Example 5.44. Conducting cylinders lie at p = 3 and p = 12 mm: both extend from z = 0 to z = 1 m. Perfect dielectrics occupy the interior region: eg = 1 for 3 < p < 6 mm. er= 4 for 6 < p < 9 mm. and ER=8 for 9 < p < 12 mm. a) Calculate C: First we know that D = (3ps/p)a, C/m². with p expressed in mm. Then. with p in mm. and Vo = -√₁₂ 12 .003 ps €0 E₁ = E₂ = 3ps 3Ps E3 = SEOP The voltage between conductors will be: -dp- Q c = %10 Vo 3Ps EOP 3Ps dp 460P In In [¹() + +¹( ) + ( )] - = 3Ps 4€0P ap V/m (3 < p < 6) -a, V/m (6 < p < 9) 6 -a, V/m (9 < p < 12) - [²30 dp] x 10-3³(m/mm) 16 -(0.830) V Now, the charge on the 1 m length of the inner conductor is Q = 27 (.003) (1) ps. The capacitance is then 27 (.003) (1) Ps (.003)p, (.830)/€0 = - .003ps €0 27740 .830 = 67 pF
Example 5.44. Conducting cylinders lie at p = 3 and p = 12 mm: both extend from z = 0 to z = 1 m. Perfect dielectrics occupy the interior region: eg = 1 for 3 < p < 6 mm. er= 4 for 6 < p < 9 mm. and ER=8 for 9 < p < 12 mm. a) Calculate C: First we know that D = (3ps/p)a, C/m². with p expressed in mm. Then. with p in mm. and Vo = -√₁₂ 12 .003 ps €0 E₁ = E₂ = 3ps 3Ps E3 = SEOP The voltage between conductors will be: -dp- Q c = %10 Vo 3Ps EOP 3Ps dp 460P In In [¹() + +¹( ) + ( )] - = 3Ps 4€0P ap V/m (3 < p < 6) -a, V/m (6 < p < 9) 6 -a, V/m (9 < p < 12) - [²30 dp] x 10-3³(m/mm) 16 -(0.830) V Now, the charge on the 1 m length of the inner conductor is Q = 27 (.003) (1) ps. The capacitance is then 27 (.003) (1) Ps (.003)p, (.830)/€0 = - .003ps €0 27740 .830 = 67 pF
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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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
5.44. Conducting cylinders lie at p = 3 and p = 12 mm: both extend from z = 0 to z = 1 m. Perfect
dielectrics occupy the interior region: eg = 1 for 3 < p < 6 mm. er= 4 for 6 < p < 9 mm. and
ER = 8 for 9 < p < 12 mm.
a) Calculate C: First we know that D = (3ps/p)a, C/m², with p expressed in mm. Then. with p in
mm.
and
€0
E₁ =
E₂ =
3ps
E3 =
SEOP
The voltage between conductors will be:
C=
3Ps
EOP
Q
Vo
a, V/m (3 < p < 6)
3ps
4E0P
6
Vo
6 = [-1₁²2 3 dp ·Sº
3ps
860€
3Ps dp
460P
112
.003 ps
· [¦ ¦ (²²) + ¦ ¹ ( ² ) + ¹ ( ² )]
In
In
In
=
-a, V/m (6 < p < 9)
ap V/m (9 < p < 12)
To
[² 30% dp]
Now, the charge on the 1 m length of the inner conductor is Q = 27 (.003) (1) ps. The capacitance
is then
2T (.003) (1) Ps
(.003)ps (.830)/€0
dp x 10-3(m/mm)
.003ps
€0
=
27740
.830
=
-(0.830) V
67 PF](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1cc218a1-4e73-4b9a-9bc8-082ad97612bc%2F91a9bffa-cef0-46a2-a262-86b30f561c9e%2Fyljyx3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example
5.44. Conducting cylinders lie at p = 3 and p = 12 mm: both extend from z = 0 to z = 1 m. Perfect
dielectrics occupy the interior region: eg = 1 for 3 < p < 6 mm. er= 4 for 6 < p < 9 mm. and
ER = 8 for 9 < p < 12 mm.
a) Calculate C: First we know that D = (3ps/p)a, C/m², with p expressed in mm. Then. with p in
mm.
and
€0
E₁ =
E₂ =
3ps
E3 =
SEOP
The voltage between conductors will be:
C=
3Ps
EOP
Q
Vo
a, V/m (3 < p < 6)
3ps
4E0P
6
Vo
6 = [-1₁²2 3 dp ·Sº
3ps
860€
3Ps dp
460P
112
.003 ps
· [¦ ¦ (²²) + ¦ ¹ ( ² ) + ¹ ( ² )]
In
In
In
=
-a, V/m (6 < p < 9)
ap V/m (9 < p < 12)
To
[² 30% dp]
Now, the charge on the 1 m length of the inner conductor is Q = 27 (.003) (1) ps. The capacitance
is then
2T (.003) (1) Ps
(.003)ps (.830)/€0
dp x 10-3(m/mm)
.003ps
€0
=
27740
.830
=
-(0.830) V
67 PF
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