General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 16, Problem 22E

(a)

To determine

To show that the electric field normal to the wire of length extending from L to +L is E=2kλLr(r2+L2)1/2.

(a)

Expert Solution
Check Mark

Answer to Problem 22E

The electric field normal to the wire of length extending from L to +L is E=2kλLr(r2+L2)1/2.

Explanation of Solution

Observe the Figure 16.18.

Write the expression for the linear charge density of the wire.

  λ=QL        (I)

Here, λ is the linear charge density, Q is the charge, and L is the length of wire.

For an elemental length dx, the charge is λdx.

From Figure 16.18a, the distance to an arbitrary point P is R, where R is given by,

  R=(r2+x2)1/2        (II)

This gives the magnitude of electric field due to the charge λdx as,

  dE=kλdxR2        (III)

Use equation (II) in (III).

  dE=kλdx(r2+x2)        (IV)

The total field can be resolved into horizontal and vertical components given by,

  dEx=dEcosθ

  dEy=dEsinθ

Among them the horizontal components the electric field due to two elementary lengths of the wire equidistant from the center cancel each other. Thus, Ex is zero and the net electric field is contributed by only the vertical component of the field.

From Figure 16.18b, sinθ can be deduced as,

  sinθ=r(r2+x2)1/2

The vertical component of the elementary field is thus obtained as,

  dEy=krλdx(r2+x2)3/2        (V)

Thus the net electric field is obtained by integrating over the length of the wire. In this case the wire extends from L to +L. Thus integrate equation (V) in this limit to find the net electric field.

  E=L+LdEy=L+Lkrλdx(r2+x2)3/2=kλL+Lr(r2+x2)3/2dx=kλrL+Lr2(r2+x2)3/2dx        (VI)

Use the standard integration formula Equation B.46 in Appendix B of the text book.

  a2(a2+t2)dt=t(a2+t2)1/2        (VII)

Use equation (VII) for determining the integral in equation (VI).

  E=kλr[x(r2+x2)1/2]L+L=kλr(L(r2+L2)1/2(L)(r2+(L)2)1/2)=kλr2L(r2+L2)1/2=2kλLr(r2+L2)1/2        (VIII)

This shows that the electric field normal to the wire of length extending from L to +L is E=2kλLr(r2+L2)1/2.

Conclusion:

Therefore, the electric field normal to the wire of length extending from L to +L is E=2kλLr(r2+L2)1/2.

(b)

To determine

To show that the equation E=2kλLr(r2+L2)1/2 reduces to E=2kλr for the infinitely long wire.

(b)

Expert Solution
Check Mark

Answer to Problem 22E

The equation E=2kλLr(r2+L2)1/2 reduces to E=2kλr for the infinitely long wire.

Explanation of Solution

For infinitely long wire the integration limits in equation (VI) becomes to +. Use equation (VII) to perform the integration.

  E=kλr[x(r2+x2)1/2]+=kλr[1(1)]=2kλr

Conclusion:

Therefore, the equation E=2kλLr(r2+L2)1/2 reduces to E=2kλr for the infinitely long wire.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Let us assume you are lifting out a 179 lb sheep. The density of the air around the balloon is 1.23 kg/m3 and the density of the air inside the balloon is 0.946 kg/m3.  If the sheep accelerates upwards at 4.84 m/s2, what is the volume of the balloon?  1 kg = 2.20 lbs
Air streams past a small airplane's wings such that speed is 50 m/s over the top surface and 30m/s past the bottom. If the plane has a wing of 9m^2. Ignoring the small height difference find 1.The pressure difference between the top and bottom of the plane's wings. 2. What would be the gravitational pull on the plane assuming the plane is moving horizontally. .
Draw a right-handed 3D Cartesian coordinate system (= x, y and z axes). Show a vector A with tail in the origin and sticking out in the positive x, y and z directions. Show the angles between A and the positive x, y and z axes, and call these angles α₁, α₂ and α3 Prove that Ax Acos α₁ Ay = Acos α₂ A₂- Acos α3

Chapter 16 Solutions

General Physics, 2nd Edition

Ch. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 49ECh. 16 - Prob. 50ECh. 16 - Prob. 51ECh. 16 - Prob. 52ECh. 16 - Prob. 53ECh. 16 - Prob. 54ECh. 16 - Prob. 55ECh. 16 - Prob. 56ECh. 16 - Prob. 57ECh. 16 - Prob. 58ECh. 16 - Prob. 59ECh. 16 - Prob. 60ECh. 16 - Prob. 61ECh. 16 - Prob. 62ECh. 16 - Prob. 63ECh. 16 - Prob. 64ECh. 16 - Prob. 65ECh. 16 - Prob. 66ECh. 16 - Prob. 67ECh. 16 - Prob. 68ECh. 16 - Prob. 69ECh. 16 - Prob. 70ECh. 16 - Prob. 72ECh. 16 - Prob. 73ECh. 16 - Prob. 74ECh. 16 - Prob. 75ECh. 16 - Prob. 76ECh. 16 - Prob. 78ECh. 16 - Prob. 81ECh. 16 - Prob. 82ECh. 16 - Prob. 83ECh. 16 - Prob. 84ECh. 16 - Prob. 85ECh. 16 - Prob. 86ECh. 16 - Prob. 87ECh. 16 - Prob. 88ECh. 16 - Prob. 89ECh. 16 - Prob. 90ECh. 16 - Prob. 91ECh. 16 - Prob. 92ECh. 16 - Prob. 93ECh. 16 - Prob. 94ECh. 16 - Prob. 95ECh. 16 - Prob. 96ECh. 16 - Prob. 97ECh. 16 - Prob. 98ECh. 16 - Prob. 99ECh. 16 - Prob. 100ECh. 16 - Prob. 101ECh. 16 - Prob. 102ECh. 16 - Prob. 103ECh. 16 - Prob. 104ECh. 16 - Prob. 105ECh. 16 - Prob. 106ECh. 16 - Prob. 107ECh. 16 - Prob. 108E
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:OpenStax College
Text book image
Classical Dynamics of Particles and Systems
Physics
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Cengage Learning
Text book image
University Physics Volume 2
Physics
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Magnets and Magnetic Fields; Author: Professor Dave explains;https://www.youtube.com/watch?v=IgtIdttfGVw;License: Standard YouTube License, CC-BY