### Understanding the Electric Field of an Electric Dipole#### Problem StatementConsider the electric dipole shown in the figure. Show that the electric field at a distant point on the y-axis is approximately given by:\[ E_x \approx 4k_e \frac{qa}{x^3} \]#### ExplanationThe provided figure illustrates an electric dipole consisting of two charges: a negative charge \(-q\) and a positive charge \(+q\), separated by a distance \(2a\) along the x-axis. The point of interest, where we aim to determine the electric field, is located on the y-axis, at a distance \(x\) from the origin.##### Figure DescriptionThe figure includes:- A coordinate system with the x-axis and y-axis.- Two point charges: - A negative charge \(-q\) positioned on the left side at \(-a\) on the x-axis. - A positive charge \(+q\) positioned on the right side at \(a\) on the x-axis.- The distance between the charges is labeled as \(2a\).#### Using Coulomb's LawTo derive the expression for the electric field at a distant point on the y-axis, Coulomb's Law will be applied. Coulomb’s Law states that the magnitude of the electric field created by a point charge \(q\) at a distance \(r\) is given by:\[ E = k_e \frac{q}{r^2} \]where:- \( E \) is the magnitude of the electric field.- \( k_e \) (also known as Coulomb's constant) is approximately \( 8.99 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \).- \( q \) is the charge.- \( r \) is the distance from the charge to the point where the electric field is being calculated.##### Contributions to the Electric FieldFor the dipole, the contributions to the electric field from both the positive and negative charges need to be considered. Due to the symmetry of the problem and the distance \(x \gg a\), the resultant field can be approximated.By evaluating the direction, magnitudes, and combining the components of the electric fields due to each charge, it can be shown mathematically that:\[ E_x \approx 4k_e \frac

The electric dipole is as shown, and the direction of electric field due to this dipole can be shown…

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Consider the electric dipole shown in the figure. Show that the electric field at a distant point on the y axis is

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

### Understanding the Electric Field of an Electric Dipole

#### Problem Statement
Consider the electric dipole shown in the figure. Show that the electric field at a distant point on the y-axis is approximately given by:

\[ E_x \approx 4k_e \frac{qa}{x^3} \]

#### Explanation
The provided figure illustrates an electric dipole consisting of two charges: a negative charge \(-q\) and a positive charge \(+q\), separated by a distance \(2a\) along the x-axis. The point of interest, where we aim to determine the electric field, is located on the y-axis, at a distance \(x\) from the origin.

##### Figure Description
The figure includes:
- A coordinate system with the x-axis and y-axis.
- Two point charges:
- A negative charge \(-q\) positioned on the left side at \(-a\) on the x-axis.
- A positive charge \(+q\) positioned on the right side at \(a\) on the x-axis.
- The distance between the charges is labeled as \(2a\).

#### Using Coulomb's Law
To derive the expression for the electric field at a distant point on the y-axis, Coulomb's Law will be applied. Coulomb’s Law states that the magnitude of the electric field created by a point charge \(q\) at a distance \(r\) is given by:

\[ E = k_e \frac{q}{r^2} \]

where:
- \( E \) is the magnitude of the electric field.
- \( k_e \) (also known as Coulomb's constant) is approximately \( 8.99 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \).
- \( q \) is the charge.
- \( r \) is the distance from the charge to the point where the electric field is being calculated.

##### Contributions to the Electric Field
For the dipole, the contributions to the electric field from both the positive and negative charges need to be considered. Due to the symmetry of the problem and the distance \(x \gg a\), the resultant field can be approximated.

By evaluating the direction, magnitudes, and combining the components of the electric fields due to each charge, it can be shown mathematically that:

\[ E_x \approx 4k_e \frac

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