Where could you place one positive charge and one negative

charge to produce the pattern of the electric field shown in Figure 13.58? (As usual, each electric field vector is drawn with its tail at the location where the electric field was measured.) Briefly explain your choices.

Short Answer

Expert verified

The direction of the electric field is always outwards from the positive charge and towards the negative charge. Hence, we will keep a positive change in the upper part and a negative charge in the lower part to produce the pattern.

Step by step solution

01

Concept used to solve the problem.

The direction of the electric field is always outward from the positive charge and towards the negative charge.

02

Finding the location of charges.

As we know that the electric field always comes out from a positive charge and from the figure, we can observe that lines are coming out from the upper parts, which indicates the existence of some positive change there.

We also, know that the electric field always goes into a negative charge, and from the figure, we can observe that lines are going into the lower part, which indicates the existence of some negative change there.

Therefore, the distribution of the charges can be given as.

Hence, we will keep a positive charge in the upper part and a negative charge in the lower part to produce the pattern.

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 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

At a given instant in time, three charged objects are located near each other, as shown in Figure 13.57. Explain why the equation

FonQbydipoleQ(14πε2qsr3)

cannot be used to calculate the electric force on the ball of charge +Q.
Figure 13.57

A dipole consists of two charges +6nCand 6nC, held apart by a rod of length 3mm, as shown in Figure 13.71. (a) What is the magnitude of the electric field due to the dipole at location A, 5cmfrom the center of the dipole? (b) What is the magnitude of the electric field due to the dipole at location B, 5cmfrom the center of the dipole?

A Π(“pi-minus”) particle, which has charge e-, is at location 7×109,4×109,5×109 m.

(a) What is the electric field at location 5×109,5×109,4×109 m, due to the π − particle?

(b) At a particular moment an antiproton (same mass as the proton, chargee- ) is at the observation location. At this moment what is the force on the antiproton, due to the Π?

A sphere with radius 2cm is placed at a location near a point charge. The sphere has a charge of -9×10-10C spread uniformly over its surface. The electric field due to the point charge has a magnitude of 470N/C at the center of the sphere. What is the magnitude of the force on the sphere due to the point charge?

A dipole is located at the origin and is composed of charged particles with charge+eand-e, separated by a distance 2×10-10malong the x axis. Calculate the magnitude of electric field due to this dipole at a location(0,3×10-8,0)m.

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.

Sign-up for free