Download Physics 9 Fall 2011 Homework 3 - Solutions

Physics 9 Fall 2011
Homework 3 - Solutions
Friday September 9, 2011
Make sure your name is on your homework, and please box your final answer. Because
we will be giving partial credit, be sure to attempt all the problems, even if you don’t finish
them. The homework is due at the beginning of class on Friday, September 16th. Because
the solutions will be posted immediately after class, no late homeworks can be accepted! You
are welcome to ask questions during the discussion session or during office hours.
1. A hollow, neutral, conducting spherical shell has inner radius a and outer radius b. A
point charge of magnitude q is placed in the hollow region in the center of the shell.
Using Gauss’s law, determine the electric field in the three regions, r ≤ a, a ≤ r ≤ b,
and r ≥ b.
————————————————————————————————————
Solution
We can use Gauss’s law,
I
~ · dA
~ = Qencl ,
E
0
where the left-hand side is the flux of the electric field through any surface (a Gaussian
surface), and Qencl is the charged enclosed inside that surface. Applying Gauss’s law
to the inside of the sphere, r ≤ a, a spherical Gaussian surface will enclose the charge
Qencl = q. The right-hand side is easy to work out since the field is constant on the
Gaussian surface, and points in the same direction as the normal to the surface. Thus,
E may be taken outside the integral,
I
I
q
~ · dA
~ = E dA = E 4πr2 = q ⇒ E =
,
E
0
4π0 r2
which is just the field of a point charge.
Now, this point charge causes a negative charge −q to be distributed on the inner surface (the negative charges, being free to move around, are pulled to the inner surface),
and the electric field inside the shell is zero. The charges inside the conductor move
around to minimize their energy and rearrange themselves until the net field inside is
zero, such that they don’t feel a force anymore.
The conductor started off neutral, and now negative charges were pulled onto the inner
surface, leaving behind a net positive charge which distributes itself uniformly over the
outer surface. So, a charge of +q is induced on the outer surface, such that a Gaussian
surface around the shell encloses a net charge of +q and the field is just that of a point
charge,
q
E=
.
4π0 r2
1
2. Show that the electric field due to due to an infinitely long, uniformly charged thin
cylindrical shell of radius a having a surface charge density η is given by the following
expressions: E = 0 for 0 ≤ r ≤ a, and E = ηa/0 r for r > a, where r is the distance
from the cylinder.
————————————————————————————————————
Solution
Inside the shell there is no enclosed charge, and so there is no electric field inside the
shell. Hence, E = 0 for 0 ≤ r ≤ a. Outside the shell we can use Gauss’s law to find
the field. Because of the cylindrical symmetry, we should take a cylindrical Gaussian
surface of radius r < a and length L. Then the field is constant on the surface of this
cylinder, and so
I
~ · dA
~ = E (2πrL) ,
E
where we have neglected the flux through the ends of the cylinder since the field is
perpendicular to these areas. Now, the enclosed charge can be written as the surface
charge density, η times the area through which the field is passing, 2πaL. Thus,
Qencl = η (2πaL), and so
E (2πrL) =
ηa
2πaLη
.
⇒E=
0
0 r
2
3. A solid, insulating sphere of radius R and total charge Q has a nonuniform charge
density that varies with r according to the expression ρ(r) = Ar2 , where A is a constant
and r ≤ R is measured from the center of the sphere.
(a) Determine the constant A in terms of the total charge Q on the sphere.
(b) In terms of Q, what is the electric field at a distance r, outside the sphere?
(c) In terms of Q, what is the electric field at a distance r, inside the sphere?
————————————————————————————————————
Solution
(a) The total charge on the sphere can be found by integrating the charge density
over the entire sphere, taking the volume element to be dV = 4πr2 dr,
Z
Z
Z R
ρr2 dr.
Q = dQ = ρdV = 4π
0
Plugging in for the charge density,
Z R
5
4πA 5
R ⇒A=
Q.
Q = 4πA
r4 dr =
5
4πR5
0
(b) Outside of the sphere the field looks just like that of a point charge, so
E (r) =
Q
.
4π0 r2
(c) We can use Gauss’s law to find the field inside the sphere. Taking our Gaussian
surface to be a sphere of radius r < R, we have
I
~ · dA
~ = E 4πr2 .
E
The enclosed charge can be found by integrating the charge density only up to
the radius r,
Z
Z r
r5
5Q r 4
2
r dr = 5 Q.
Qencl = 4π
ρr dr = 5
R 0
R
0
(Notice that when r = R we get back Qencl = Q, as we should expect.) Thus, the
field inside is
Qr5
Q
E 4πr2 =
⇒E=
r3 .
5
0 R
4π0 R5
3
4. Because the formulas for Coulomb’s law and Newton’s law of gravity have the same
inverse-square law dependence on distance, a formula analogous to the formula for
Gauss’s law can be found for gravity. The gravitational field ~g at a location is the force
per unit mass on a test mass m0 placed at that location. (Then, for a point mass m
at the origin, the gravitational field g at some position ~r is ~g = − (GN m/r2 ) r̂.)
(a) Compute the flux of the gravitational field through a spherical surface of radius r
centered at the origin, and verify that the gravitational analog of Gauss’s law is
I
~ = −4πGN mencl .
Φg = ~g · dA
(b) Using this result, determine the gravitational field at a point a distance r from
the center of the Earth, assuming that the Earth’s density is uniform.
————————————————————————————————————
Solution
(a) The gravitational field from a spherical mass distribution is again spherical. Hence,
the field will be constant on a spherical Gaussian surface. Thus, taking our Gaussian surface as a sphere of radius r, we have
I
~ = g 4πr2 .
~g · dA
If this Gaussian surface encloses a point mass m, then mencl = m, and so
GN m
g 4πr2 = −4πGN m ⇒ g = − 2 ,
r
which is the correct field.
(b) We can determine the gravitational field in this case in exactly the same way as
we would determine the electric field inside a uniformly charged
H sphere. Taking
~ = g (4πr2 ).
our Gaussian surface as a sphere of radius r, again, we have ~g · dA
Then, in terms of the density, ρ = ME /VE , where ME is the mass of the Earth,
3
3
ρ.
and VE = 4πRE
/3 is the volume, the enclosed mass can be written mencl = 4πr
3
Thus,
4πr3
4πGN ρ
2
g 4πr = −4πGN
ρ⇒g=−
r,
3
3
or in terms of the mass, ME ,
g=−
4
GN ME
r.
3
RE
5. Consider a solid sphere of radius R with charge Q distributed uniformly. Suppose that
a point charge q of mass m, with sign opposite to that of Q, is free to move within the
solid sphere. Charge q is placed at rest on the surface of the solid sphere and released.
Describe the subsequent motion. In particular, what is the period of this motion, and
what is the total energy of the point charge?
————————————————————————————————————
Solution
As we’ve seen, the electric field inside a uniformly-charged sphere depends linearly on
distance away from the center, and can be found by Gauss’s law,
I
~ · dA
~ = Qencl .
E
0
Taking a spherical Gaussian surface, because of the symmetry of the charge distribution, the left-hand side becomes E (4πr2 ), while the right-hand side can be expressed
in terms of the charge density, ρ, as Qencl /0 = 4πρr3 /30 . Solving for the electric field
gives
Q
ρ
r=
r.
E=
30
4π0 R3
Thus, the net force on the oppositely-charged point charge is
qQ
F = −qE = −
r,
4π0 R3
which is the harmonic oscillator expression with angular frequency
r
qQ
.
ω0 =
4π0 mR3
Thus, the point charge experiences simple harmonic motion, moving from the surface,
down through the center, to the opposite surface, and back again. So, the position of
the point charge is
x(t) = R cos ω0 t.
The period of this motion is T = 2π/ω, or
s
T = 2π
4π0 mR3
.
qQ
Now, the total energy of a harmonic oscillator is E = 12 mv 2 + 21 mω02 x2 , where v = ẋ =
−ω0 R sin ω0 t is the velocity of the particle. So,
1
1
1
E = m (−ω0 R sin ω0 t)2 + mω02 (R cos ω0 t)2 = mω02 R2 ,
2
2
2
which depends only on the frequency and amplitude. This can be expressed in terms
of the charges by plugging in for ω02 as
1
qQ
E = mω02 R2 =
.
2
8π0 R
5