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
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
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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electromagnetic field)I want a detailed solution because my teacher changes the numbers. I want a detailed solution. Understand the solution
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92cb0cd2-c69f-44bc-bda6-8cfd57ceffd7%2F4c79486f-d29d-448f-ba88-fda147a22500%2F0q1g8ma_processed.jpeg&w=3840&q=75)
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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