/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 93 A dipole consists of two equal a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A dipole consists of two equal and opposite point charges \((\pm q)\) separated by a distance \(d\) (a) Write an expression for the magnitude of the electric field at a point \((x, 0)\) a large distance \((x \gg d)\) from the midpoint of the charges on a line perpendicular to the dipole axis. [Hint: Use small angle approximations.] (b) Give the direction of the field for \(x>0\) and for \(x<0\).

Short Answer

Expert verified
Answer: \(E \approx \frac{2kq}{x^2}\), where \(k\) is Coulomb's constant, \(q\) is the charge, and \(x\) is the distance from the midpoint of the dipole along the horizontal axis.

Step by step solution

01

Electric field from a point charge

The electric field from a point charge is given by the expression: \(E = \frac{k|q|}{r^2}\), where \(k\) is the Coulomb's constant (\(8.99\times10^{9} Nm^2/C^2\)), \(q\) is the charge, and \(r\) is the distance from the point charge to the point of interest.
02

Determine the electric field from each charge

Let's consider the dipole with positive charge \(q\) located at \((0, d/2)\) and negative charge \(-q\) located at \((0, -d/2)\). The distance from each charge to the point \((x, 0)\) is given by: \(r_+ = \sqrt{x^2 + (d/2)^2}\) for the positive charge \(r_- = \sqrt{x^2 + (d/2)^2}\) for the negative charge Since the distances are the same, the magnitudes of the electric fields from each charge at point \((x, 0)\) are also equal.
03

Determine the electric field vectors

The electric field vectors from each charge to the point \((x, 0)\) can be written as: \(\vec{E}_+ = E_+ \cos\theta \hat{i} - E_+ \sin\theta \hat{j}\) \(\vec{E}_- = -E_- \cos\theta \hat{i} - E_- \sin\theta \hat{j}\) Where \(E_+ = E_- = \frac{k|q|}{r^2}\), and \(\theta\) is the angle between the electric field vector and the horizontal axis.
04

Sum the electric field vectors

Adding the electric field vectors, we obtain the net electric field: \(\vec{E} = (\frac{2kq\cos\theta}{r^2})\hat{i} - (\frac{2kq\sin\theta}{r^2})\hat{j}\)
05

Apply the small angle approximation

Since \(x \gg d\), the angle \(\theta\) is very small. The small angle approximations are: \(\cos\theta \approx 1\) and \(\sin\theta \approx \frac{d/2}{r}\). Substituting these approximations, we get: \(\vec{E} = (\frac{2kq}{r^2})\hat{i} - (\frac{kd}{r^3})\hat{j}\). But \(r^2 \approx x^2\), so the magnitude of the electric field at \((x, 0)\) is approximately: \(E = \sqrt{(\frac{2kq}{x^2})^2 + (\frac{kd}{x^3})^2}\) (a) Since \(x^2 \gg d^2\), we can ignore the second term in the magnitude under the square root sign, and obtain: \(E \approx \frac{2kq}{x^2}\) (b) The direction of the electric field depends on the sign of \(x\). For \(x > 0\), the electric field vector \(\vec{E}\) points to the right and downward (positive \(\hat{i}\) direction and negative \(\hat{j}\) direction). For \(x < 0\), the electric field vector \(\vec{E}\) points to the left and downward (negative \(\hat{i}\) direction and negative \(\hat{j}\) direction).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Some forms of cancer can be treated using proton therapy in which proton beams are accelerated to high energies, then directed to collide into a tumor, killing the malignant cells. Suppose a proton accelerator is \(4.0 \mathrm{m}\) long and must accelerate protons from rest to a speed of $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} .$ Ignore any relativistic effects (Chapter 26) and determine the magnitude of the average electric field that could accelerate these protons.
Two metal spheres of radius \(5.0 \mathrm{cm}\) carry net charges of $+1.0 \mu \mathrm{C}\( and \)+0.2 \mu \mathrm{C} .$ (a) What (approximately) is the magnitude of the electrical repulsion on either sphere when their centers are \(1.00 \mathrm{m}\) apart? (b) Why cannot Coulomb's law be used to find the force of repulsion when their centers are \(12 \mathrm{cm}\) apart? (c) Would the actual force be larger or smaller than the result of using Coulomb's law with \(r=12 \mathrm{cm} ?\) Explain.
Two metal spheres separated by a distance much greater than either sphere's radius have equal mass \(m\) and equal electric charge \(q .\) What is the ratio of charge to mass \(q / m\) in C/kg if the electrical and gravitational forces balance?
A conducting sphere that carries a total charge of \(-6 \mu \mathrm{C}\) is placed at the center of a conducting spherical shell that carries a total charge of \(+1 \mu \mathrm{C}\). The conductors are in electrostatic equilibrium. Determine the charge on the outer surface of the shell. [Hint: Sketch a field line diagram.]
(a) Use Gauss's law to prove that the electric field outside any spherically symmetric charge distribution is the same as if all of the charge were concentrated into a point charge. (b) Now use Gauss's law to prove that the electric field inside a spherically symmetric charge distribution is zero if none of the charge is at a distance from the center less than that of the point where we determine the field.
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.