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Homework 1
chapter 23: 27, 34, 41, 51
Problem 23.27
A uniformly charged insulating rod of length 14.0 cm is bent into the shape of a
semicircle as in figure P23.33. The rod has a total charge of -7.5 mC. Find the
magnitude and direction of the electric field at O, the center of the semicircle.
We can assume, with a good approximation, that the rod is a linear
object. The linear charge density (λ) of this object is related to the
charge (Q) of the object and its size (length L).
dq Q
λ=
=
dl L
In order to find the electric field at a certain location we must add
(vectorially) electric fields created by all "point charges" in the body.
A contribution d E to the electric field, due
to a differential fragment dl, can be found
from Coulomb's law.
y
dl
r
φ
x
dE
dE =
kdq
[sin φ,− cos φ]
r2
From the mathematical point of view, it is
convenient to relate that differential field
vector with the differential of angle dφ as
marked in the figure.
dE =
k dq dl
k Q
[
]
d
sin
,
cos
⋅
⋅
⋅
φ
⋅
φ
−
φ
=
⋅ ⋅ r ⋅ dφ ⋅ [sin φ,− cos φ]
r 2 dl dφ
r2 L
We can eliminate the unknown radius from its relation to the length of
the rod. Hence
πkQ
d E = 2 [sin φ,− cos φ]dφ
L
In order to find the net electric field we must integrate (add) the
differential fields over the entire body. In terms of the variable φ, it
requires integration in limits from 0 to π.
π
πkQ
πkQ ⎡π
⎤
E = ∫ d E = ∫ 2 [sin φ,− cos φ]dφ = 2 ⎢ ∫ sin φdφ, ∫ − cos φdφ⎥ =
L ⎣0
⎦
object
0 L
0
π
(
)
Nm 2
π ⋅ 9 ⋅ 10
⋅ − 7.5 ⋅ 10 −6 C
2
πkQ
N
7
C
[
]
= 2 [2,0] =
=
−
⋅
2
,
0
2
.
16
10
,
0
C
L
(0.14m )2
9
[
]
Problem 23.34
Figure P23.40 shows the electric field lines from two point charges separated by a
small distance. (a) Determine the ratio q1/q2. (b) What are the signs of q1 and q2?
a) We should realize that, in principle,
there is an infinite number of electric field
lines originating from particle q2 and an
infinite number of electric field lines
terminating on particle q1. The representation
q2
of an electric field by drawing some of the
field lines has a descriptive rather than an
accurate character. In a region of smaller
q1
separation between the lines the electric field
is stronger than in a region where the lines
are farther apart. (The strength is compared
in terms of magnitude of the electric field vector at these two locations.)
Also the number of lines crossing a surface is proportional to the
electric flux through that surface. One can decide to choose a factor α
that relates the number of drawn electric field lines N through a surface
with the electric flux Φ through that surface.
(1)
Φ = αN
In this particular problem such an assumption is made. We can
therefore find the ratio of electric flux through two surfaces by dividing
the numbers of lines crossing the surfaces. We can draw such a
conclusion from equation (1):
(2)
Φ1 αN1 N1
=
=
Φ 2 αN 2 N 2
Here the subscripts indicate two different surfaces.
Let's consider two Gaussian surfaces: one (1) enclosing particle
q1 only, and the second (2) enclosing particle q2 only. From Gauss' law
one knows that the electric flux through a Gaussian surface is
proportional to the net charge enclosed by the surface. If we apply this
law to the ratio between the two charges together with relation (2), we
obtain:
q1 − ε 0Φ1
N
6
1
=
=− 1 =− =−
ε 0Φ 2
q2
N2
18
3
The minus sign in the numerator results from the direction of the electric
field lines around particle q1. (In Gauss' law we should consider the
flux directed outwards not inwards)
b) According to the generally accepted convention the direction of the
electric field lines at each point coincides with the direction of the
electric field vector. Therefore they originate on a positive charge and
terminate on a negative charge. Therefore charge q1 is negative while
charge q2 is positive.
Problem 23.47
A proton moves at 4.50 × 105 m/s in the horizontal direction. It enters a uniform
vertical electric field of 9.60 × 103 N/C. Ignoring any gravitational effects, find
(a) the time it takes the proton to travel 5.0 cm horizontally, (b) its vertical
displacement after it has traveled 5.0 cm horizontally, and (c) the horizontal and
vertical components of the proton's velocity after it has traveled 5.0 cm
horizontally.
v0
For a simpler description let's choose the x-direction along the velocity
r
v 0 of the proton as it enters the electric field, the y-direction along the
electric field vector and the reference time t0 as the instant when the
proton enters the field. As long as the proton is in the uniform electric
field the electric force exerted on the proton is constant. For all
practical purposes we can assume that the other forces are small (in
magnitude) when compared with the electric force exerted by the
electric field described in the problem. Therefore the net force exerted
on the proton is approximately equal to the electric force and is also
constant. Consistent with Newton's second law the particle moves with a
constant acceleration. Using the definition of electric field vector we can
find that acceleration force in terms of the given physical quantities:
(1)
a=
Fnet Fel q Eel q
≈
=
= [0, E ]
m
m
m
m
Recall that in a motion with a constant acceleration the position is a
quadratic function of time, and velocity is a linear function of time. With
the values for the initial position and initial velocity (consistent with our
choice of the reference frame and the reference time)
r0 = [0,0] ,
v 0 = [ v0 ,0]
the following function represents the position and the velocity of the
proton.
q
[
0, E ]t 2
2
at
qE 2 ⎤
⎡
(1) r (t ) = r0 + v 0 t +
t ⎥
= [0,0] + [v 0 ,0]t + m
= ⎢v 0 t,
2
2
2m ⎦
⎣
(2)
v(t ) = v 0 + a t = [v 0 ,0] +
q
[0, E]t = ⎡⎢v 0 , qE t ⎤⎥
m
m ⎦
⎣
a) In this question we have to find how much time elapsed from the
chosen reference time (instant) t0 = 0s to the time (instant) t1 when the
proton reaches the given location with the x-component (x1 = 5cm).
From the general expression (1) for position we obtain
(3)
v 0 t 1 = x1
Hence the elapsed time is
Δt = t1 − t 2 =
x1
0.05m
− t0 =
− 0s = 1.1 ⋅ 10 − 7 s
m
v0
4.5 ⋅ 105
s
b) Again using equation (1) we can determine the vertical component of
the position at instant t1 and use that result to find the displacement from
the initial position.
qE 2
Δy = y1 − y 0 =
t1 − y 0 =
2m
N⎞
⎛
1.6 ⋅ 10−19 C ⋅ ⎜ 9.6 ⋅ 103 ⎟
C⎠
−7 2
⎝
=
⋅
1
.
1
⋅
10
s − 0m = 5.67 ⋅ 10− 3 m
− 27
2 ⋅ 1.67 ⋅ 10 kg
(
(
)
)
(
)
c) Using the function representing velocity (2), we can find its value at
instant t1:
⎤
⎡
⎛
3 N⎞
−19
1
.
6
10
C
9
.
6
10
⋅
⋅
⋅
⎟
⎜
⎥
C⎠
⎡ qE ⎤ ⎢
5 m
−7
⎝
t1 = ⎢4.5 ⋅10
,
v1 = ⎢ v 0 ,
⋅ 1.1 ⋅10 s ⎥ =
m ⎥⎦ ⎢
s
1.67 ⋅10 − 27 kg
⎣
⎥
⎥⎦
⎢⎣
m
m⎤
⎡
= ⎢4.5 ⋅105 ,1.02 ⋅105 ⎥
s
s⎦
⎣
(
(
)
)
(
)
Problem 23.51
Identical thin rods of length 2a carry equal charges Q uniformly distributed along
their lengths. The rods lie along the x-axis which their centers separated by a
distance b > 2a. Show that the magnitude of the force exerted by the left rod on the
⎞
⎛ kQ 2 ⎞ ⎛ b 2
⎟.
right one is F = ⎜⎜ 2 ⎟⎟ ln⎜⎜ 2
2⎟
4
a
b
4
a
−
⎠
⎠ ⎝
⎝
dx1
-a
dx2
b
x1
a
x2
b-a
x
b+a
Solution 1
The symmetry of the arrangement implicates that all relevant
vector quantities are aligned with the x-axis. The components of the
vectors in the direction transverse to this axis are zero and only the
x-component must be calculated. From Coulomb’s law (electric field
produced by a particle), the contribution to the (x-component of the)
electric field vector at location x2 by the differential segment of the left
rod at location x1 is equal to
dE1 =
kQ ⋅ dx1
2a ⋅ ( x2 − x1 )2
Calculating the electric field produced by the entire (left) rod requires
integration with respect to x1
a
E1 ( x2 ) = ∫ dE1 = ∫
rod
x1 = a
kQ ⋅ dx1
- a 2a ⋅
−1
kQ
=−
⋅
2a x2 − x1
( x2 − x1 )
2
x1 = a
=
x1 = -a
=− ∫
x1 = -a
kQ ⋅ d( x2 − x1 )
2a ⋅ ( x2 − x1 )
kQ ⎛ 1
1 ⎞
⎟⎟
⋅ ⎜⎜
−
2 a ⎝ x 2 − a x2 + a ⎠
2
=
From the definition of the electric field vector, the (x-component of
the) electrostatic force exerted on the differential fragment of the right
rod at location x2 is
Q ⋅ dx 2
kQ 2 ⎛ 1
1 ⎞
⎟ ⋅ dx2
dF =
⋅ E1 ( x2 ) = 2 ⋅ ⎜⎜
−
2a
4a ⎝ x2 − a x2 + a ⎟⎠
Calculating the electrostatic force on the rod requires integration with
respect to x2
b+a
⎞
kQ 2 ⎛ b + a 1
1
⋅ dx2 − ∫
⋅ dx2 ⎟⎟ =
F = ∫ dF = 2 ⋅ ⎜⎜ ∫
4 a ⎝ b - a x2 − a
b-a
b - a x2 + a
⎠
b+a
x2 = b + a
⎞
kQ 2 ⎛⎜ x 2 = b + a 1
1
⎟=
= 2⋅
⋅
−
−
⋅
+
x
a
x
a
(
)
(
)
d
d
∫
∫
2
2
⎜
⎟
4 a ⎝ x 2 = b - a x2 − a
x 2 = b - a x2 + a
⎠
[
]
kQ 2
b+a
b+a
= 2 ⋅ ln( x2 − a ) b - a − ln( x2 + a ) b - a =
4a
kQ 2
= 2 ⋅ [ln b − ln(b − 2a ) − ln(b + 2a ) + ln b] =
4a
kQ 2
kQ 2
b2
b2
= 2 ⋅ ln
= 2 ⋅ ln 2
−
+
(
)(
)
2
2
b
a
b
a
4a
4a
b − 4a 2
Solution 2
According to Coulomb’s law, the (differential) electrostatic force
exerted by the differential fragment of the left rod on the differential
fragment of the right rod is equal to
⎞
⎞⎛ Q
⎛Q
k ⋅ ⎜ ⋅ dx1 ⎟⎜ ⋅ dx2 ⎟
kQ 2 dx1dx2
2a
2a
⎠
⎠
⎝
⎝
dF" =
= 2⋅
2
4a ( x2 − x1 )2
( x2 − x1 )
Integrating the above expression with respect to x2 results in the
(differential) force exerted by the differential fragment of the left rod on
the entire right rod
kQ 2 dx1dx2
kQ 2 dx1 x 2 = b + a d( x2 − x1 )
⋅
=
⋅ ∫
=
dF' = ∫ dF" = ∫
2
2
2
2
4
4
a
a
−
−
(
)
(
)
x
x
x
x
x2 =b - a
rod
x2 =b - a
2
1
2
1
x2 =b + a
b+a
⎞
−1
kQ 2 ⋅ dx1
kQ 2 ⎛
1
1
⎜
⎟ ⋅ dx1
=
⋅
=
⋅
−
x2 − x1 b - a 4a 2 ⎜⎝ b − a − x1 b + a − x1 ⎟⎠
4a 2
Calculating the force exerted by entire left rod requires integration of
the above expression with respect to x1
⎞
kQ 2 ⎛
1
1
⎜
⎟⎟ ⋅ dx1 =
⋅
−
F = ∫ dF' = ∫
2 ⎜
−
−
+
−
b
a
x
b
a
x
rod
− a 4a
⎝
1
1⎠
a
a
kQ 2 ⎛ a
dx1
dx1 ⎞
⎟=
= 2 ⋅ ⎜⎜ ∫
− ∫
4a ⎝ − a b − a − x1 − a b + a − x1 ⎟⎠
kQ 2 ⎛⎜ x1 = a d(b − a − x1 ) x1 = a d(b + a − x1 ) ⎞⎟
= 2⋅ − ∫
+ ∫
=
⎜
⎟
−
−
+
−
b
a
x
b
a
x
4 a ⎝ x1 = − a
x1 = − a
1
1 ⎠
[
]
kQ 2
a
a
= 2 ⋅ − ln (b − a − x1 ) − a + ln (b + a − x1 ) − a =
4a
kQ 2
kQ 2
b2
= 2 ⋅ [− ln(b − 2a ) + ln b + ln b − ln(b − 2a )] = 2 ⋅ ln 2
4a
4a
b − 4a 2
Solution 3a
One may find the electrostatic force by simultaneous addition
(integration) of the forces exerted by differential fragments of the left
rod on the differential fragments of the ring rod.
( x 2 = b + a ) ( x1 = a )
kQ 2 dx1dx2
⋅
=
F = ∫∫ dF" =
∫
∫
2
2
( x 2 = b − a ) ( x1 = − a ) 4a ( x2 − x1 )
rods
kQ 2 ( x2 = b + a ) ⎡ ( x1 = a ) d ( x2 − x1 )⎤
= 2 ⋅ ∫ ⎢− ∫
⎥ dx2 =
4a ( x 2 = b − a ) ⎢⎣ ( x1 = − a ) ( x2 − x1 )2 ⎥⎦
−1
kQ 2 ( x2 = b + a ) ⎡
= 2 ⋅ ∫ ⎢−
4a ( x 2 = b − a ) ⎢⎣ x2 − x1
⎤
⎥ dx 2 =
x1 = − a ⎥
⎦
x1 = a
kQ 2 ( x2 = b + a ) ⎡ 1
1 ⎤
= 2⋅ ∫ ⎢
−
⎥ dx 2 =
4 a ( x 2 = b − a ) ⎣ x2 − a x2 + a ⎦
[
]
kQ 2
b+a
b+a
= 2 ⋅ ln ( x2 − a ) b − a − ln ( x2 + a ) b − a =
4a
kQ 2
kQ 2
b2
= 2 ⋅ [ln b − ln(b − 2a ) − ln(b + 2a ) + ln b] = 2 ⋅ ln 2
4a
4a
b − 4a 2
Solution 3b
It does not matter in what order the addition (integration) was
performed
( x1 = a ) ( x 2 = b + a )
kQ 2 dx2 dx1
kQ 2
b2
F = ∫∫ dF" = ∫
⋅
= ... = 2 ⋅ ln 2
∫
2
2
a
4
4a
b − 4a 2
( x2 − x1 )
rods
( x1 = − a ) ( x 2 = b − a )