Download Physics 213 — Problem Set 5 (Due before Feb. 26) Spring 1998

Physics 213 — Problem Set 5 (Due before Feb. 26)
Spring 1998
The first Prelim in P213 will be on Thursday Feb. 26, 7:30 pm – 9 pm (room assignments will be made in
lecture). The Prelim will cover material from Problem Sets 1–4, Lab 1, Lectures 1–8, and Serway 23.1–8,
24.1–4, 25.1–3, 27.1–2, 27.6, and 28.1–3.
1. Reading Assignment
Serway Sections 25.3–6, 26.1–5.
2. Serway 25.68A
A ring of radius R carries a uniformly distributed positive charge, as in Figure P25.68 of your text. The
linear charge density of the ring is λ, and an electron is located a distance d above the plane of the ring
on the central perpendicular axis. If this electron is released from rest, what is its speed when it reaches
the center of the ring?
SOLUTION: This problem can be solved by conservation of energy, which means we must first compute
the electric potential. By superposition, the potential is given by
Z
V =
ring
ke dq
= ke
r
Z
ring
dq
√
x2 + R 2
as a function of x along the axis. However, x and R are constants in this integration, so
ke
V = √
2
x + R2
Z
ke Q
dq = √
.
x2 + R 2
ring
The total charge Q is related to the charge density λ by Q = λ(2πR), so
ke (2πλR)
.
V = √
x2 + R 2
Now we can find ∆V = Vf inal − Vinitial :
R
∆V = V (x = 0) − V (x = d) = 2πke λ 1 − √
d2 + R 2
,
and the change in the kinetic energy:
∆K = me (v 2 − (0)2 )/2.
By conservation of energy
0
=
∆E = ∆U + ∆K = qe (∆V ) + ∆K
=
R
1/2me v 2 − |qe |(2πke λ) 1 − √
d2 + R 2
Solving for v gives
v=
4πke |qe |λ
me
R
1− √
d2 + R 2
.
1/2
.
3. Serway 25.36
How much work is required to assemble eight identical point charges, each with charge q, at the corners
of a cube of side s?
SOLUTION: To determine the energy needed to assemble the charge distribution, we imagine starting with
a vacuum and bringing one charge at a time in from ∞. Since no charges are present when we start, V (r)
for the first charge is a constant, and since ∆U = qi ∆V , and ∆V = 0, no work is needed. For every other
charge, however, there will be a potential difference :
∆V = Vf inal (r) − V (∞) V (∞) = 0
1
so for each additional particle, we just need to calculate the potential at the rest position. Label the eight
vertices of the cube as follows: number the four vertices on one face 1 through 4, and the remaining vertices
5 through 8. We denote by Vi,j the potential at vertex i due to a charge at vertex j. Then
W1
=
0
W2
=
q[V1,2 ] = ke
W3
=
q[V1,3 + V2,3 ] = ke
W4
=
q[V1,4 + V2,4 + V3,4 ] = 2ke
W5
=
q[V1,5 + · · · + V4,5 ] = ke
W6
=
W7
=
W8
=
ET ot
=
q2
s
q2
q2
+ ke √
s
2s
q2
q2
+ ke √
s
2s
q2
q2
q2
+ 2ke √ + ke √
s
2s
3s
q2
1
1
(2 · 1 + 2 · √ + √ )
s
2
3
1
q2
1
[V1,7 + · · · + V6,7 ] = ke (2 · 1 + 3 · √ + √ )
s
2
3
q2
1
1
[V1,8 + · · · + V7,8 ] = ke (3 · 1 + 3 · √ + √ )
s
2
3
W1 + W2 + · · · W8
√
q2
q2
1
4√
1
ke (12 + 12 √ + 4 √ ) = ke (12 + 6 2 +
3).
s
s
3
2
3
q[V1,6 + · · · + V5,6 ] = ke
=
4. Serway 25.38
The electric potential in a certain region is V = 4xz − 5y + 3z 2 V. Find the magnitude of the electric field
at (+2, −1, +3), where all distances are in meters.
SOLUTION: From Serway Equation 25.16 :
Ex = −
∂V
∂x
Ey = −
∂V
∂y
Ez = −
∂V
.
∂z
Working out all the derivatives :
Ex = −4z
Ez = −4x − 6z.
Ey = 5
For the point (x, y, z) = (2, −1, 3) :
Ex = −12 V/m
Ey = 5 V/m
Ez = −26 V/m.
5. Serway 25.82
A disk of radius R has a nonuniform surface charge density σ = Cr, where C is a constant and r is
measured from the center of the disk (see Figure P25.82 in your text). Find (by direct integration) the
potential at point P in the figure.
SOLUTION: The potential at P is given by
Z
V = ke
dq
r0
√
where r0 = distance from point P to the small element of charge dq. For this case, r 0 = x2 + r2 where x
is the distance from the disk and r is the distance the charge element dq is from the axis of the disk. Now,
choose dq to be a ring of radius r and thicknes dr. The amount of charge on this ring is :
dq = (σ)(2πrdr) = (Cr)(2πrdr) = 2πCr2 dr.
2
Substituting dq and r0 into the equation for V gives
Z
V
=
R
ke
h
i
p
= πke C R R2 + x2 − x2 ln R + R2 + x2 + x2 ln|x|
p
0
=
iR
p
2πCr2 dr
2πke C p 2
√
=
r r + x2 − x2 ln r + r2 + x2 2
x2 + r 2
h p
πke C R
R2 + x2 − x2 ln
|x|
√
R + R 2 + x2
0
.
(This integral can be found in standard integral tables.)
6. Serway 25.76
Consider two thin, conducting, spherical shells as in Figure P25.76 of your text. The inner shell has
a radius r1 = 15 cm and a charge of 10 nC. The outer shell has a radius r2 = 30 cm and a charge of
−15 nC. Find (a) the electric field E and (b) the electric potential V in regions A, B, and C of the figure.
(Assume V = 0 at infinity.) Sketch plots of both the magnitude of the electric field and the potential
versus radius r.
SOLUTION:
a)Because of the symmetry in this problem, let’s choose spherical Gaussian surfaces and use Gauss’s Law
~ :
to find E
I
Qenc
~ · d~a.
E
=Φ=
0
Because |E| is constant and normal to the spherical surface,
I
Φ = |E|
dA = |E|(4πr2 ).
Therefore |E| = Qenc /(4π0 r2 ) for all r where Qenc is the charge inside the Gaussian sphere going through
the point at r.
|E| = 0 (no charge enclosed)
For Region A (r < r1 )
2
)
(10 nC)(9 × 109 Nm
90
C2
= 2
2
r
r
2
)
(10 nC − 15 nC)(9 × 109 Nm
−45
C2
|E| =
= 2 ,
r2
r
Region B (r1 ≤ r ≤ r2 )
|E| =
Region C (r2 < r)
~
because it encloses both shells. Here |E| is in units of N/C if r is measured in meters. The direction of E
is radially outward.
RB
~ · d~s = V (A) − V (B). We need to start at
b)To determine the potential, we need to use ∆V = − A E
infinity and integrate in because we know V (∞) = 0.
Z
Z
r
r
~ · d~s = −
E
V (r) = −
|E|ds
∞
∞
~ and d~s point radially. For Region C :
because E
Z
r
V (r) = −
∞
h
−45
−45
dr =
r2
r
ir
∞
−45
45
−0=−
r
r
=
(in units of volts for r in meters). For Region B, we need to break up the integral into two pieces :
Z
(0.3 m)
V (r) = −
∞
−45
dr −
r2
Z
r
(0.3 m)
−90
dr.
r2
However, the first integral is V(0.3 m) for Region C :
h
V (r) =
−45
90
+
0.3
r
ir
=
0.3
−45
90
90
90
+
−
= −450 +
.
0.3
r
0.3
r
3
For Region A V (r) − V (0.15) = −
Rr
0.15
0 dr, so
V (r) = V (0.15) = −450 +
90
= 150 V.
0.15
c)See Figures 1 and 2 at the right.
7. Serway 26.6+
Consider a (conducting) solid metal sphere.
a) By what factor does the capacitance of the metal shpere increase if its volume is tripled?
b) If the sphere ends up at radius R, how much energy is stored when the sphere has charge Q? Do this
two ways: 1) use U = Q2 /2C, and 2) integrate the electric field’s energy density over all of space.
SOLUTION:
a)The potential at the surface of a conducting sphere of radius R and charge Q is V = 4πQ0 R . The
capacitance is
Q
C=
= 4π0 R
V
√
To triple the volume you have to multiply the radius by a factor of 3 3 which increases the capacitance by
the same factor.
b)
Q2
Q2
U=
=
2C
8π0 R
Alternately we integrate the energy density of the electric field, uE = 12 0 E 2 , over the entire space. Inside of
the conducting sphere the electric field is zero. Outside, for r ≥ R, E(r) = Q/4π0 r2 . The energy contained
in a spherical shell of radius r and thickness dr is dU = uE 4πr2 dr . Integrating
Z
U=
dU =
1
0
2
Q2
=
8π0
Z
Z
∞
E(r)2 4πr2 dr
R
∞
R
4
1
Q2
.
dr
=
r2
8π0 R