Download Physics 213—Problem Set 8—Solutions Fall 1997

Physics 213—Problem Set 8—Solutions
Fall 1997
1. Reading Assignment
a) This problem set covers Serway 25.3-25.8, 26.1-26.2 b) For next week, we will cover material in
chapters 26.3-26.7, 28.4-28.6.
2. In example 25.11, Serway shows the potential everywhere in space for a uniformly charged sphere of
radius R and total charge Q. Clearly it takes work to build up such a sphere, in that each charge will
be repelling all the other charges in the sphere. If you suddenly turned off whatever ’glue’ is holding
the charges together, they would certainly fly apart. How much electrostatic potential is stored in such
a sphere? Start with what the potential would be if you had a sphere of the same charge density but
smaller radius and then tried to add a little charge dq; how much work would that take?
SOLUTION: Potential energy is stored in the sphere, not potential. The work required to assemble it is
equal to the potential energy. Assume you had built such a sphere up to a radius r, with charge density ρ.
The charge in this sphere is q = ρ 43 πr 3 . At its surface, the potential is the same as if the charge q were a
point charge located at the center. V (r) = k qr = kρ 43 πr2 . Bringing an additional charge dq to the surface
of your sphere requires an amount of work dW = V (r)dq. Let dq be in the form of a thin shell around the
1
sphere of volume 4πr2 dr and charge density ρ. Then, with k = 4π
0
4
4
dW = V dq = kρ πr 2 ρ4πr2 dr =
πρ2 r4 dr
3
30
The work to assemble an entire sphere of radius R is calculated by integrating dW over r from 0 to R. This
work is equal to the total potential energy stored in the sphere,
Z
U=
The charge Q of this sphere is Q =
dW =
ρ 43 πR3 .
4
πρ2
30
Z
R
r4 dr =
0
4
πρ2 R5
150
In these terms, the electrostatic energy stored in the sphere is
Q2 3
4π0 5R
U=
3. Serway 26-6 and extensions
Consider a simple, conducting, solid metal sphere.
a) By what factor does the capacitance of the metal sphere increase if its volume is tripled?
b) If the sphere in part (a) ends up at radius R, how much energy is stored by having such a conductor
at charge Q? Do this two ways: use U = Q2 /2C and integrate the energy density over the entirety of
space.
c) Is there more or less stored energy in this conducting solid sphere or in a solid insulating sphere that
is uniformly charged throughout its volume with the same total charge Q? Is your answer corroborated
by your previous work on the solid insulating sphere?
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)
U=
Q2
Q2
=
2C
8π0 R
1
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=
1
dU = 0
2
=
Q2
8π0
Z
Z
∞
E(r)2 4πr2 dr
R
∞
R
1
Q2
dr =
2
r
8π0 R
c)More energy is stored in the uniformly charged solid sphere than in the conducting sphere, because on
average charges are closer to each other. The answer to this question is obvious in the field picture, because
the electric field and thus uE is not zero inside the uniformly charged sphere. The electric field inside, for
r ≤ R, is E = Qr/4π0 R3 (see Serway p.691). If we integrate the energy density of the electric field inside
the sphere we get an additional term
U=
1
0
2
Z
R
E(r)2 4πr2 dr =
0
=
Q2
8π0 R6
Z
R
r4 dr
0
Q2 R 5
Q2 1
=
6
8π0 R 5
8π0 5R
Adding the energy stored in the electric field inside the sphere to the energy outside from part b) we get
Usolid =
Q2 1
1
Q2 3
( +
)=
8π0 R
5R
4π0 5R
which is equal to the potential energy found in the preceding problem.
4. Serway 26-17A
A variable air capacitor used in tuning circuits is made of N semicircular plates each of radius R and
positioned a distance d from each other. A second identical set of plates that is free to rotate is enmeshed
with the first set (see Figure P26.17 in your text). Determine the capacitance as a function of the angle
of rotation θ, where θ = 0 corresponds to the maximum capacitance.
SOLUTION: As seen in Figure P26.17, the distance between adjacent plates of the two sets is d/2. The
plates face each other completely when θ = 0 and have no overlap when θ = π. Correspondingly, the
2
capacitance of two semicircular plates of radius R and distance d/2 varies from 0 to Cmax = 0 πR
(the 2’s
d
of the area and the distance cancel). At some intermediate angle θ the fraction of the area of the plates
facing each other is π−θ
π . For two sets of N plates there are 2N − 1 gaps, and thus 2N − 1 parallel capacitors,
with an equivalent capacitance
C = (2N − 1)
π−θ
R2
Cmax = (2N − 1)0 (π − θ)
π
d
Alternately,C can be calculated from the energy stored in the electric field if a battery of voltage V is
connected to the two sets of plates. Since the voltage across each gap of width d/2 is V , the electric field
between them has magnitude E = 2V /d, and its energy density is uE = 12 0 E 2 = 20 V 2 /d2 . At angle θ, the
2
π−θ
electric field exists in a volume (2N − 1) d2 πR
. Multiplying uE with this volume we find the energy U
2
π
stored in the electric field which must be equal to U = V 2 /2C
U = 0
V2
1
(2N − 1)R2 (π − θ) = CV 2
2d
2
Solving for C, we get the same result as above.
5. Serway 26-22
A cylindrical capacitor has outer and inner conductors whose radii are in the ratio of b/a = 4/1. The
inner conductor is to be replaced by a wire whose radius is one-half the radius of the original inner
conductor. By what factor should the length be increased in order to obtain a capacitance equal to that
of the original capacitor?
2
SOLUTION: See Serway Example 26.2. The capacitance of a length l of two such cylinders is
C=
l
2π0 l
=
2ke ln b/a
ln b/a
Change the radius of the inner conductor to a0 = a/2 and find the necessary length l0 from the condition
that the capacitance remain unchanged
C=
2π0 l
2π0 l0
=
= C0
ln b/a
ln b/a0
ln b/a0
ln 2b/a
l0
ln 8
3 ln 2
=
=
=
=
= 1.5
l
ln b/a
ln b/a
ln 4
2 ln 2
6. Serway 25-33
Show that the amount of work required to assemble four identical point charges of magnitude Q at the
corners of a square of side s is 5.41keQ2 /s.
SOLUTION: W = QV is the amount of work required to bring a point charge Q from infinity, (where the
potential is zero) to a location at which the electrostatic potential is V . We will bring in the charges to their
assigned places one by one, and each time the potential at that place will be given by the other charges
already at their place. Let us start by bringing the first charge to the top left corner of the square, then the
second charge goes to the top right corner, third charge to the bottom left and fourth charge to the bottom
right.
- First charge: No other charge around yet, W1 = 0.
- Second charge: Needs to be a distance s from the first, V2 = ke Q/s, W2 = ke Q2 /s.
√
- Third charge:
Needs to be a distance
s from the second and a distance 2s from the first, V3 =
√
√
ke Q/s + ke Q/ 2s, W3 = ke Q2 /s + ke Q2 / 2s.
√
- Fourth charge: Needs to be a distance
a distance
2s from the second, and a distance
√ s from the first,
√
s from the third. V4 = 2ke Q/s + ke Q/ 2s, W4 = 2ke Q2 /s + ke Q2 / 2s.
The amount of work required is the sum of all four terms
W = W1 + W2 + W3 + W4 = ke Q2
4
2
+√
s
2s
= 5.41
ke Q2
.
s
7. Serway 25-47 and 25-48
A rod of length L as shown in Figure P25.47 in your text lies along the x axis with its left end at the
origin and has a nonuniform charge density λ = αx (where α is a positive constant).
a) What are the units of α?
b) Calculate the electric potential at A.
c) Calculate the electric potential at point B that lies on the perpendicular bisector of the rod a distance
b above the x axis.
~ at the points A and B from the potentials in parts (b) and (c).
d) Calculate the electric field E
SOLUTION: Note: In this solution the coordinates of points on the charged rod will be denoted with x0 ,
and points on the xy−plane with x and y.
a)
charge
The units of α are
(length)2
b)Each segment dx0 of the rod has a charge dq = λdx0 = αx0 dx0 and contributes dV = ke dq/(d + x0 ) =
ke αx0 dx0 /(d + x0 ) to the potential at x = −d, the location of Point A. The potential at Point A due to the
entire charged rod is
Z
Z L 0 0
αx dx
V (−d) =
dV = ke
d + x0
0
3
Substituting u = d + x0
Z
L+d
V (−d) = ke α
d
(u − d)du
= ke α [u]L+d
− ke α [d ln u]L+d
d
d
u
L+d
= ke α L − d ln
d
c)More generally than for point B, let us calculate the potential at a point P with coordinates
x and y. The
p
contribution dV (x, y) due to a segment dx0 of the charged rod is dV = ke αx0 dx0 / x0 − x)2 + y 2 and the
potential at P
Z L
x0 dx0
V = ke α
p
(x0 − x)2 + y 2
0
Substituting u = x0 − x
Z
V
−x+L
(u + x)du
p
ke α
=
Z
−x+L
u2
−x
Z
xdu
p
ke α
=
(7.1)
u2 + y 2
−x
+
−x+L
+ ke α
y2
p
udu
(7.2)
u2 + y 2
−x
Both of these integrals can be found in the appendix on p.A30 of Serway. We have
h
V (x, y) = ke αx ln(u +
= ke αx ln
L−x+
−x +
iL−x
p
u2 + y 2 )
p
(L − x)2 + y 2
p
(−x)2
+
y2
h
−x
iL−x
p
u2 + y 2
+ ke α
p
p
(L − x)2 + y 2 − ke α
+ ke α
−x
(−x)2 + y 2
(7.3)
Substituting x = −d and y = 0 into (7.3) we get the potential at point A from part a) of this problem.
Inserting the coordinates of point B, x = L/2 and y = b, we find that the second and third term in (7.3)
cancel. The potential at point B is
p
(L/2)2 + b2 + L/2
L
L
V ( , b) = ke α ln p
2
2
(L/2)2 + b2 − L/2
d)To find the components of the electric field at any point in the xy−plane (outside of the rod) we have to
∂V
calculate the partial derivatives ∂V
∂x and ∂y and evaluate them at the point of interest. Starting with the
y-component,
1 ∂V (x, y)
ke α
∂y
√
=
(L−x)2 +y 2
x
+
p
=
p
−x +
y
p
p
L−x+
(L − x)2 + y 2
(L −
+
y
x2 + y 2 − √
x2 +y 2
p
(L −
+
y2
y
(L − x)2 + y 2
x)2
L−x+
p
(L −
x)2
+
L−x+
−x +
+ y2
y
− p
x2 + y 2
xy
x)2
y
y2
p
(L − x)2 + y 2
p
x2 + y 2
−p
x2
+
y2
xy
−x +
p
x2 + y 2
y
− p
x2 + y 2
At point A, since y = 0, Ey = 0 as one would expect from symmetry.
At point B, (x = L/2, y = b), the y-component of the electric field turns out to be, after a little algebra
Ey = −
∂V
∂y
= ke α
x=L/2,y=b
4
p
2b
L2
(L/2)2 + b2
To find Ex , differentiate (7.3) analog to the above. The result is, after more algebra
Ex
=
=
∂V (x, y)
∂x
p
L − x + (L − x)2 + y 2
ke αL
p
−kα ln
+p
2
2
−x + (−x) + y
(L − x)2 + y 2
−
(7.4)
(7.5)
In (7.5) the term with the logarithm is part of the derivative of the first term in (7.3), the second term in
(7.5) is the remainder of this derivative plus the derivatives of the second and third term in (7.3).
At point A, with x = −d and y = 0 inserted in (7.5), we find
Ex = ke α
L
L+d
− ln
L+d
d
At point B, with x = L/2 and y = b
Ex = ke α
p
L
(L/2)2 + b2
− ln
5
(L/2) +
p
(L/2)2 + b2
p
−(L/2) + 2
(L/2)2 + b2
!