Download Make-up Midterm Solutions

Physics 7D Makeup Midterm Solutions
Problem 1
A hollow, conducting sphere with an outer radius of 0.253 m and an inner radius of 0.194
m has a uniform surface charge density of 6.96 × 10−6 C/m2 . A charge of −0.510 µC is
now introduced into the cavity inside the sphere. (a) What is the new charge density on the
outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere.
(c) What is the electric flux through a spherical surface just inside the inner surface of the
sphere?
Solution
(a) Because there is no electric field inside a conductor the sphere must acquire a charge of
+0.510 µC on its inner surface. Because there is no net flow of charge onto or off of the
conductor this charge must come from the outer surface, so that the total charge of the outer
2
) − 0.510µC = 5.09µC. Now we divide again by the
surface is now Qi − 0.510µC = σ(4πrout
surface area to obtain the new charge density.
σnew
5.09 × 10−6 C
=
= 6.32 × 10−6 C/m2
2
4π(0.253m)
(1)
(b) The electric field just outside the sphere can be determined by Gauss’ law once we
determine the total charge on the sphere. We’ve already accounted for both the initial surface
charge and the newly introduced point charge in our expression above, which tells us that
Qenc = 5.09 × 10−6 C. Gauss’ law then tells us that
E=
Qenc
= 7.15 × 105 N/C
2
4π0 rout
(2)
(c) Finally we want to find the electric flux through a spherical surface just inside the
sphere’s cavity. From Gauss’ law we have that the electric flux is just Qenc /0 . In this case
the enclosed charge is just the new charge −0.510µC, so our electric flux is
ΦE =
−0.510µC
= −5.76 × 104 Nm2 /C
0
(3)
Problem 2
A point charge q1 = 2.0µC is located at (−0.11m, −0.06m), a second point charge q2 =
−5.0µC is located at (0.15m, 0.11m), and a third point charge q3 = 3.0µC is located at
(0.7m, −0.30m). Calculate the work required to assemble such a system.
1
Solution
To assemble this system we start with the first point charge q1 . In the absence of any external
potential it takes no work to place this charge anywhere. Next we bring in the second point
charge from infinity. This takes negative work because the second point charge is attracted
to the first. The potential energy in question is
U2 =
k(2.0µC)(−5.0µC)
kq1 q2
=p
= −0.289 J
r12
(0.15 + 0.11)2 + (0.11 + 0.06)2 m
(4)
Now that our first two charges are situated we bring in the third charge. It has two
contributions to its energy: the potential energy caused by the interaction between it and
the first charge and that due to the second charge
kq1 q3 kq2 q3
+
r13
r23
k(−5.0µC)(3.0µC)
U3 =
k(2.0µC)(3.0µC)
=p
+p
(0.7 + 0.11)2 + (−0.3 + 0.06)2 m
(0.7 − 0.15)2 + (−0.30 − 0.11)2 m
= −0.133 J
(5)
Therefore the work that was done to assemble this configuration is
W = 0.422 J
(6)
Problem 3
The disk with a circular hole at its center (an annulus) has inner radius R1 and outer radius
R2 . The disk has a uniform surface charge density σ. The annulus lies in the yz−plane with
its center at the origin. Determine the electric field for an arbitrary point on the x-axis.
Solution
To solve this problem we use superposition. First I parameterize the annulus: any point lying
on the annulus is completely described by its distance r from the x-axis and an angle θ from
some arbitrary starting point. A differential piece of the annulus has charge dq = σr dr dθ,
and we can write the electric field magnitude due to this piece of the annulus as
σr dr dθ
4π0 (r2 + x2 )
2
(7)
By symmetry considerations the z and y components of the electric field will cancel, so
we need only consider the x component. We extract this from the magnitude given above
by multiplying by a trigonometric factor.
dEx =
σr dr dθ
x
√
2
2
2
4π0 (r + x ) x + r2
(8)
Now we evaluate the total electric field by integrating over the annulus
Z
R2
Z
2π
σrx
4π0 (r2 + x2 )3/2
0
R1
Z
rdr
σx R2
=
2
20 R1 (r + x2 )3/2
R 2
σx
1
=
− 2
20
(r + x2 )1/2 R1
"
#
1
σx
1
p
−p 2
=
20
R12 + x2
R2 + x2
Ex =
dr
dθ
(9)
Problem 4
A cylindrical air capacitor with a length of 17.0 m stores an amount of energy equal to
3.60×10−9 J when the potential difference between the two conductors is 4.80 V. (a) Calculate
the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the
inner and outer conductors.
Solution
(a) For this part we don’t need to know anything about the capacitor. We start with
U = Q2 /2C = QV /2. Then solving for Q we have
Q=
2U
7.20 × 10−9 J
=
= 1.5 nC
V
4.80 V
(10)
(b) Now we do need to know something about cylindrical capacitors. If we place a charge
of Q on the inner cylinder then the electric field in the space between the two conductors
will be (ignoring fringing effects and taking the dielectric constant of air to be negligible)
E=
Q/`
2π0 r
3
(11)
where r is the distance from the cylinder’s axis and ` is the length of the capacitor. The
potential difference will then be
b
Z
|∆V | =
Edr =
a
Q
b
log
2π0 `
a
(12)
where a and b are the radii of the cylinders, so that the capacitance is
C=
Q
2π0 `
=
∆V
log(b/a)
(13)
Now we use our other expression for the energy
U=
CV 2
2U
⇒C= 2
2
V
(14)
and we can finally solve for the ratio of the radii.
2
V π0 l
b
(4.80V)2 π0 (17.0m)
= exp
= exp
= 20.6
a
U
3.6 × 10−9 J
4
(15)