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 68PQ

(a)

To determine

The magnitude of electric field along the disk’s axis at a distance of 2.50cm.

(a)

Expert Solution
Check Mark

Answer to Problem 68PQ

The electric field at distance 2.50cm is 1.25×109N/C.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    E=2πkσ(1yR2+y2)

Here, E is the electric field, k is the coulomb’s constant, σ is the surface charge density, y is the distance of the point from the center, and R is the radius of the disk.

Conclusion:

  Substitute 8.99×109Nm2/C2 for k, 2.45×102C/m2 for σ, 0.250m for R, 2.50cm for y to

determine the electric field.

    E=2π(8.99×109Nm2/C2)(2.45×102C/m2)(12.50cm(0.250m)2+(2.50cm)2)=2π(8.99×109Nm2/C2)(2.45×102C/m2)(12.50×102m(0.250m)2+(2.50×102m)2)=1.25×109N/C

The electric field at distance 2.50cm is 1.25×109N/C.

(b)

To determine

The magnitude of electric field along the disk’s axis at a distance of 25.0cm.

(b)

Expert Solution
Check Mark

Answer to Problem 68PQ

The electric field at distance 25.0cm is 4.05×108N/C.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    E=2πkσ(1yR2+y2)

Here, E is the electric field, k is the coulomb’s constant, σ is the surface charge density, y is the distance of the point from the center, and R is the radius of the disk.

Conclusion:

  Substitute 8.99×109Nm2/C2 for k, 2.45×102C/m2 for σ, 0.250m for R, 25.0cm for y to

determine the electric field.

    E=2π(8.99×109Nm2/C2)(2.45×102C/m2)(125.0cm(0.250m)2+(2.50cm)2)=2π(8.99×109Nm2/C2)(2.45×102C/m2)(125.0×102m(0.250m)2+(25.0×102m)2)=4.05×108N/C

The electric field at distance 25.0cm is 4.05×108N/C.

(c)

To determine

The magnitude of electric field along the disk’s axis at a distance of 50.0cm.

(c)

Expert Solution
Check Mark

Answer to Problem 68PQ

The electric field at distance 50.0cm is 1.46×108N/C.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    E=2πkσ(1yR2+y2)

Here, E is the electric field, k is the coulomb’s constant, σ is the surface charge density, y is the distance of the point from the center, and R is the radius of the disk.

Conclusion:

  Substitute 8.99×109Nm2/C2 for k, 2.45×102C/m2 for σ, 0.250m for R, 50.0cm for y to

determine the electric field.

  E=2π(8.99×109Nm2/C2)(2.45×102C/m2)(150.0cm(0.250m)2+(50.0cm)2)=2π(8.99×109Nm2/C2)(2.45×102C/m2)(150.0×102m(0.250m)2+(50.0×102m)2)=1.46×108N/C

The electric field at distance 50.0cm is 1.46×108N/C.

(d)

To determine

The magnitude of electric field along the disk’s axis at a distance of 5.00m.

(d)

Expert Solution
Check Mark

Answer to Problem 68PQ

The electric field at distance 5.00m is 1.73×106N/C.

Explanation of Solution

Write the formula for the electric field created by a disk along the axis of the disk.

    E=2πkσ(1yR2+y2)

Here, E is the electric field, k is the coulomb’s constant, σ is the surface charge density, y is the distance of the point from the center, and R is the radius of the disk.

Conclusion:

  Substitute 8.99×109Nm2/C2 for k, 2.45×102C/m2 for σ, 0.250m for R, 5.00m for y to

determine the electric field.

    E=2π(8.99×109Nm2/C2)(2.45×102C/m2)(15.00m(0.250m)2+(5.00m)2)=1.73×106N/C

The electric field at distance 5.00m is 1.73×106N/C.

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Chapter 24 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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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