PHGN200: All Sections
Recitation 3
February 13, 2007
1. We want to design a spherical vacuum capacitor of a given radius a for the outer sphere, which will be able
to store the greatest amount of electrical energy, subject to the constraint that the electric field strength at
the surface of the inner sphere may not exceed Eo .
(a) What radius b should be chosen for the inner spherical conductor?
Solution: Let Q be the amount of positive charge on the inner sphere; the electric field between the
two spheres is given by (use Gauss’s Law to show this or look at the Recitation 2 problems)
E=
Q
,
4πo r2
b < r < a,
(1)
where r is the radius drawn from the origin of the coordinate system. The electric field must not
exceed Eo at b, thus, letting E = Eo , r = b in (1), and solving for Q yields
Q = 4πo b2 Eo .
The electric potential difference, Vb − Va , is given by (see Recitation 2 problems)
Z a
~ · d~`
Vb − Va =
E
b
Z a
~ · dxı̂ı̂
=
E
b
Z a
=
Edx
b
Z a
1
Q
dx
=
4πo b x2
Q(a − b)
=
,
4πo ab
(2)
(3)
where Q is given by (2). The amount of electrical energy stored in a capacitor is
1
U = Q4V
2
1
= Q(Vb − Va )
2
Q(a − b)
1
2
=
4πo b Eo
2
4πo ab
4πo 2 3
=
E b (a − b).
2a o
(4)
We want to find b that maximizes U ; thus, the derivative of U with respect to (w.r.t) b must vanish.
Differentiating (4) w.r.t b yields
3
b = a.
(5)
4
PHGN200: All Sections
Recitation 3
February 13, 2007
(b) How much energy can be stored?
Solution: Substituting (5) into (4) yields
27a3 Eo2
U = 4πo
.
512
2. An experimentalist observes a particle of charge q, mass m, and energy qVo moving in a circular orbit of
radius ro inside a cylindrical capacitor with inner radius a and outer radius b, see Fig. (1).
b
ro
a
Figure 1: A particle of charge q, mass m, and energy qVo moving inside a cylindrical capacitor is shown.
(a) Assume the cylindrical capacitor is very long compared to the space between its walls, i.e., b−a
1.
`
Furthermore, assume that the moving particle has no effect on the cylindrical capacitor. Find the
capacitance of the cylindrical capacitor.
Solution: Let Q be the amount of positive charge on the inner cylinder; the electric field between
the two cylinders is given by (use Gauss’s Law to show this or look at the recitation two problems)
E=
Q
,
2π`o r
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a < r < b,
(6)
PHGN200: All Sections
Recitation 3
February 13, 2007
The electric potential difference, Va − Vb , is given by (see Recitation 2 problems)
Z
b
~ · d~`
E
Va − Vb =
a
Z
b
~ · dxı̂ı̂
E
=
a
Z
b
=
Edx
Z b
1
Q
dx
=
2π`o a x
b
Q
ln
.
=
2π`o
a
a
The capacitance is given by
Q
4V
Q
=
Va − Vb
2π`o
=
ln (b/a)
C=
(7)
(b) The experimentalist quickly observes that the particle only moves in a circular orbit with radius ro when
Va − Vb has a very special value. Find this value. (No, it’s not zero.)
Solution: The electric field between the walls of the capacitor is (see Recitation 2 problems)
E=
Substituting C =
Q
Va −Vb
Q
.
2π`o r
(8)
into (8) and using (7) yields
E=
Va − Vb
.
r ln (b/a)
(9)
Setting the particle’s kinetic energy equal to qVo yields
2qVo
,
m
where v is the particle’s speed. Finally, using (9), (10), and Newton’s Second Law yields
v2 =
v2
Va − Vb
q
=m
ro ln (b/a)
ro
b
Va − Vb = 2Vo ln
.
a
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PHGN200: All Sections
Recitation 3
February 13, 2007
3. Find the equivalent capacitance, Ceq , of the circuit shown in Fig. (2). Data: C1 = 1, C2 = 2, C3 = 3, C4 =
4, C5 = 5, C6 = 6, C7 = 7.
C1
Va
C2
C3
C4
C6
Vb
C7
C5
Figure 2: Find Ceq for this circuit. Hint: use color pencils/pens.
Solution:
Va
C1
C2
C3
C4
C6
Vb
C7
C5
Figure 3: Capacitors in series are denoted by ⊥, and capacitors in parallel are denoted by k.
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PHGN200: All Sections
Va
Recitation 3
February 13, 2007
C1
C8 = C2 ⊥ C3
C4
C6
Vb
C7
C5
Figure 4: C8 = C2 ⊥ C3
Va
C1
C9 = C4kC8
C6
Vb
C7
C5
Figure 5: C9 = C4 kC8
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PHGN200: All Sections
Recitation 3
February 13, 2007
C1
Va
C6
Vb
C7
C10 = C5 ⊥ C9
Figure 6: C10 = C5 ⊥ C9
Va
C1
C11 = C6kC10
Vb
C7
Figure 7: C11 = C6 kC10
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PHGN200: All Sections
Recitation 3
Va
February 13, 2007
C12 = C1 ⊥ C11 ⊥ C7
Vb
Figure 8: C12 = C1 ⊥ C10 ⊥ C7
Recall that capacitors in series obey
Ceq =
N
X
!−1
Ci−1
,
(11)
i=1
and capacitors in parallel obey
Ceq =
N
X
Ci .
i=1
Using Fig. (3)-(8), (11), and (12) yields
Ceq =
3052
≈ 0.794
3845
4. Prove that a network of capacitors cannot always be grouped into series and parallel combinations.
Solution: We prove the above by finding a counter example, see Fig. 9.
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PHGN200: All Sections
Recitation 3
C1
February 13, 2007
C2
Va
Vb
C3
C5
C4
Figure 9: The above circuit cannot be reduced by grouping capacitors into series and parallel combinations.
Page 8