Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 24, Problem 42P

(a)

To determine

The volume density of electric charge at x=0.300m.

(a)

Expert Solution
Check Mark

Answer to Problem 42P

Volume density of electric charge at x=0.300m is 31.9C/m3.

Explanation of Solution

Calculation:

Write the expression for electric flux.

    ϕ=EA                                                                                                                       (I)

Here, ϕ is the electric flux, A is the surface area and E is the electric field.

As given, consider a box which extends from x=0.300m to x=0.300m+dx thickness. Then calculate the electric field at x=0.300m

Write the expression for net flux through the box.

    ϕnet=ϕin+ϕout                                                                                                           (II)

Here, ϕnet is the net flux.

Write the expression for the net flux from Gauss law.

    ϕnet=ρAdxε0

Here, ρ is the volume charge density, ε0 is permittivity of free space.

Rewrite the above equation for ρ.

    ϕnet=ρAdxε0ρ=ε0ϕnetAdx                                                                                                            (III)

Conclusion:

Substitute 0.300m for x and (6.00×103)x2Nm2/C for E in Equation (I) to calculate ϕin.

    ϕin=((6.00×103)x2Nm2/C)(0.3m)2A=(540N/C)A

Substitute (0.300m+dx) for x and (6.00×103)x2Nm2/C for E in Equation (I) to calculate ϕout.

    ϕout=((6.00×103)Nm2/C)(0.3m+dx)2A=((6.00×103)Nm2/C)((0.3m)2+(0.6m)dx+(dx)2)A=((6.00×103)Nm2/C)((0.3m)2+(0.6m)dx)A=(540N/C)A+(3600Nm/C)dxA

Substitute 540N/CA for ϕin and [(540N/C)A+(3600Nm/C)dxA] for ϕout in Equation (II) to calculate ϕnet.

    ϕnet=540N/CA+[(540N/C)A+(3600Nm/C)dxA]=(3600Nm/C)dxA

Substitute 8.85×1012C2/Nm2 for ε0, (3600Nm/C)dx×A for ϕnet in Equation (III) to calculate ρ.

    ρ=(8.85×1012C2/Nm2)((3600Nm/C)dxA)Adx=(3600Nm/C)(8.85×1012C2/Nm2)=31.9C/m3

Therefore volume charge density at x=0.300m is 31.9C/m3.

(b)

To determine

Whether the region of space could be inside the conductor.

(b)

Expert Solution
Check Mark

Answer to Problem 42P

The region of space is not inside the conductor.

Explanation of Solution

The electric field inside a conductor should be zero but the region from x=0.300m to x=0.300m+dx contains electric field of (6×103)x2i^Nm2/C. Hence electric field can’t be zero for the value of x=0.300m and all the above values as this region can’t be inside the conductor.

Conclusion:

Therefore, this region of space is not inside the conductor.

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Chapter 24 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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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