Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

FIGURE shows a thin rod of length Lwith total charge Q.

a. Find an expression for the electric field strength at point Pon the axis of the rod at distance rfrom the center.

b. Verify that your expression has the expected behavior if rL.

c. Evaluate Eat r=3.0cmif L=5.0cmand Q=3.0nC.

Short Answer

Expert verified

Parta

aThe expression for the electrical field, E=14πε0Qr2-L24i^.

Partb

bThe expression reduces to the expression for the point charge, E=14πε0Qr2i^.

Partc

cThe field strength is E=9.8·104N/C.

Step by step solution

01

Find the expression (part a)

As illustrated in the illustration, move the xaxis all along rod with origin at the end. The linear charge density is calculated as follows:

λ=QL

It has a charging aspect to it.

dQ=λdx=QLdx

Because the wire section is so small, it can be viewed as a point charge. As a result, the field element produced just at point away from the wire along the axis is given by

dE=14πε0dqr+L2-x2i^=14πε0QLdxr+L2-x2i^

The equation is integrated as,

E=0LdE

Remove everything in integral that is constant.

E=i^14πε0QL0Ldxr+L2-x2

02

Explained by diagram (part a)

Denote,

Int=0Ldxr+L2-x2

Introduce a substitution,

ξ=x-r-L2,dξ=dx

The new integration bounds are

x=0ξ=-r-L2;x=Lξ=-r+L2

Int=-r-L2-r+L2dξξ2=-1ξ-r-L2-r+L2=1r-L2-1r+L2=Lr2-L24

E=14πε0Qr2-L24i^

03

The expression reduces to the expression for the point charge (part b)

In this case, rLthen L24can be neglected by r2. so,

E=14πε0Qr2i^

This turns out to be an expression for the expected charge far away from the point; the entire wire appears to be a point charge.

04

Field strength (part c)

The vector sum of all forces exerted by a field at a particular position within the field on a unit mass, unit charges, unit magnetic pole, and so on from the figure as

E=9.8·104N/C.(constant)

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

You have a summer intern position with a company that designs and builds nanomachines. An engineer with the company is designing a microscopic oscillator to help keep time, and you’ve been assigned to help him analyze the design. He wants to place a negative charge at the center of a very small, positively charged metal ring. His claim is that the negative charge will undergo simple harmonic motion at a frequency determined by the amount of charge on the ring.

a. Consider a negative charge near the center of a positively charged ring centered on the z-axis. Show that there is a restoring force on the charge if it moves along the z-axisbut stays close to the center of the ring. That is, show there’s a force that tries to keep the charge at z=0. b. Show that for small oscillations, with amplitude <<R, a particle of mass mwith charge-qundergoes simple harmonic motion with frequency f=12πqQ4πε0mR3,RandQare the radius and charge of the ring.

c. Evaluate the oscillation frequency for an electron at the center of a 2.0μmdiameter ring charged to 1.0×10-13C.

Show that the on-axis electric field of a ring of charge has the expected behavior when zRand whenzR.

An electric dipole is formed from two charges, ±q, spaced1.0cm apart. The dipole is at the origin, oriented along the y-axis. The electric field strength at the point x,y=(0cm,10cm)is 360N/C.

a. What is the charge q? Give your answer in nC.

b. What is the electric field strength at the pointx,y=10cm,0cm?

A parallel-plate capacitor has 2.0cm×2.0cmelectrodes with surface charge densities ±1.0×106C/m2. A proton traveling parallel to the electrodes at 1.0×106m/s enters the center of the gap between them. By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor.

A20cm×20cmhorizontal metal electrode is uniformly charged to +80nC. What is the electric field strength 2.0mmabove the center of the electrode?

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