Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 24, Problem 21PQ

Often we have distributions of charge for which integrating to find the electric field may not be possible in practice. In such cases, we may be able to get a good approximate solution by dividing the distribution into small but finite particles and taking the vector sum of the contributions of each. To see how this might work, consider a very thin rod of length L = 16 cm with uniform linear charge density λ = 50.0 nC/m. Estimate the magnitude of the electric field at a point P a distance d = 8.0 cm from the end of the rod by dividing it into n segments of equal length as illustrated in Figure P24.21 for n = 4. Treat each segment as a particle whose distance from point P is measured from its center. Find estimates of EP for n = 1, 2, 4, and 8 segments.

Chapter 24, Problem 21PQ, Often we have distributions of charge for which integrating to find the electric field may not be

FIGURE P24.21

Expert Solution & Answer
Check Mark
To determine

The magnitudes of electric fields at P for the segments n=1, n=2, n=3 and n=4.

Answer to Problem 21PQ

The magnitude of electric fields n=1 is 2.81×103N/C, n=2 is 3.40×103N/C, n=4 is 3.64×103N/C and n=4 is 3.72×103N/C.

Explanation of Solution

Write the expression to calculate the electric field.

  EP=kq4r12+kq4r22+kq4r32+kq4r42

Here, EP is the electric field at P, k is the coulomb constant, r1 is the distance from the center of the first segment to P, r2 is the distance from the center of the second segment to P, r3 is the distance from the center of the third segment to P and r4 is the distance from the center of the fourth segment to P and q4 is the one quarter of total charge.

Write the expression to calculate the charge in each segment.

  q4=λL4

Here, λ is the linear charge density and L is the length of the rod

Substitute the above equation in the expression for EP to rewrite.

  EP=(λkL4)(1r12+1r22+1r32+1r42)                                                                           (I)

Write the expression to calculate r1.

  r1=7L8+d                                                                                                                       (II)

Here, d is the distance of the point P from the end of the rod.

Write the expression to calculate r2.

  r2=5L8+d                                                                                                                     (III)

Write the expression to calculate r3.

  r3=3L8+d                                                                                                                     (IV)

Write the expression to calculate r4.

  r4=L8+d                                                                                                                     (V)

Substitute the equations (II), (III), (IV) and (V) in (I) to rewrite.

  EP=(λkL4)(1(7L8+d)2+1(5L8+d)2+1(3L8+d)2+1(L8+d)2)

Conclusion:

Substitute 50.0nC/m for λ, 8.99×109Nm2/C2 for k, 16cm for L and 8cm for d in the above equation to calculate EP.

  EP=((50.0nC/m(109C1nC))(8.99×109Nm2/C2)(16cm(102m1cm))4)(1(7((16cm(102m1cm)))8+(8cm(102m1cm)))2+1(5((16cm(102m1cm)))8+(8cm(102m1cm)))2+1(3((16cm(102m1cm)))8+(8cm(102m1cm)))2+1((16cm(102m1cm))8+(8cm(102m1cm)))2)3.64×103N/C

Similarly, by following the same concepts the electric field for n=1 is 2.81×103N/C, n=2 is 3.40×103N/C, n=4 is 3.64×103N/C and n=4 is 3.72×103N/C.

Therefore, the magnitude of electric fields n=1 is 2.81×103N/C, n=2 is 3.40×103N/C, n=4 is 3.64×103N/C and n=4 is 3.72×103N/C.

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Chapter 24 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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