SOLUTIONS: PROBLEM SET 1
ELECTRIC FORCE AND ELECTRIC FIELD
PART A: CONCEPTUAL QUESTIONS
C. Charges in the uncharged sphere respond to the electric field produced by the charged sphere. The two halves of
the uncharged sphere become equally and oppositely charged, with the side close to the charged sphere having
the opposite charge to the charged sphere. The side closer to the charged sphere gets attracted and the side
further gets repelled, but the attractive force is greater (F ∝ 1/r2, r smaller for the closer side). Therefore, the net
force is attractive.
1
The instant the sphere touch, charges flow from one sphere into the other (they are metal and conduct charges
easily), making them both charged similarly. Like charges repel, therefore they get pushed apart.
We know that afterward both spheres are charged with the same charge, but we have no way of knowing if they
are positive or negative. If the charged sphere had been positive, then electrons would have flowed into it from
the uncharged sphere, making it less positive and the other sphere positive. If the charged sphere had initially
been negative, electrons would have flowed from it into the uncharged sphere, making it less negative and the
other sphere negative as well.
D.
a) Different clothing fabrics have different affinities for electric charges. When rubbed together, some clothing
items tend to grab electrons from others, making some negatively charged and some positively charged. As a
result, the oppositely charged items attract each other. By the way, since charged objects will also be attracted to
neutral objects (see question C), charged clothing will also be attracted to human legs, arms, etc. (Oh, the
heartbreak of static cling!)
b) The effect would be much smaller, since there would be less tendency for charges to be transferred if all the
material had the same affinity for electrons. (Try this at home!)
c) For the same reason discussed in a), bits of clothing become charged as they rub together. If the charge buildup is sufficient, there might be a spark as the electrons jump back across space. (This is referred to as dielectric
breakdown, when the electric field magnitude becomes stronger than the dielectric strength of the medium between
the charges. In this case that medium is air; dry air has a dielectric strength of 3x106 N/C. Descriptively, what
happens is this: the electric field is so strong that electrons are pulled free from the negatively charged piece of
material. These electrons crash into molecules of air, ionizing them and liberating even more electrons.
Essentially, we end up with an avalanche of charge, and the air conducts electrons from the negative to the
positive material. This happens with amazing quickness. The air is suddenly heated and expands, causing the
sound. As the electrons recombine with the ionized atoms in the air, light is emitted. If this description sounds a
lot like thunder and lightning, that’s essentially what it is, on a tiny scale.)
d) Water molecules are polar. Warm, humid air contains a large number of these polar molecules. The polar
molecules tend to steal excess charge from a charged object, making it difficult for any material to accumulate a
significant amount of excess charge.
E. a)
F. b)
G. d)
H. c)
I.
-
2
PART B: NUMERICAL QUESTIONS
QUESTION 1
a) q = −3e = −4.8 ×10−19 C .
Since the charge is at equilibrium, the electric force and gravitational force are equal and opposite.
FE = FG = mg
= 3.92 ×10−13 N
r
r FE
E=
q
=
3.92 x10−13 N
= −8.17 ×105 ˆj N / C
−4.8 ×10−19 C
b) −9.81 ˆj N / kg down
3
QUESTION 2
a) The magnitude of the electric force is given by:
r
kq q
FE = 12 2
r
(9 ×109 )(1.6 ×10−19 )2
=
= 2.3 ×10−8 N
(10−10 )2
(attraction)
b) the magnitude of the gravitational force is given by:
r
Gm1m2
FG =
r2
(6.67 ×10−11 )(9.1×10−31 )(1.67 ×10−27 )
=
= 1×10−47 N
−10 2
(10 )
Note that FE >> FG
r
c) Halving the distance r increases the force F by a factor of 4.
Doubling one charge q , doubles the electric force
Therefore, the new electric force:
FEnew = 8FE
= 1.8 ×10−7 N
Doubling one mass (i.e 2 protons) doubles the force.
FGnew = 8FG
= 8 ×10−47 N
A helium nucleus (2 protons and 2 neutrons) has 4 times the original mass.
FGnew = 16 FG
= 1.6 ×10−46 N
4
QUESTION 3
a) If q1 = q then q2 = 5 ×10−5 − q
We can isolate the unknown charge q in the force equation.
F=
kq1q2 kq (5 ×10−5 − q )
=
r2
r2
Fr 2 = 5 ×10−5 (k )(q ) − (k )q 2
⇒
9 ×109 q 2 − 4.5 ×105 q + 4 = 0
4.5 ×105 ± (4.5 ×105 ) 2 − 4(9 ×109 )(4)
q=
1.8 ×1010
q = 3.85 ×10−5 C
or
q = 1.15 ×10−5 C
q1 is either of these; q2 is the other ( q2 = 5 ×10−5 − q )
Parts b) and c)
For the sphere with 3.85 ×10−5 C :
Number of electrons removed:
3.85 ×10−5 C
= 2.47 ×1014 e
−19
1.6 ×10 C / e
Mass of these electrons: 2.47 ×1014 (9.11×10−31 ) = 2.25 ×10−16 kg
% of mass:
2.25 ×10−16
×100 = 7.50 ×10−12 %
−3
3 ×10
For the sphere with 1.15 ×10−5 C
1.15 ×10−5 C
Number of electrons removed:
= 7.19 ×1013 e
−19
1.6 ×10 C / e
Mass of these electrons: 7.19 ×1013 (9.11×10−31 ) = 6.55 ×10−17 kg
% of mass:
6.55 ×10−17
×100 = 2.18 ×10−12 %
−3
3 ×10
5
QUESTION 4
For the net force to be zero, q3 (6µC ) must be placed on line passing through q1 and q2 .
r
r
r
r
r
Fnet = F1on3 + F2on3 = 0 ⇒ F1on3 = − F2on3 (equal and opposite)
Three situations to consider:
•
Situation 1: q3 at the left of q1 :
•
Situation 2: q3 between q1 and q2 :
•
Situation 3: q3 at the right of q2 :
In situations 1 and 3, the net force is not equal to zero. So q3 must be placed somewhere between q1 and q2 . Note
that this would not be the case if q1 and q2 were oppositely charged (try it!).
6
According to situation 2:
r
r
| F1on3 | = | F2on3 |
kq1q3
kq2q3
=
2
x
(6 − x) 2
q1 (6 − x) 2 = q2 x 2
(q1 − q2 ) x 2 − 12q1x + 36q1 = 0
2 ×10−6 x 2 − 3.6 ×10−5 x + 1.08 ×10−4 = 0
Using the quadratic equation (again… it might be a good idea for you to remember it!)
x=
3.6 ×10−5 ± (3.6 ×10−5 ) 2 − 4(2 ×10−6 )(1.08 ×10−4 )
4 ×10−6
x = 3.80 cm
or
x = 14.2cm
We’ll choose the x = 3.00 cm because the other one ( x − 14.2 cm ) puts q3 to the right of q2 . Which of course
r
makes no sense! The reason it came about is that the equation doesn’t consider the fact that F2 on 3 changes direction
as x gets bigger than 6 cm.
So the charge q3 must be placed between the other charges, 3.80 cm from the 3µC charge.
N.B. This problem could have been solved equivalently by finding the point P at which the electric field due to q1
r
r
and q2 is zero ( E1P + E2 P = 0 ). Why?? Try it!
7
QUESTION 5
The electric field of a point charged is given by:
r kq
E = 2 rˆ
r
a) For the point located at -0.03 m on the x-axis:
r k (4 ×10−6 )
E=
(−iˆ) = −4.00 ×107 iˆ N / C
2
(0.03)
b) the point located at 0.03 m on the y-axis:
r k (4 ×10−6 )
ˆj = 4.00 ×107 ˆj N / C
E=
2
(0.03)
c) for the point located at (0.04; 0.03)m:
r
the distance at which this point is located from the charge: r = 0.032 + 0.042 = 0.05 m
r
The difference between the point P and the charge is r = (0.04iˆ + 0.03 ˆj ) m
r
r 0.04iˆ + 0.03 ˆj
rˆ = r =
= (0.8iˆ + 0.6 ˆj ) or find the angle
r
0.05
r k (4 ×10−6 )
E=
(0.8iˆ + 0.6 ˆj )
(0.05)2
or with the angle:
⎛ 0.03 ⎞
⎟ = 36.8°
⎝ 0.04 ⎠
θ = tan −1 ⎜
r k (4 ×10−6 )
E=
(cosθ iˆ + sin θ ˆj )
(0.05)2
r
E = (1.15 ×107 iˆ + 8.64 ×106 ˆj ) N / C
Before you go: are the directions of these vectors what you expected? What would happen if the charge is negative?
Try it!
8
QUESTION 6
The force and the electric field are related together by the following equation:
r
r
F = qE
r
a) F = 2 ×10−8 (4.00 ×107 ˆj ) = +0.800 ˆj N
r
b) F = 2 ×10−8 (−4.00 ×107 iˆ) = −0.800iˆ N
r
c) F = −2 ×10−8 (1.15 ×107 iˆ + 8.64 × 106 ˆj ) = (−0.230iˆ − 0.173 ˆj ) N
Before you go: are the directions of these vectors what you expected?
9
QUESTION 7
r
r
a) According to Newton’s second law (yes, physics NYA is still valid!!) : F = ma
That is equal to the electric force experience by a charged in an electric field:
r
r
r
r
r qE
F = ma = qE ⇒ a =
m
r (−1.6 ×10−19 )(0iˆ + 2 ×102 ˆj )
a=
= (0iˆ − 3.51×1013 ˆj ) m / s 2
−31
9.11×10
b) Remember the projectile motion in physics NYA….
x-direction
vox = 2 ×106 m / s (constant)
y-direction
voy = 3 ×106 m / s
Δt = 2 ×10−7 s
Δt = 2 ×10−7 s
a = −3.51×1013 m / s2
Δy = ?
Δx = ?
Kinematics equations:
1
Δy = vot + at 2
2
Δy = −0.103m
Δx = vt
Δx = 0.4m
r
the initial position: ro = (0,0)m
r
the final position: r = (0.400, −0.103)m
c) The velocity is a vector, so we have to find the x and y components:
In the x-direction:
vx = 2 ×106 m / s (constant)
In the y-direction
v y = vo + at
v y = −4.02 ×106 m / s
r
The velocity: v = (2.00 ×106 iˆ − 4.02 ×106 ˆj ) m / s
10
QUESTION 8
We first have to find the x and y components of the velocity vector:
r
v = (6.00 ×106 cos 45iˆ + 6.00 ×106 sin 45 ˆj ) = (4.24iˆ + 4.24 ˆj ) ×106 m / s
a) In the y-direction:
Fy = ma y = qE y
⇒
ay =
qE y
m
(−1.6 ×10−19 )(2 ×103 )
ay =
= −3.51×1014 ˆj m / s 2
−31
9.11×10
voy = 4.24 ×106 m / s
Δy = 0.02m
Δt = ?
1
Δy = voyt + a yt 2 solve for the time t…..
2
1
2 ×10−2 = 4.24 ×106 t + (−3.51×1014 )t 2
2
1.76 ×1014 t 2 − 4.24 ×106 t + 2 × 10−2 = 0
Solve using the quadratic equation…
4.24 ×106 ± (4.24 ×106 ) 2 − 4(1.76 ×1014 )(2 ×10 −2 )
t=
2(1.76 ×1014 )
t = 6.46 ×10−9 s
or
t = 1.77 ×10−8 s
The time t = 6.46 ×10−9 s corresponds to the time taken for the charge to travel y = 2 cm on its way up. The time
t = 1.77 ×10−8 s corresponds to the time taken for the charge to travel y = 2 cm on its way down. So it hits the top
plate after 6.46 ×10−9 s .
How far has it moved horizontally in that time?
In the x-direction
vox = 4.24 ×106 m / s
ax = 0
r
( Ex = 0 )
Δt = 6.46 ×10−9 s
Δx = vox Δt
4.24 ×106 (6.46 ×10−9 ) = 2.74 ×10−2 m
11
b)
The electron just misses the top plate. It leaves the
field at the point [4; 2] cm
x-direction
y-direction
voy = 4.24 ×106 m / s
Δy = 0.02 m
6
vox = 4.24 ×10 m / s
Δx = 0.04 m
a=?
Δx = vΔt ⇒ Δt =
& Δt = ?
Δx
= 9.43 × 10−9 s
v
1
Δy = vot + at 2
2
ay =
2(Δy − vot )
= −4.50 × 1014 ˆj m / s 2
2
t
The force acting on the charge is related to the acceleration of the charge and the electric field in which the charge
travels.
r
r
r
F = ma = qE
r
ma y (9.11×10−31 )(−4.50 ×1014 )
Ey =
=
q
−1.6 ×10−19
= 2.56 ×103 ˆj N / C
r
Emin = (0iˆ + 2.56 ×103 ˆj ) N / C
12
QUESTION 9
Visualize the situation:
a)
p = qd
= (1e ) ( 4 ×10−10 m )
= 4 ×10−10 e ⋅ m
r
p = ( 4.00 ×10−10 e ⋅ m, 35°)
b)
τ = pE sin θ
= ( 4 ×10−10 e ⋅ m )( 4.8 ×10−16 N/e ) sin 35
= 1.10 ×10−24 N ⋅ m
r
τ = −1.10 ×10−25 kˆ N ⋅ m
(clockwise)
13
QUESTION 10
You must draw a free body diagram. Both charges will have the same f.b.d. so let’s draw it for q2 . Four forces are
present: the tension [T], the weight [mg], the electric force between the two charges [F1on2] and, the electric force
due to the charge placed in the external electric field [q2E].
1) ΣFx = q2 E − T sin θ − F1on 2 = 0
with F1on 2 =
2) ΣFy = T cos θ − m2 g = 0
kq1q2
r2
⇒T =
m2 g
cos θ
⎡ m g ⎤
ΣFx = q2 E − ⎢ 2 ⎥ sin θ − F1on 2 = 0
⎣ cos θ ⎦
= q2 E − m2 g tan θ − F1on 2 = 0
The distance r between the charges can be obtained by geometry.
d = L sin θ
r = 2d = 3.42 cm
Solve for the electric field from the above
equation:
ΣFx = q2 E − m2 g tan θ − F1on 2 = 0
⇒E=
F1on 2 + m2 g tan θ
q2
k (5 ×10−8 ) 2
+ 2.00 × 10−3 ⋅ 9.81tan10
2
0.0342
=
5 ×10−8
= 4.54 × 105 N / C
14
15