Physics 9 Fall 2009
Homework 3 - Solutions
1. Chapter 28 - Exercise 8.
The cube in the figure contains no
net charge. The electric field is constant over each face of the cube.
Does the missing electric field vector on the front face point in or out?
What is the field strength?
————————————————————————————————————
Solution
Because the electric field is constant, and because the cube contains no net charge,
the total flux through the cube has to be zero. The flux into the cube is 15 × A +
20 × A = 35A, where A is the area of the cubes face. The flux out of the the cube is
10 × A + 15 × A + 15 × A = 40A. This means that more flux is leaving the cube than
entering it. Because the fluxes in have to balance the fluxes out, we see that the flux
on the front face has to point in, and have a magnitude of 5 N/C.
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2. Chapter 28 - Exercise 11.
The electric flux through the surface shown in the figure is 25
Nm2 /C. What is the electric field
strength?
————————————————————————————————————
Solution
~ ·A
~ =
Since the electric field is constant, the net flux through the surface is ΦE = E
EA cos θ. The electric field makes an angle of 60◦ with the surface of the area, meaning
that it makes an angle of θ = 30◦ with respect to the normal of the surface, which is
what we want. Now, ΦE = EA cos θ ⇒ E = ΦE /A cos θ. With numbers we find
E=
ΦE
25
=
= 2890 N/C.
2
A cos θ
(0.1) cos 30◦
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3. Chapter 28 - Exercise 13.
A 2.0 cm × 3.0 cm rectangle lies in the xz−plane. What is the electric flux through
the rectangle if
~ = 50î + 100k̂ N/C?
(a) E
~ = 50î + 100ĵ N/C?
(b) E
————————————————————————————————————
Solution
The rectangle lies in the xz−plane, and so its normal points along the y−direction.
~ = Ay ĵ, where Ay = 0.02 × 0.03 = 6 × 10−4 m2 . Thus, A
~ = 6 × 10−4 ĵ.
So, A
~ · A,
~ and since î · ĵ = î · k̂ = 0, then the net flux in part
(a) Since the flux is ΦE = E
(a) is zero!
~ · A,
~ and since î · ĵ = 0, and ĵ · ĵ = 1, we have
(b) Again, ΦE = E
~ ·A
~ = Ey Ay = 100 × 6 × 10−6 = 0.06 N/m2 .
ΦE = E
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4. Chapter 28 - Exercise 29.
The figure shows four sides of a 3.0
cm × 3.0 cm × 3.0 cm cube.
(a) What are the electric fluxes
Φ1 to Φ4 through sides 1 to
4?
(b) What is the net flux through
these four sides?
————————————————————————————————————
Solution
~ and the normal
(a) Again, the flux is ΦE = AE cos θ. On side 1, the angle between E
◦
◦
◦
is θ1 = 150 . For side 2, it’s θ2 = 60 , for side 3 it’s θ3 = 30 , and for side 4 it’s
θ4 = 120◦ . Thus,
Φ1
Φ2
Φ3
Φ4
=
=
=
=
EA cos θ1
EA cos θ2
EA cos θ3
EA cos θ4
= (500) (.03)2 cos 150◦
= (500) (.03)2 cos 60◦
= (500) (.03)2 cos 30◦
= (500) (.03)2 cos 120◦
= −0.39 N/Cm2 .
= 0.225 N/Cm2 .
= 0.39 N/Cm2 .
= −0.225 N/Cm2 .
(b) The net flux through the four sides is just the sum of the individual fluxes,
Φ = Φ1 + Φ2 + Φ3 + Φ4 = 0,
which we know has to be the case, since there is no enclosed net charge.
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5. Chapter 28 - Problem 39.
A hollow metal sphere has inner radius a and outer radius b. The hollow sphere has
charge +2Q. A point charge +Q sits at the center of the hollow sphere.
(a) Determine the electric fields in the three regions r ≤ a, a < r < b, and r ≥ b.
(b) How much charge is on the inside surface of the hollow sphere? On the exterior
surface?
————————————————————————————————————
Solution
The setup is seen in the figure
to the right. There is a single
point charge, Q at the center of
the sphere. The inner radius of the
sphere is a, and the outer radius is
b. We need to determine the fields
in each of the regions using Gauss’s
law.
(a) Gauss’s law says that
I
~ · dA
~ = Qencl .
E
0
Inside the sphere, for r ≤ a, the net enclosed charge is entirely from the point
charge at the center. So, Qencl = +Q. Now, the spherical symmetry of the charge
says that we should take a sphere of radius r < a as our Gaussian surface. Everywhere on that sphere, the electric field is constant. Furthermore, the direction of
the electric field is perpendicular to
to the Gaussian
H the direction
H of the normal
H
~
~
surface (both point radially). So, E · dA = E dA = E dA = EA, where A
is the area of the surface. Since the surface area of sphere is 4πr2 , we find that
1 Q
E (r) = 4π
2 , as expected.
0 r
Next, let’s look outside the sphere, where r ≥ b. In this case the net enclosed
charge is +3Q, from the center point charge, and the charged sphere. We again
1 3Q
want a spherical Gaussian surface. Proceeding as before we find E (r) = 4π
2 .
0 r
Finally, what happens inside the metal itself? This is a static collection of charges.
The charges inside the metal sphere feel the positive charge inside the shell. They
move around due to the electrical force until they have reached a configuration
where all the force cancels out – in other words, until the electric field is zero.
What happens? Since there is a positive charge inside, negative charges in the
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metal shell are attracted to it and move to “coat” the inner surface with a negative
charge, −Q, which is just enough to cancel the positive point charge.
This leaves a net positive charge in the shell which then pushes away from itself,
and onto the surface of the sphere. So, this charge is now on the surface, in
addition to the +2Q that was there before! Thus, the net charge on the surface
of the shell is +3Q!
So, to recap,

1 Q

 4π0 r2 r ≤ a
E (r) = 0
a<r<b

 1 3Q
r ≥ b.
4π0 r2
(b) As discussed above, there is −Q smeared over the inside surface, and +3Q on the
outside.
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6. Chapter 28 - Problem 44.
A positive point charge q sits at the center of a hollow spherical shell. The shell, with
radius R and negligible thickness, has net charge −2q. Find an expression for the
electric field strength (a) inside the sphere, r < R, and (b) outside the sphere, r < R.
In what direction does the electric field point in each case?
————————————————————————————————————
Solution
(a) Again, let’s start by drawing a picture, seen to the right. Taking a
Gaussian surface inside the shell,
r < R, encloses only the positive
charge +q. So, by Gauss’s law,
I
~ · dA
~ = E 4πr2 = q ,
E
0
~ = 1 q2 r̂, for
which says that E
4π0 r
r ≤ R, and points radially outward.
(b) Now, outside the shell, a Gaussian surface encloses both charges, and Hso the net
~ ·dA
~=
enclosed charge is Qencl = −2q +q = −q. So, Gauss’s theorem says that E
q
q
1
~ =−
r̂, for r ≥ R, and points
− 0 . Evaluating the integral as before gives E
4π0 r2
radially inward.
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7. Chapter 28 - Problem 45.
Find the electric field inside and outside a hollow plastic ball of radius R that has
charge Q uniformly distributed on its outer surface.
————————————————————————————————————
Solution
Inside the sphere the field is zero – there’s no charge inside the ball; it’s all on the
surface. So, inside E = 0, for r < R. Outside the ball, we do see a net charge of
Q. Since the ball is spherically symmetric, we choose a spherical Gaussian surface of
radius r. As before, the electric field is constant everywhere on the surface because of
the symmetry. So, as we’ve seen before,
I
~ · dA
~ = EA = Qencl = Q ,
E
0
0
and so we just find the ordinary Coulomb’s law, E =
like an ordinary point charge, as we’ve seen before.
8
1 Q
.
4π0 r2
So, the field looks just
8. Chapter 28 - Problem 53.
A spherical shell has inner radius Rin and outer radius Rout . The shell contains total
charge Q, uniformly distributed. The interior of the shell is empty of charge and
matter.
(a) Find the electric field outside of the shell, r ≥ Rout .
(b) Find the electric field in the interior of the shell, r ≤ Rin .
(c) Find the electric field within the shell, Rin ≤ r ≤ Rout .
(d) Show that your solutions match at both the inner and outer boundaries.
(e) Draw a graph of E versus r.
————————————————————————————————————
Solution
Once again, we will use Gauss’s Law to find the electric field in each region.
(a) Outside of the shell, when r ≥ Rout , the field looks like that of a point charge,
~ = 1 Q2 r̂.
since any spherical Gaussian surface encloses the full charge. So, E
4π0 r
(b) Inside the shell, where r ≤ Rin , there is no charge and hence nothing to source
~ = 0.
the electric field. So, the electric field is zero, E
(c) Now, this gets a bit tougher. Now, we want to use a Gaussian surface of radius
Rin ≤ r ≥ Rout ; i.e., the surface is inside the metal itself. Since the total charge on
the shell is uniformly distributed, the Gaussian surface encloses a certain amount
of charge. The charge density is constant, so ρ = Q/V , where V is the volume
of the charge distribution. The total volume of the sphere is just the difference
3
3
). Thus, ρ =
− Rin
between two spheres of radii Rout and Rin , V = 34 π (Rout
3Q
3 −R3 .
4π (Rout
in )
3
Now the Gaussian surface encloses a volume
Vencl= 43 π (r3 − Rin
), which gives an
enclosed charge of Qencl = ρVencl = Q
H
~ · dA
~ = EA = E (4πr2 ) =
find that E
~ =
E
1 Q
4π0 r2
3
r3 −Rin
3
3
Rout −Rin
Qencl
, and
0
so
3
r3 − Rin
3
3
Rout
− Rin
Again, we find the following field values



0
3 3 r −R
1 Q
~
E (r) = 4π0 r2 R3 −Rin3 r̂
out
in


 1 Q r̂
4π0 r2
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. Again, using Gauss’s law we
r̂
r ≤ Rin
Rin ≤ r ≤ Rout
r ≥ Rout .
(d) Now, the boundaries are at the values r = Rin and Rout . We want to look
at the field inside the metal,
when Rin ≤ r ≤ Rout . When r = Rin we get
3 −R3
R
Q
1
in
in
~ (Rin ) =
r̂ = 0. This matches the value inside the sphere. On
E
2
3 −R3
4π0 Rin
Rout
in
3
3
Rout −Rin
Q
1
~ = 1 Q2
the other boundary, when r = Rout , then E
r̂ = 4π
3
3
2 r̂,
4π0 Rout
Rout −Rin
0 Rout
which exactly matches our solution for r ≥ Rout on the surface. So, we can trust
our solutions!
The electric field is plotted
in the graph to the right.
The field starts out at zero
at the origin, and remains so
until it reaches the inner surface at r = Rin . Then the
field begins to rise, not quite
(e)
linearly, going as r−r−2 , until it reaches the outer surface, r = Rout . This is where
it reaches its maximum, after which it begins to fall off
with the usual r−2 Coulomb
law.
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9. Chapter 28 - Problem 54.
An early model of the atom, proposed by Rutherford after his discovery of the atomic
nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of
radius R with uniformly distributed negative charge −Ze. Z is the atomic number, the
number of protons in the nucleus and the number of electrons in the negative sphere.
(a) Show that the electric field inside this atom is
1
Ze
r
.
Ein =
−
4π0 r2 R3
(b) What is E at the surface of the atom? Is this the expected value? Explain.
(c) A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field strength
at r = 21 R?
————————————————————————————————————
Solution
The atom is seen in the figure
to the right. It contains a positively charged nucleus of charge
+Ze, which is embedded inside a
uniform sphere of charge −Ze. We
are interested in the field inside the
sphere, so we take a spherical gaussian surface of radius r.
(a) Since the electric field is radial, and everywhere
constant on the Gaussian surface,
H
~
~ = EA = E (4πr2 ) = Qencl .
Gauss’s law gives the usual result that E · dA
0
Now, what’s Qencl ? This problem is very similar to the last one, except with
the addition of +Ze at the center. So, the total enclosed charge is Qencl =
+Ze − negative charge.
is just the charge density, times the
The negative charge
4π 3
r3
−Ze
enclosed volume, 4 πR3 × 3 r = −Ze R3 . Thus,
3
r3
= Ze 1 − 3
R
Qencl
,
which, using Gauss’s law as discussed above, gives
Ze
1
r
Ein =
−
.
4π0 r2 R3
(b) On the surface of the atom, r = R, and so the electric field is zero. This is to be
expected, since on the surface the enclosed charge is zero (+Ze − Ze), and so the
electric flux is zero from Gauss’s law. The atom looks neutral outside the surface.
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(c) When r = R/2, then
R
Ze
4
1
7 Ze
.
E
=
−
=
2
2
2
4π0 R
2R
2 4π0 R2
Plugging in the numbers gives
7 Ze
R
7 92 × 1.602 × 10−19 × 9 × 109
=
= 4.64 × 1013 N/C.
E
=
2
−9
2
2 4π0 R2
2
(0.1 × 10 )
So, E = 4.64 × 1013 N/C, which is a very strong electric field!
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10. Chapter 28 - Problem 58.
A sphere of radius R has total charge Q. The volume charge density (C/m3 ) within
the sphere is
r
.
ρ = ρ0 1 −
R
This charge density decreases linearly from ρ0 at the center to zero at the edge of the
sphere.
(a) Show that ρ0 = 3Q/πR2 .
Hint: You’ll need to do a volume integral.
(b) Show that the electric field inside the sphere points radially outward with magnitude
Qr r
E=
4
−
3
.
4π0 R3
R
(c) Show that your result of part b has the expected value at r = R.
————————————————————————————————————
Solution
(a) Now, the charge density is not constant!
So,
R
R we can’t just say that Q = ρV .
Instead, we have to integrate, Q = dq = ρdV . As we’ve seen before, for a
sphere dV = 4πr2 dr, as you can see by imagining a series of thin shells of thickness
dr piled on top of each other. So,
3
R
Z
Z R
r 2
r4
r
π
Q = ρdV = 4πρ0
1−
r dr = 4πρ0
−
= ρ0 R 3 ,
R
3
4R 0
3
0
and so ρ0 =
3Q
.
πR3
(b) HNow, we take a Gaussian surface of radius r inside the sphere. From Gauss’s law,
~ · dA
~ = Qencl , the left-hand side is E (4πr2 ) because the field is radial inside
E
0
the sphere. What’s Qencl ? Again, we integrate as before, only now out to a radius
r, corresponding to our surface.
3
Z
Z r
4πρ0 3 3r4
r 2
r
r4
Qencl = ρdV = 4πρ0
1−
r dr = 4πρ0
−
=
r −
.
R
3
4R
3
4R
0
Plugging for ρ0 gives
Qencl
4πρ0 3 3r4
r3 h
ri
=
r −
=Q 3 4−3
.
3
4R
R
R
Thus, from Gauss’s law
Qr r
4
−
3
.
4π0 R3
R
R
(c) At r = R, then E (R) = 4πQR
4
−
3
= 4πQ0 R2 . So, the electric field at r = R
3
R
0R
is that of a point charge, which is exactly we should expect!
E (r) =
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