2 (q1 q2 ) .
The force doubles because the product of
the charge doubles.
Conceptual 16 Q02
008 10.0 points
If you double the distance between two
charged objects, by what factor is the electric force affected?
Correct answer: 0.25.
Explanation:
Fe = ke
q1 q2
1
∝ 2
2
r
r
The factor is
1. zero.
2. undetermined.
3. attractive. correct
4. repulsive.
Explanation:
1
1
1
=
= 0.25 2 .
2
2
(2 r)
4r
r
Induction and Grounding
009 10.0 points
1) Two uncharged metal balls, X and Y, each
stand on a glass rod and are touching.
Version 001 – Electostatics I – tubman – (2014B2)
Y
X
2) A third ball carrying a negative charge, is
brought near the first two.
−
Y
X
3) While the positions of these balls are fixed,
ball X is connected to ground.
−
Y
X
4) Then the ground wire is disconnected.
−
Y
X
5) While X and Y remain in touch, the ball
carring the negative charge is removed.
Y
X
6) Then ball X and Y are separated.
Y
X
After these procedures, what are the signs
of the charge qX on X and qY on Y?
1.
qX is negative and qY is neutral
2.
qX is negative and qY is negative
3.
qX is neutral and qY is positive
4
4.
qX is positive and qY is positive correct
5.
qX is negative and qY is positive
6.
qX is neutral and qY is negative
7.
qX is positive and qY is negative
8.
qX is neutral and qY is neutral
9.
qX is positive and qY is neutral
Explanation:
When the ball with negative charge is
brought nearby, the free charges inside X and
Y rearrange themselves. The positive charges
are attracted and go to the right (i.e., move to
Y), leaving negative charges on the left hand
side of the system X Y, i.e., in X.
When we ground X, electrons flow from the
ground to X (making it neutral), whereas the
positive charges in Y are still held enthralled
by the negative charge on the third ball. We
break the ground.
Now we remove the third ball with negative
charge. The charge on Y is redistributed in
the system X Y; i.e., they share the positive
charge (equally if identical).
Finally we separate X and Y. The signs
of the charge on X and that on Y are both
positive.
Force Between Electrons
010 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:
Let :
d = 1.6 × 10−10 m and
ke = 9 × 109 N · m2 /C2 .
The force between the electrons is
ke qe2
ke q1 q2
= 2
F =
d2
d
Version 001 – Electostatics I – tubman – (2014B2)
=
9 × 109 N · m2 /C2
1.6 × 10−10 m2
× (1.602 × 10−19 C)2
= 9.02251 × 10−9 N .
5