Download Physics 213 — Problem Set 2 — Solutions Spring 1998

Physics 213 — Problem Set 2 — Solutions
Spring 1998
1. Reading Assignment
Serway Sections 23.5–23.8, 24.1–24.2
2. Serway 23.56A
Two small spheres each of mass m are suspended by light strings of length L. (See Figure P23.56 in text.)
A uniform electric field is applied in the x direction. If the spheres have charges −q and +q, determine
the electric field that enables the spheres to be in equilibrium at an angle θ.
SOLUTION: We first consider a free body diagram of either one of the two spheres. Consider the positively
charged sphere (which, in the book’s diagram, is pushed towards the +x direction). There are 4 forces acting
on it : (a) the tension T along the string (b) its weight W , acting downwards, (c) a force FE = qE, due
to the electric field pushing it in the +x direction, and (d) the attractive force Fq,−q due to the negativley
charged sphere, in the −x direction. Now since the distance between the two spheres is 2(L sin θ), we have :
Fq,−q = |
ke q(−q)
ke q 2
|
=
(2L sin θ)2
(2L sin θ)2
Since the system is in equilibrium the sum of the forces in the x and y directions should each be zero. Hence
we have,
(x − dir)
−T sin θ − Fq,−q + FE
=0
(y − dir)
T cos θ − mg
= 0.
¿From equation (2) we have T = mg/ cos θ. Substituting this into (1), we get :
or
−mg tan θ − Fq,−q + FE
=
0
ke q 2
+ qE
− mg tan θ −
(2L sin θ)2
=
0.
Solving for E, we get,
E=
ke q
mg
tan θ +
q
(2L sin θ)2
3. Serway 23.30
A line of charge starts at x = +x0 and extends to positive infinity. If the linear charge density is
λ(x) = λ0 x0 /x, determine the electric field at the origin.
SOLUTION: The charge on a segment dx of the line is dq = λdx. If dq is at a distance x from the origin
its contribution to the electric field at the origin is
dE = ke
Integrating from x0 to ∞ we get
Z
∞
E=
ke
x0
λdx
λ0 x 0
= ke 3 dx
x2
x
∞ ke λ0
λ0 x0
1
dx = − ke λ0 x0 x−2 x =
0
x3
2
2x0
For positive λ the electric field points in the negative x-direction.
4. A uniformly charged insulating rod of length L is bent into the shape of a semicircle as in Figure P23.39
in your text. If the rod has total charge Q, what is the magnitude and direction of the electric field at
the center of the semicircle (point O in the figure)?
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SOLUTION: Direction: By symmetry, the y-components of the electric field cancel. The net electric field
has to point to the left, toward the negative charges. The length of the rod is l = πR, where R is the radius
of the semicircle, and the charge density is λ = Q/l. Take O as the origin of the coordinate system and φ as
the angle with respect to the positive x-axis. The electric field at O due to a segment R dφ of the semicircle
is
λRdφ
λdφ
dE = ke
= ke
R2
R
with an x-component
ke λ
cos φ dφ
dEx =
R
Integrating over φ gives the field due to the entire rod
Ex =
ke λ
R
Z
3π/2
cos φdφ =
π/2
ke λ
ke λ
ke Q π
3π/2
[sin φ]π/2 = 2
=2 2 .
R
R
l
5. Serway 23.46
An electron moves at 3 × 106 m/s into a uniform electric field of magnitude 1000 N/C. The field is parallel
to the electron’s velocity and acts to decelerate the electron. How far does the electron travel before it is
brought to rest?
SOLUTION: Note here that the field E is pointing in the same direction that the electron is moving. Since
the electron is negatively charged, it feels a force qe E in the opposite direction and hence the field causes it
to decelerate. The deceleration is a = qe E/me , where me is the electron’s mass.
We could solve for the stopping distance by solving the equation of motion (F = ma). Instead, in this
case it is simpler to use conservation of energy. Since the decelerating electric field is a constant force, it gives
rise to a potential energy proportional to the distance, d, the electron has moved in the field: P.E. = (qe E)d.
Initially the electron has all its energy as kinetic energy K.E. = me v 2 /2 and no potential energy. After it
has stopped, all its energy is all potential energy. Equating them gives me v 2 /2 = qe Ed. Solving for d we
get,
me v 2
d=
.
2qe E
With v = 3 × 106 m/s, qe = 1.6 × 10−19 C, me = 9.1 × 10−31 kg, and E = 1000N/C, we get d = 0.026m.
6. Serway 23.74
An electric dipole in a uniform electric field is displaced slightly from its equilibrium position, as in
Figure P23.74, where θ is small. The moment of inertia of the dipole is I. If the dipole is released form
this position, show that it exhibits simple harmonic motion. Find the frequency of this motion.
SOLUTION: See also Figures 26.15 and 26.16 in your text. Let 2a be the distance between the negative
and posititive charge of the dipole. The dipole moment is then p = 2a q where a points from −q to +q.
If the dipole is rotated with respect to the direction of the electric field through a small angle θ, where
sin θ ' θ, it experiences a torque of magnitude
| p × E |= 2aqE sin θ ' 2aqE θ
In the figure, the angle is measured counter-clockwise, whereas the torque is clockwise. The torque will
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always be opposed to the angular displacement. Using α = ddt2θ for the angular acceleration we can write
d2 θ
= −2aqE θ
dt2
2
d θ
2aqE
=−
θ = −ω 2 θ
dt2
I
Iα = I
(6.1)
(6.2)
Equation (6.2) describes a simple harmonic motion of angular frequency ω. The frequency of this oscillation
is
r
ω
1
2qaE
f=
=
2π
2π
I
The dipole moves in the electric field like a torsional pendulum.
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7. Serway 24.2
An electric field of magnitude 3.5×103 N/C is applied along the x axis. Calculate the electric flux through
a rectangular plane 0.35 m wide and 0.70 m long if the plane (a) is parallel to the yz plane, (b) is parallel
to the xy plane, and (c) contains the y axis and its normal makes an angle of 40 degrees with the x axis.
SOLUTION: For this problem we just have to remember that, for a uniform electric field, the flux through
flat area A, is given by Φ = EA cos θ, where θ is the angle between the direction of the field, and the normal
to the area. (Recall that a normal vector is one which is perpendicular to the surface.) Here the area of the
plane is A = 0.35 × 0.7 m2 = 0.245 m2 and E=3.5 ×103 î N/C.
a)Since the plane is parallel to the yz plane, the angle between the field and the normal to the rectangle is
0o , and cos 0◦ = 1. So Φ = 860 Nm2 /C.
b)Here the rectangle lies in the xy plane, thus its normal must be perpendicular to the field. Since cos 90◦ = 0,
Φ = 0. In other words, there are no electric field lines that go through the rectangle.
c)Here θ = 40◦ . So Φ = EA cos 40◦ = 660 Nm2 /C.
8. Serway 24.12
If the constant electric field in Figure P24.12 has magnitude E0 , calculate the total electric flux through
the paraboliodal surface shown in the figure.
SOLUTION: For this problem note that there are no charges inside the closed suface, which consists of the
paraboloid and the circular base. Hence the net flux through it has to be zero at all times. Thus all electric
field lines entering the volume must exit it too. Based on the orientation in the figure, it is clear that field
penetrates the surface through the circular base, perpendicular to the plane of the base. Hence the flux
through the base is ΦEntering = −E0 (πr 2 ). (The minus sign arises due to the fact that the field is entering a
closed surface.) All this flux must leave the volume through remaining part, i.e., the paraboloidal surface.
Hence the flux through the paraboloidal surface is ΦExiting = −ΦEntering = E0 πr2 .
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