EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684668
Author: Katz
Publisher: VST
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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

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

Ch. 24 - A sphere with a charge of 3.50 nC and a radius of...Ch. 24 - Is it possible for a conducting sphere of radius...Ch. 24 - Prob. 7PQCh. 24 - For each sketch of electric field lines in Figure...Ch. 24 - Prob. 9PQCh. 24 - Two large neutral metal plates, fitted tightly...Ch. 24 - Given the two charged particles shown in Figure...Ch. 24 - Prob. 12PQCh. 24 - Prob. 13PQCh. 24 - A particle with charge q on the negative x axis...Ch. 24 - Prob. 15PQCh. 24 - Figure P24.16 shows three charged particles...Ch. 24 - Figure P24.17 shows a dipole. If the positive...Ch. 24 - Find an expression for the electric field at point...Ch. 24 - Figure P24.17 shows a dipole (not drawn to scale)....Ch. 24 - Figure P24.20 shows three charged spheres arranged...Ch. 24 - Often we have distributions of charge for which...Ch. 24 - Prob. 22PQCh. 24 - A positively charged rod with linear charge...Ch. 24 - A positively charged rod of length L = 0.250 m...Ch. 24 - Prob. 25PQCh. 24 - Prob. 26PQCh. 24 - A Find an expression for the position y (along the...Ch. 24 - The electric field at a point on the perpendicular...Ch. 24 - Prob. 29PQCh. 24 - Find an expression for the magnitude of the...Ch. 24 - What is the electric field at point A in Figure...Ch. 24 - A charged rod is curved so that it is part of a...Ch. 24 - If the curved rod in Figure P24.32 has a uniformly...Ch. 24 - aA plastic rod of length = 24.0 cm is uniformly...Ch. 24 - A positively charged disk of radius R = 0.0366 m...Ch. 24 - A positively charged disk of radius R and total...Ch. 24 - A uniformly charged conducting rod of length =...Ch. 24 - Prob. 38PQCh. 24 - Prob. 39PQCh. 24 - Prob. 40PQCh. 24 - Prob. 41PQCh. 24 - Prob. 42PQCh. 24 - What are the magnitude and direction of a uniform...Ch. 24 - An electron is in a uniform upward-pointing...Ch. 24 - Prob. 45PQCh. 24 - Prob. 46PQCh. 24 - A very large disk lies horizontally and has...Ch. 24 - An electron is released from rest in a uniform...Ch. 24 - In Figure P24.49, a charged particle of mass m =...Ch. 24 - Three charged spheres are suspended by...Ch. 24 - Figure P24.51 shows four small charged spheres...Ch. 24 - Prob. 52PQCh. 24 - A uniform electric field given by...Ch. 24 - A uniformly charged ring of radius R = 25.0 cm...Ch. 24 - Prob. 55PQCh. 24 - Prob. 56PQCh. 24 - A potassium chloride molecule (KCl) has a dipole...Ch. 24 - Prob. 58PQCh. 24 - Prob. 59PQCh. 24 - Prob. 60PQCh. 24 - A total charge Q is distributed uniformly on a...Ch. 24 - A simple pendulum has a small sphere at its end...Ch. 24 - A thin, semicircular wire of radius R is uniformly...Ch. 24 - Prob. 64PQCh. 24 - Prob. 65PQCh. 24 - Prob. 66PQCh. 24 - Prob. 67PQCh. 24 - Prob. 68PQCh. 24 - A thin wire with linear charge density =0y0(14+1y)...Ch. 24 - Prob. 70PQCh. 24 - Two positively charged spheres are shown in Figure...Ch. 24 - Prob. 72PQCh. 24 - Prob. 73PQCh. 24 - Prob. 74PQCh. 24 - A conducting rod carrying a total charge of +9.00...Ch. 24 - Prob. 76PQCh. 24 - A When we find the electric field due to a...Ch. 24 - Prob. 78PQCh. 24 - Prob. 79PQCh. 24 - Prob. 80PQCh. 24 - Prob. 81PQCh. 24 - Prob. 82PQ
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