40 - Excercises And Problems Derive Equation 23.11聽for the f... [FREE SOLUTION]

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Chapter 23: 40 - Excercises And Problems (page 655)

Derive Equation $23.11$ for the field ${\overset{鈬赌}{E}}_{d i p o l e}$ in the plane that bisects an electric dipole.

Short Answer

Expert verified

The Equitorial time at field point is $\frac{P}{4 蟺蔚_{0} r^{3}}$ .

Step by step solution

Step: 1 Electric field:

The electric field radial component by

$\begin{array}{l} E_{r} = \frac{鈭� 鈭�}{鈭� r} \\ E_{r} = \frac{鈭� 鈭�}{鈭� r} (\frac{P \cos 鈦� 胃}{4 蟺蔚_{0} r^{2}}) \\ E_{r} = \frac{鈭� P \cos 鈦� 胃}{4 蟺蔚_{0}} (\frac{鈭� L}{r^{3}}) \\ E_{r} = \frac{2 P \cos 鈦� 胃}{4 蟺蔚_{0} r^{3}} \end{array}$

Step; 2 Transverse electric field:

The component transverse of electric field by

$\begin{array}{l} E_{胃} = \frac{鈭� 1}{r} (\frac{鈭� E}{鈭� 胃}) \\ E_{胃} = \frac{鈭� 1}{r} \frac{鈭�}{鈭� 胃} (\frac{P \cos 鈦� 胃}{4 蟺蔚_{0} r^{2}}) \\ E_{胃} = \frac{P \sin 鈦� 胃}{4 蟺蔚_{0} r^{3}} \end{array}$

Step: 3 Resultant field:

The resultant field by,

$\begin{array}{l} E = \sqrt{{(\frac{2 P \cos 鈦� 胃}{4 蟺蔚_{0} r^{3}})}^{2} + {(\frac{P \sin 鈦� 胃}{4 蟺蔚_{0} r^{3}})}^{2}}] \\ E = \frac{P}{4 蟺蔚_{0} r^{3}} \sqrt{4 \cos^{2} 鈦� 胃 + \sin^{2} 鈦� 胃} \\ E = \frac{P}{4 蟺蔚_{0} r^{3}} \sqrt{1 + 3 \cos^{2} 鈦� 胃} 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� \end{array}$

Step; 4 Field angle:

The angle at field,

$\begin{array}{l} \tan 鈦� 蠒 = \frac{E_{胃}}{E_{r}} \\ \tan 鈦� 蠒 = [\frac{P \sin 鈦� 胃}{4 蟺蔚_{0} r^{3}}] \\ \tan 鈦� 蠒 = [\frac{2 P \cos 鈦� 胃}{4 蟺蔚_{0} r^{3}}] \\ \tan 鈦� 蠒 = \frac{1}{2} \frac{\sin 鈦� 胃}{\cos 鈦� 胃} 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� 鈥� . \end{array}$

Because of magnitude and direction in dipole,the angle on equitorial time at $胃 = 90^{鈭�}$

Therefore,the equitorial time on field point is $\frac{P}{4 蟺蔚_{0} r^{3}} .$

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91影视

Derive Equation $23.11$ for the field ${\overset{鈬赌}{E}}_{d i p o l e}$ in the plane that bisects an electric dipole.

Short Answer

Step by step solution

Step: 1 Electric field:

Step; 2 Transverse electric field:

Step: 3 Resultant field:

Step; 4 Field angle:

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91影视

Derive Equation 23.11for the field E鈬赌dipolein the plane that bisects an electric dipole.

Short Answer

Step by step solution

Step: 1 Electric field:

Step; 2 Transverse electric field:

Step: 3 Resultant field:

Step; 4 Field angle:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Waves Physics

Measurements

Kinematics Physics

Quantum Physics

Space Physics

Study anywhere. Anytime. Across all devices.

Derive Equation $23.11$ for the field ${\overset{鈬赌}{E}}_{d i p o l e}$ in the plane that bisects an electric dipole.