Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 24: Q71P (page 715)

Starting from Eq. 24-30, derive an expression for the electric field due to a dipole at a point on the dipole axis.

Short Answer

Expert verified

E=12πεo⋅pr3

Step by step solution

01

Understanding the concept

After reading the question, consider an electric dipole of charge ±q to be separated by ‘d’.

The electric potential at a distance ‘r’ with angle θfrom the middle of the electric dipole is given as:

V=14πεo⋅p cos θr2,wherep=qd

If θ=0,cos 0=1

V=14πεo⋅pr2

02

Step 2: Derive an expression for the electric field due to a dipole at a point on the dipole axis

According to the definition of a potential gradient,

E=−∂V∂rE=−p4πεod(r−2)drE=+2×p4πεor−3E=12πεo⋅pr3

Hence, the expression isE=12πεo⋅pr3

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Question: Figure 24-45 shows a thin rod with a uniform charge density of2.00μc/m. Evaluate the electric potential at point Pifd=D=L/4.00. Assume that the potential is zero at infinity.

As a space shuttle moves through the dilute ionized gas of Earth’s ionosphere, the shuttle’s potential is typically changed by -1.0 Vduring one revolution. Assuming the shuttle is a sphere of radius 10 m, estimate the amount of charge it collects.

The thin plastic rod of length L = 10.0cmin Fig. 24-47 has a non-uniform linear charge densityλ=cx, wherec=49.9pC/m2. (a) With V= 0 at infinity, find the electric potential at point P2 on the yaxis at y = D = 3.56cm. (b) Find the electric field component at P2. (c) Why cannot the field component Exat P2 be found using the result of (a)?

Figure 24-37 shows a rectangular array of charged particles fixed in place, with distance a = 39.0 cmand the charges shown as integer multiples of q1 = 3.40 pCand q2 = 6.00 pC. With V = 0at infinity, what is the net electric potential at the rectangle’s center? (Hint:Thoughtful examination of the arrangement can reduce the calculation.)

Question: In Fig. 24-41a, a particle of elementary charge +eis initially at coordinate z = 20 nmon the dipole axis (here a zaxis) through an electric dipole, on the positive side of the dipole. (The origin of zis at the center of the dipole.) The particle is then moved along a circular path around the dipole center until it is at coordinate z = -20 nm, on the negative side of the dipole axis. Figure 24-41bgives the work done by the force moving the particle versus the angle u that locates the particle relative to the positive direction of the z-axis. The scale of the vertical axis is set byWas=4.0×10-30J.What is the magnitude of the dipole moment?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Wave Optics

Read Explanation

Waves Physics

Read Explanation

Quantum Physics

Read Explanation

Fluids

Read Explanation

Scientific Method Physics

Read Explanation
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.

Sign-up for free