Download Physics 213 — Problem Set 3 — Solutions Spring 1998

Physics 213 — Problem Set 3 — Solutions
Spring 1998
The course Web page can be found at www.physics.cornell.edu/p213. It contains copies of course handouts
and useful information like the office hours for the Physics 213 staff.
1. Reading Assignment
Serway Sections 24.1–24.4. Also Sections 27.1–3,6 and 28.1–2 are useful preparation for the first lab.
2. Serway 24.20A
A charge Q is at the center of a cube of side L.
a) Find the total flux through each face of the cube.
b) Find the flux through the whole surface of the cube.
c) Would your answers to the previous parts of this problem change if the charge were not at the center
of the cube? Explain.
SOLUTION: Imagine that we put a single charge Q at the center of a cube of side L (see Fig. P24.18 with
only charge Q inside the cube.)
a)The flux through each side is Φside = (1/6)Φtotal by symmetry, where Φtotal is the total flux through the
entire cube.
b)If we treat the cube as a Gaussian surface, by Gauss’s law the total flux through the cube is
I
~ · d~a = Q .
E
o
Φtotal
c)If Q were not at the center of the cube, Gauss’s Law tells us that the total flux remains
the same as (b)
H
~ · d~a = Q changes
since the enclosed charge is the same; the flux through each face differs from (a) since E
o
when the position of the charge changes.
3. Serway 24.30
A sphere of radius a carries a volume charge density of ρ = ρ0 (r/a)2 for r < a. Calculate the electric field
inside and outside the sphere.
SOLUTION: The electric field from this charge distribution is spherically symmetric and radially outward
(for a positive charge distribution). Since we will use Gauss’s law to calculate the electric field, let’s first
calculate the charge enclosed by a spherical Gaussian surface of radius r centered around the center of a
2
sphere of radius a carrying a volume charge density ρ = ρ0 ar for r < a:
Z
Qenc
r<a
=
Z
r
ρ4πr̃2 dr̃.
ρdV =
0
(See Problem 24.54; the volume element dV for a spherical shell of radius r̃ and thickness dr̃ is 4πr̃ 2 dr̃.)
Z
Qenc
=
r<a
2
r
r̃
a
ρ0
0
4πρ0
a2
=
r̃5
5
4πr̃2 dr̃ =
r
=
0
4πρ0
a2
Z
r
r̃4 dr̃
0
4πρ0 5
r .
5a2
If r > a, the enclosed charge is :
Z
Qenc
r>a
=
a
ρ0
0
r 2
a
4πr2 dr =
4πρ0 3
4πρ0 a5
=
a
a2 5
5
a)Electric field inside: use Gaussian sphere with r < a. By Gauss’ law
I
~ · d~a = Qenc r>a ,
E
0
1
and by the definition of electric flux
I
I
~ · d~a =
E
I
Eda = E
so
E4πr2 =
or,
da = E4πr2 ,
4πρ0 5
r ,
50 a2
ρ0 3
r .
50 a2
E=
b)Electric field outside; use Gaussian sphere with r < a:
I
~ · d~a = E4πr2 = 4πρ0 a3 ,
E
50
so
E=
ρ 0 a3 1
.
50 r2
4. Serway 24.34
An infinitely long cylindrical insulating shell of inner radius a and outer radius b has a uniform volume
charge density ρ C/m3 . A line of charge denisty λ C/m is placed along the axis of the shell. Determine
the electric field everywhere.
SOLUTION: The electric field from this charge distribution has a cylindrical symmetry (see Example 24.7).
To calculate E we’ll use a cylindrical Gaussian surface of radius r and length l centered on the axis of the
charge distribution (similar to that shown in Fig, 24.14a). Find the charge enclosed by the Gaussian surface
for :
H
0<r≤a
a<r≤b
r>b
Qenc = λl
Qenc = λl + ρlπ(r2 − a2 )
Qenc = λl + ρlπ(b2 − a2 )
~ · d~a = Qenc /0 ; the flux through the end caps of the Gaussian surface is zero (since E
~ is
By Gauss’s law E
perpendicular to d~a), so the integral is over the curved portion of the Gaussian cylinder (see Example 24.7).
The electric field is :
H
R
~ · d~a = E da = E2πrl = λ`/0
E
0<r≤a
1 λ
so, E = 2π
0 r
a<r≤b
E2πrl = (λl + ρlπ(r2 − a2 ))/0
2
2
1 (λ+ρπ(r −a ))
E = 2π
r
0
r>b
E2πrl = (λl + ρlπ(b2 − a2 ))/0
2
2
1 (λ+ρπ(b −a ))
E = 2π
.
r
0
5. Serway 24.44
A hollow conducting sphere is surrounded by a larger concentric, spherical, conducting shell. The inner sphere has a charge −Q, and the outer sphere has a charge 3Q. The charges are in electrostatic
equilibrium. Using Gauss’s law, find the charges and the electric fields everywhere.
SOLUTION: By symmetry all electric field vectors E have to be radial. The value of E on a Gaussian
sphere of radius r and concentric with the two conducting spheres depends on the charge enclosed by it,
Qencl , by Gauss’ law:
I
Qencl
E · dA = 4πr2 E =
.
0
2
Let us call the radius of the inner conducting sphere of the problem R1 , and the inner and outer radii of the
conducting shell R2 and R3 , respectively, and use the above relation to find E
0 < r < R1
R1 < r < R2
Qencl = 0
E=0
Qencl = −Q
E=−
R2 < r < R3
Q
4π0 r2
E=0
Qencl = −Q + 3Q = 2Q
R3 < r
E=
2Q
Q
=
4π0 r2
2π0 r2
The field for R2 < r < R3 is zero because the spherical shell is a conductor. The field for R1 < r < R2 is
directed inward, toward the center, and the field outside the conducting shell (r > R3 ) is directed outward.
6. Serway 24.58
Two infinite, nonconducting sheets of charge are parallel to each other as in Figure P24.58 of your text.
The sheet on the left has a uniform surface charge density σ, and the one on the right has a uniform
charge density −σ. Calculate the value of the electric field at points (a) to the left of, (b) in between,
and (c) to the right of the two sheets. (Hint: See Example 24.8 in your text.)
SOLUTION: Let us orient the two sheets parallel to the yz-plane in a coordinate system where the x-axis
goes from left to right. As shown in Example 24.8, the electric field of a single sheet of surface charge
+σ points away from the sheet and is 2σ0 (x̂) everywhere to the right of the sheet and 2σ0 (−x̂) on its left.
Similarly, a negatively charged sheet (−σ) has an electric field pointing toward it, 2σ0 (x̂) on its left and
σ
20 (−x̂) on its right. The net field produced by two parallel sheets is the superposition of the fields of the
two single sheets.
a)To the left of both sheets, the electric field is
σ
σ
(−x̂) +
(x̂) = 0.
20
20
Eleft =
b)We are to the right of the sheet with +σ and to the left of the sheet with −σ, and obtain
Ebetween =
σ
σ
σ
(x̂) +
(x̂) = (x̂).
20
20
0
c)To the right of both sheets, the electric field vectors cancel again, Eright = 0.
7. On a clear sunny day there is a vertical electric field of 130 N/C pointing down over a calm lake. Assuming
that the lake is a conductor, compute the surface charge density (sign and magnitude) on the lake.
SOLUTION: If the electric field points vertically downward toward the flat surface of a conducting lake,
the charge density on that surface must be uniform; assume it is σ (C/m2 ). To calculate the relationship
between and σ use a Gaussian pillbox similar to that shown in Fig. 24.18. Since the lake is a conductor,
~ = 0 everywhere inside the lake. The flux through the pillbox is then
E
I
~ · d~a = −EA
E
~ points down and dA
~ points up.
where the only contribution to the integral is from the top end cap where E
The enclosed charge is Qenc = σA, so by Gauss’ law −EA = σA/0 , so
σ
8.85 × 10−12
=
−0 E =
=
−1.2 × 10−9 C/m2
3
C2
Nm2
−130
N
C
8. Serway Question 28.28
A series circuit consists of three identical lamps connected to a battery as in Figure 28.29 of your text.
When the switch S is closed, what happens (a) to the intensities of lamps A and B; (b) to the intensity
of lamp C; (c) to the current in the circuit; and (d) to the voltage drop across the three lamps? (e) Does
the power dissipated in the circuit increase, decrease, or remain the same?
SOLUTION: We will assume the battery is an ideal voltage source that delivers E volts regardless of the
load placed on it.
When the switch is open (see Fig. 28.29) :
VA = VB = VC = E/3
When the switch is closed :
VA = VB = E/2, VC = 0
a)The intensity of lamps A and B increases, since each of them now has half of the battery voltage across
it, rather than one third as before.
b)The intensity of lamp C is zero, since there is no voltage across it.
c)The current in the circuit increases, except there is no current in lamp C.
d)Half the battery voltage is across each of lamps A and B, no voltage across C.
e)The power dissipated in the circuit increases; P = E 2 /RT ot , switch open RT ot = RA + RB + RC ; when
the switch is closed RT ot = RA + RB .
(NOTE: if we account for the temperature dependent resistivity of the lamp filament (see Lab Unit 1), the
resistance of the lamps A & B will change when the switch is closed, which would affect the quantitative
calculations of the current in the circuit, and the total power dissipated. Also note, however, the voltages
should remain the same.)
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