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 79PQ

(a)

To determine

The expression for the electric field at point A located at a distance l above the mid-point of the rod.

(a)

Expert Solution
Check Mark

Answer to Problem 79PQ

The expression for the electric field at point A located at a distance l above the mid-point of the rod is Etot=(0.628)kQl2j^.

Explanation of Solution

Sketch the diagram showing the five charges.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 24, Problem 79PQ

The x component of the electric field is zero based on the geometry.

Write the expression for the y component of the electric field.

    E=kqr2r^=kqr2sinθj^                                                                                                  (I)

Here, E is the electric field, k is the coulomb constant, q is the charge and r is the distance between the charge and the point.

Write the equation for the total electric field.

    Etot=E1+E2+E3+E4+E5                                                                       (II)

Conclusion:

Substitute Q5 for q, l2+l2 for r and ll2+l2 for sinθ in equation (I) to find E1.

    E1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^                                                                         (III)

Substitute Q5 for q, l2+l2 for r and ll2+l2 for sinθ in equation (I) to find E2.

    E2=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^                                                                (IV)

Substitute Q5 for q and l for r in equation (I) to find E3.

    E3=k(Q5)l2j^=kQ5l2j^                                                                                               (V)

Substitute Q5 for q, l2+l2 for r and ll2+l2 for sinθ in equation (I) to find E4.

    E4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^                                                                (VI)

Substitute Q5 for q, l2+l2 for r and ll2+l2 for sinθ in equation (I) to find E5.

    E5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^                                                                         (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find Etot.

    Etot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^

Thus, the expression for the electric field at point A located at a distance l above the mid-point of the rod is Etot=(0.628)kQl2j^.

(b)

To determine

The electric field at point A located at a distance l above the mid-point of the rod using the exact expression.

(b)

Expert Solution
Check Mark

Answer to Problem 79PQ

The electric field at point A located at a distance l above the mid-point of the rod using the exact expression is Etot=kQ2l2j^.

Explanation of Solution

Write the exact expression for the total electric field.

    E=kqy1(l2+y2)j^                                                                                    (VIII)

Here, E is the electric field, k is the coulomb constant, q is the charge, y is the perpendicular distance between the rod and the point and l is the length of the rod.

Conclusion:

Substitute Q for q, and l for y in equation (VIII) to find Etot.

    Etot=kQl1(l2+l2)j^=kQ2l2j^

Thus, the electric field at point A located at a distance l above the mid-point of the rod using the exact expression is Etot=kQ2l2j^.

(c)

To determine

Compare the approximate result with the exact result.

(c)

Expert Solution
Check Mark

Answer to Problem 79PQ

The approximate result is 0.888 times the exact result.

Explanation of Solution

Find the ratio of the approximate result with the exact result.

    Etot,aEtot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888

Conclusion:

Thus, the approximate result is 0.888 times the exact result.

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Chapter 24 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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