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Appendix A3: Solution to Equation 11.25
Equation 11.25 is reproduced and renumbered as below:
− = −
e
m
E
dU
dt
U
d
dt
r
r
φ
φ
(A3.1a)
− = +
e
m
E U
d
dt
dU
dt
rφ
φ
φ
(A3.1b)
− =
e
m
E
dU
dt
z
z
(A3.1c)
U U a U a U a
dr
dt
a
rd
dt
a
dz
dt
a
r r z z r z
= + + = + +
φ φ φ
φ
(A3.1d)
Since d
ϕ
/dt =
ω
, U
ϕ
= r
ω
, Equation A3.1 gets modied to
− = −
e
m
E
d r
dt
r
r
2
2
2
ω
(A3.2a)
− = + =
e
m
E
dr
dt
d r
dt r
d
dt
r
φ
ω
ω
ω
( )
( )
1
2
(A3.2b)
− =
e
m
E
d z
dt
z
2
(A3.2c)
In this case the behaviour of a charged particle in the E-eld is illustrated
through a cylindrical diode shown in Figure 11.6. The two concentric cylin-
ders with radii ‘a’ and ‘b’ are maintained at V = 0 and V = V
0
voltages, respec-
tively. Since this case ...