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

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

Short Answer

Expert verified

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

Step by step solution

Step: 1 Electric field:

The electric field radial component by

$\begin{array}{l} E_{r} = \frac{- \partial}{\partial r} \\ E_{r} = \frac{- \partial}{\partial r} (\frac{P \cos θ}{4 π ε_{0} r^{2}}) \\ E_{r} = \frac{- P \cos θ}{4 π ε_{0}} (\frac{- L}{r^{3}}) \\ E_{r} = \frac{2 P \cos θ}{4 π ε_{0} r^{3}} \end{array}$

Step; 2 Transverse electric field:

The component transverse of electric field by

$\begin{array}{l} E_{θ} = \frac{- 1}{r} (\frac{\partial E}{\partial θ}) \\ E_{θ} = \frac{- 1}{r} \frac{\partial}{\partial θ} (\frac{P \cos θ}{4 π ε_{0} r^{2}}) \\ E_{θ} = \frac{P \sin θ}{4 π ε_{0} r^{3}} \end{array}$

Step: 3 Resultant field:

The resultant field by,

$\begin{array}{l} E = \sqrt{{(\frac{2 P \cos θ}{4 π ε_{0} r^{3}})}^{2} + {(\frac{P \sin θ}{4 π ε_{0} r^{3}})}^{2}}] \\ E = \frac{P}{4 π ε_{0} r^{3}} \sqrt{4 \cos^{2} θ + \sin^{2} θ} \\ E = \frac{P}{4 π ε_{0} r^{3}} \sqrt{1 + 3 \cos^{2} θ} \dots \dots \dots \dots \dots \dots \end{array}$

Step; 4 Field angle:

The angle at field,

$\begin{array}{l} \tan ϕ = \frac{E_{θ}}{E_{r}} \\ \tan ϕ = [\frac{P \sin θ}{4 π ε_{0} r^{3}}] \\ \tan ϕ = [\frac{2 P \cos θ}{4 π ε_{0} r^{3}}] \\ \tan ϕ = \frac{1}{2} \frac{\sin θ}{\cos θ} \dots \dots \dots \dots \dots \dots \dots . \end{array}$

Because of magnitude and direction in dipole,the angle on equitorial time at $θ = 90^{\circ}$

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

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Derive Equation $23.11$ for the field ${\overset{⇀}{E}}_{d i p o l e}$ 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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Derive Equation 23.11for the field E⇀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

Quantum Physics

Circular Motion and Gravitation

Physics of Motion

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Study anywhere. Anytime. Across all devices.

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