Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 24, Problem 60AP

(a)

To determine

The electric field for r<a.

(a)

Expert Solution
Check Mark

Answer to Problem 60AP

The electric field for r<a is (2λker) directed radial outwards.

Explanation of Solution

The diagram for the cylindrical shell with inner radius a and the outer radius b is shown below.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 60AP

Figure (1)

Here, a is the inner radius, b is the outer radius and λ is the linear charge density.

Write the expression for the linear charge density.

    λ=qenL                                                                                                                      (I)

Here, L is the length of the cylindrical shell and qen is the enclosed charge.

Rearrange equation (I) to get the expression for qen.

    qen=λL

Write the expression to calculate the electric field for the region r<a.

    EA=qenε0                                                                                                                  (II)

Here, E is the electric field for the region r<a, ε0 is the permittivity of free space and A is the area of cylinder.

Substitute λL for qen and 2πrL for A in equation (II) to solve for E.

    E(2πrL)=λLε0E=λ2πrε0×12×2E=14πε0(2λr)                                                                                       (III)

Substitute ke for 14πε0 in equation (III).

    E=(2λker)

Positive sign indicate the electric field is directed radially outward.

Conclusion:

Therefore, the electric field for r<a is (2λker) directed radial outward.

(b)

To determine

The electric field for a<r<b.

(b)

Expert Solution
Check Mark

Answer to Problem 60AP

The electric field for a<r<b is ke(2(λ+ρ(π(r2a2)))r) directed radially outward.

Explanation of Solution

Write the expression for the volume charge density.

    ρ=qvV                                                                                                                    (IV)

Here, ρ is the volume charge density and V is the volume of the enclosed charge.

Rearrange equation (IV) to obtain the expression for qen.

    qv=ρV                                                                                                                   (V)

Here, qv is the volumetric charge.

Write the expression to calculate the volume of the enclosed charge.

    V=VrVa                                                                                                               (VI)

Here, Vr is the volume of cylinder with radius r and Va is the volume of cylinder with radius a.

Substitute πr2L for Vr and πa2L for Va in equation (VI) to solve for V.

    V=πr2Lπa2L=πL(r2a2)

Write the expression for total charge inside the cylinder.

    qen=qL+qV                                                                                                          (VII)

Here, qL is the linear charge.

Substitute λL for qL and ρV for qv in equation (VII) to solve for qen.

    qen=λL+ρV

Write the expression to calculate the electric field for the region a<r<b.

    EA=qenε0.                                                                                                             (VIII)

Here, E is the electric field for the region a<r<b.

Substitute λL+ρV for qen, 2πrL for A and πL(r2a2) for V in equation (V) to solve for E.

    E(2πrL)=λL+ρ(πL(r2a2))ε0E=λ+ρ(π(r2a2))2πrε0×22=14πε0(2(λ+ρ(π(r2a2)))r)                                                             (IX)

Substitute ke for 14πε0 in equation (IX).

    E=ke(2(λ+ρ(π(r2a2)))r)

Positive sign indicates the electric field is directed radially outward.

Conclusion:

Therefore, the electric field for a<r<b is ke(2(λ+ρ(π(r2a2)))r) directed radially outward.

(c)

To determine

The electric field for r>b.

(c)

Expert Solution
Check Mark

Answer to Problem 60AP

The electric field for r>b is ke(2(λ+ρ(π(b2a2)))r) directed radially outward.

Explanation of Solution

Write the expression to calculate the volume of the enclosed charge.

    V=VbVa                                                                                                               (X)

Here, Vb is the volume of cylinder with radius b and Va is the volume of cylinder with radius a.

Substitute πb2L for Vb and πa2L for Va in equation (X) to solve for V.

    V=πb2Lπa2L=πL(b2a2)

Write the expression to calculate the electric field for the region r>b.

    EA=qenε0.                                                                                                                (XI)

Here, E is the electric field for the region r>b.

Substitute λL+ρV for qen, 2πrL for A and πL(b2a2) for V in equation (XI) to solve for E.

    E(2πrL)=λL+ρ(πL(b2a2))ε0E=λ+ρ(π(b2a2))2πrε0×22=14πε0(2(λ+ρ(π(b2a2)))r)                                                          (XII)

Substitute ke for 14πε0 in equation (XII).

    E=ke(2(λ+ρ(π(b2a2)))r)

Positive sign indicate the electric field is directed radially outward.

Conclusion:

Therefore, the electric field for r>b is ke(2(λ+ρ(π(b2a2)))r) directed radially outward.

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Chapter 24 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 24 - Prob. 9OQCh. 24 - Prob. 10OQCh. 24 - Prob. 11OQCh. 24 - Prob. 1CQCh. 24 - Prob. 2CQCh. 24 - Prob. 3CQCh. 24 - Prob. 4CQCh. 24 - Prob. 5CQCh. 24 - Prob. 6CQCh. 24 - Prob. 7CQCh. 24 - Prob. 8CQCh. 24 - Prob. 9CQCh. 24 - Prob. 10CQCh. 24 - Prob. 11CQCh. 24 - A flat surface of area 3.20 m2 is rotated in a...Ch. 24 - A vertical electric field of magnitude 2.00 104...Ch. 24 - Prob. 3PCh. 24 - Prob. 4PCh. 24 - Prob. 5PCh. 24 - A nonuniform electric field is given by the...Ch. 24 - An uncharged, nonconducting, hollow sphere of...Ch. 24 - Prob. 8PCh. 24 - Prob. 9PCh. 24 - Prob. 10PCh. 24 - Prob. 11PCh. 24 - A charge of 170 C is at the center of a cube of...Ch. 24 - Prob. 13PCh. 24 - A particle with charge of 12.0 C is placed at the...Ch. 24 - Prob. 15PCh. 24 - Prob. 16PCh. 24 - Prob. 17PCh. 24 - Find the net electric flux through (a) the closed...Ch. 24 - Prob. 19PCh. 24 - Prob. 20PCh. 24 - Prob. 21PCh. 24 - Prob. 22PCh. 24 - Prob. 23PCh. 24 - Prob. 24PCh. 24 - Prob. 25PCh. 24 - Determine the magnitude of the electric field at...Ch. 24 - A large, flat, horizontal sheet of charge has a...Ch. 24 - Prob. 28PCh. 24 - Prob. 29PCh. 24 - A nonconducting wall carries charge with a uniform...Ch. 24 - A uniformly charged, straight filament 7.00 m in...Ch. 24 - Prob. 32PCh. 24 - Consider a long, cylindrical charge distribution...Ch. 24 - A cylindrical shell of radius 7.00 cm and length...Ch. 24 - A solid sphere of radius 40.0 cm has a total...Ch. 24 - Prob. 36PCh. 24 - Prob. 37PCh. 24 - Why is the following situation impossible? A solid...Ch. 24 - A solid metallic sphere of radius a carries total...Ch. 24 - Prob. 40PCh. 24 - A very large, thin, flat plate of aluminum of area...Ch. 24 - Prob. 42PCh. 24 - Prob. 43PCh. 24 - Prob. 44PCh. 24 - A long, straight wire is surrounded by a hollow...Ch. 24 - Prob. 46PCh. 24 - Prob. 47PCh. 24 - Prob. 48APCh. 24 - Prob. 49APCh. 24 - Prob. 50APCh. 24 - Prob. 51APCh. 24 - Prob. 52APCh. 24 - Prob. 53APCh. 24 - Prob. 54APCh. 24 - Prob. 55APCh. 24 - Prob. 56APCh. 24 - Prob. 57APCh. 24 - An insulating solid sphere of radius a has a...Ch. 24 - Prob. 59APCh. 24 - Prob. 60APCh. 24 - Prob. 61CPCh. 24 - Prob. 62CPCh. 24 - Prob. 63CPCh. 24 - Prob. 64CPCh. 24 - Prob. 65CPCh. 24 - A solid insulating sphere of radius R has a...Ch. 24 - Prob. 67CPCh. 24 - Prob. 68CPCh. 24 - Prob. 69CP
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