PHGN200
Quiz 2
February 13, 2007
Name:
Question Points Score
CWID:
Calculators Not Allowed
No Work = No Credit
Write Legibly
1
10
2
10
Total:
20
1. 10 points An experimentalist observes a particle of charge q, mass m, and energy qVo moving in a circular
orbit of radius ro inside a spherical 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 spherical capacitor is shown.
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. Start with Gauss’s Law.
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
a < r < b,
(1)
PHGN200
Quiz 2
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
=
a
b
Edx
Z b
1
Q
dx
4πo a x2
b−a
Q
.
=
4πo
ab
=
Thus,
Q=
4πo ab
(Va − Vb ) .
(b − a)
(2)
Setting the particle’s kinetic energy equal to qVo yields
v2 =
2qVo
,
m
(3)
where v is the particle’s speed. Finally, using (1), (2), (3), and Newton’s Second Law yields
v2
qE = m
ro
m 2qVo
4πo ab
1
=
q
(Va − Vb )
(b − a)
4πo ro2
ro m
2Vo ro (b − a)
Va − Vb =
ab
Table 1: For Grader Use Only: Rough grading criteria are given below.
Computation of E field
Computation of Va − Vb
2
Energy of the particle, mv2 = qVo
Newton’s Law and algebra
Page 2
3
3
2
2
pts
pts
pts
pts
PHGN200
Quiz 2
Va
February 13, 2007
C1
C2
C3
C8
C4
C6
Vb
2.
C7
C5
Figure 2: A capacitor network is shown.
(a) 7 points Find the equivalent capacitance, Ceq , of the circuit shown in Fig. (2). You may use ⊥, k
symbolism.
Solution:
C9
C10
C11
C12
C13
C14
Ceq
= C2 ⊥ C3
= C9 kC4
= C5 ⊥ C10
= C6 kC11
= C1 ⊥ C12
= C8 kC13
= C14 ⊥ C7
(4)
(5)
(6)
(7)
(8)
(9)
(10)
Table 2: For Grader Use Only: Rough grading criteria are given below.
Each numbered equation 1 pt
(b) 3 points Give formulas for computing a ⊥ b and akb.
Solution:
a ⊥ b = a−1 + b−1
akb = a + b
Page 3
−1
(11)
(12)
PHGN200
Quiz 2
February 13, 2007
Table 3: For Grader Use Only: Rough grading criteria are given below.
Each numbered equation 1.5 pts
Page 4