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

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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
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