Physics for Scientists and Engineers, Volume 1
Physics for Scientists and Engineers, Volume 1
9th Edition
ISBN: 9781133954156
Author: Raymond A. Serway
Publisher: CENGAGE L
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Chapter 23, Problem 23.41P

Example 23.3 derives the exact expression for the electric field at a point on the axis of a uniformly charged disk. Consider a disk of radius R = 3.00 cm having a uniformly distributed charge of +5.20 μC. (a) Using the result of Example 23.3, compute the electric field at a point on the axis and 3.00 mm from the center. (b) What If? Explain how the answer to part (a) compares with the field computed from the near-field approximation E = σ/2ϵ0. (We derived this expression in Example 23.3.) (c) Using the result of Example 23.3, compute the electric field at a point on the axis and 30.0 cm from the center of the disk. (d) What If? Explain how the answer to part (c) compares with the electric field obtained by treating the disk as a +5.20-μC charged particle at a distance of 30.0 cm.

(a)

Expert Solution
Check Mark
To determine

The electric field at a point on the axis at a distance 3.00mm from the center of the disk.

Answer to Problem 23.41P

The electric field at a point on the axis and 3.00mm from the center is 9.36×107N/C away from the center of the disk.

Explanation of Solution

 The value of the charge is +5.20μC, radius of the disk is 3.00cm, distance from center of the disk is 3.00mm.

The formula to calculate the surface charge density is,

    σ=qA                                                                                        (1)

The formula to calculate the area is,

    A=πR2                                                                                     (2)

Substitute πR2 for A from equation (2) in equation (1).

    σ=qπR2                                                                                 (3)

Substitute +5.20μC for the value of q, 3.00cm for R in equation (3) to calculate σ.

    =5.20μC×106C1μC3.14×(3.00cm×10-2m1cm)2=0.184005662×102C/m2

The formula to calculate the electric field at a point from the center of a uniformly charged disk is,

    E=σ2ε(1xR2+x2)                                                                (4)

Here, E is the electric field at that point, σ is the surface charge density, εo is the permittivity of the free space, x is the distance of the point from the center of the disk and R is the radius of the disk.

Conclusion:

Substitute 0.184005662×102C/m2 for σ, 8.85×1012C2/Nm2 for εo, 3.00cm for R , 3.00mm for x in equation (4).

    E=0.184005662×102C/m22×8.85×1012C2/Nm2×(13mm×103m1mm((3mm×103m1mm)2+(3cm×102m1cm)2))=9.36×107N/C

Therefore, the electric field at a point on the axis and 3.00mm from the center is 9.36×107N/C away from the center of the disk.

(b)

Expert Solution
Check Mark
To determine

The change in electric field when calculated using near field approximation.

Answer to Problem 23.41P

The magnitude of electric field when computed from near field approximation is 1.04×108N/C and this value is 11% higher.

Explanation of Solution

The formula to calculate the electric field at a point from the center of a uniformly charged disk is,

    E=σ2ε                                                                                                   (5)

Here, E is the electric field at that point, σ is the surface charge density and εo is the permittivity of the free space.

Conclusion:

Substitute 0.184005662×102C/m2 for σ, 8.85×1012C2/Nm2 for εo, in equation (5) as,

    E=0.184005662×102C/m22×8.85×1012C2/N.m2=1.04×108N/C

The value of electric field for uniformly charged disk for a point 3mm from the center is 9.36×107N/C.

The formula to calculate the percentage change with respect to the field computed from near field approximation is,

    % change=(1.04×1089.36×107)9.36×107×100%11% higher

Therefore, the magnitude of electric field when computed from near field approximation is 1.04×108N/C and this value is 11% higher.

(c)

Expert Solution
Check Mark
To determine

The electric field at a point on the axis at a distance 30.0cm from the center of the disk.

Answer to Problem 23.41P

The electric field at a point on the axis and 30cm from the center is 5.16×105N/C.

Explanation of Solution

 The value of the charge is +5.2μC, radius of the disk is 3.00cm, distance from center of the disk is 30cm.

The formula to calculate the surface charge density is,

    σ=qA                                                                                        (6)

The formula to calculate the area is,

    A=π×R2                                                                                  (7)

Substitute A=π×R2 from equation (2) in equation (1).

    σ=qπ×R2                                                                                 (8)

Substitute +5.20μC for the value of q, 3.00cm for R in equation (8) to calculate σ.

    =5.20μC×106C1μC3.14×(3.00cm×10-2m1cm)2=0.184005662×102C/m2

The formula to calculate the electric field at a point from the center of a uniformly charged disk is,

    E=σ2ε(1xR2+x2)                                                                 (9)

Here, E is the electric field at that point, σ is the surface charge density, εo is the permittivity of the free space, x is the distance of the point from the center of the disk and R is the radius of the disk.

Conclusion:

Substitute 0.184005662×102C/m2 for σ, 8.85×1012C2/Nm2 for εo, 3.00cm for R , 30cm for x in equation (9).

    E=0.184005662×102C/m22×8.85×1012C2/Nm2×(130cm×102m1cm((30cm×102m1cm)2+(3cm×102m1cm)2))=5.16×105N/C

Therefore, the electric field at a point on the axis and 30cm from the center is 5.16×105N/C.

(d)

Expert Solution
Check Mark
To determine

The change in electric field at 30cm away from the center of the disc when the disc is considered to be a charge of +5.20μC.

Answer to Problem 23.41P

The magnitude of electric field obtained by treating the disk as a +5.20μC charged particle at a distance of 30cm is 5.19×105N/C and this value is 0.7% higher.

Explanation of Solution

The formula to calculate the electric field of a charged particle is,

    E=keQR2

Here, ke is the electrostatic constant, Q is the electric charge and R is the radius of the hollow sphere.

Substitute 8.99×109Nm2/C for ke, 5.20μC for the value of Q, 30.0cm for R, in the above formula.

    E=keQR2=((8.99×109Nm2/C)(5.20μC×106C1μC)30cm×(102m1cm))=5.19×105N/C

the percentage change when part (c) is compared with the field obtained by treating the disk as a +5.20μC charged particle.

The value of electric field for uniformly charged disk at a point 30cm from the center is 5.16×105N/C.

The value of the electric field while approximating the disc to be a point charge is 5.19×105N/C.

Conclusion:

The formula to calculate the percentage change when part (c) is compared with the field obtained by treating the disk as a +5.20μC charged particle is,

    % change=(5.195.16)×1055.16×105×100=0.03×1055.16×105×100=0.6% higher

Therefore, the magnitude of electric field obtained by treating the disk as a point charge at a distance of 30cm is 5.19×105N/C and this value is 0.6% higher.

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Chapter 23 Solutions

Physics for Scientists and Engineers, Volume 1

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