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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)
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 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
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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