Example 4: Find the total charge contained in a 2-cm length of the electron beam? z = 4 cm -Pu=-Se-10³pz μC/m³ p=1 cm z = 2 cm A

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Example 4: Find the total charge contained in a 2-cm length
of the electron beam?
Solution:
Finally,
e-40
= 10~10π
Q =
| (-24
Pv=-5 x 10-6e-10³ pz C/m²
The volume differential in cylindrical coordinates is given in Section 1.8; therefore,
0.04 2 0.01
Q =
=
Jo.02 Jo
We integrate first with respect to Ø because it is so easy,
-0.04
Q = = 1000* (0.01 -10-5e-105 pz pdpdz
0.02 Jo
and then with respect to z, because this will simplify the last integration with respect
to p.
0.01-10-5
-105 p
0.01
-Pu
e-20
-4000 -2000
p=1 cm
Q = 10-10
e-105pz
10-¹0(e-4000p
-Se-10%pz
z = 4 cm
-5 x 10-6e-10 pz pdpdødz
1
-4000
Z=0.04
z = 2 cm
Z=0.02
0² μC/m³
y
pdp
e-4000p
e-2000p 0.01
-4000 -2000
e-2000p) dp
1
-2000)]=
= 0.0785 pC
Transcribed Image Text:Example 4: Find the total charge contained in a 2-cm length of the electron beam? Solution: Finally, e-40 = 10~10π Q = | (-24 Pv=-5 x 10-6e-10³ pz C/m² The volume differential in cylindrical coordinates is given in Section 1.8; therefore, 0.04 2 0.01 Q = = Jo.02 Jo We integrate first with respect to Ø because it is so easy, -0.04 Q = = 1000* (0.01 -10-5e-105 pz pdpdz 0.02 Jo and then with respect to z, because this will simplify the last integration with respect to p. 0.01-10-5 -105 p 0.01 -Pu e-20 -4000 -2000 p=1 cm Q = 10-10 e-105pz 10-¹0(e-4000p -Se-10%pz z = 4 cm -5 x 10-6e-10 pz pdpdødz 1 -4000 Z=0.04 z = 2 cm Z=0.02 0² μC/m³ y pdp e-4000p e-2000p 0.01 -4000 -2000 e-2000p) dp 1 -2000)]= = 0.0785 pC
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