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

Question

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Answer 1

Question

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

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 $E→tot=(0.628)kQl2j^$ .

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.

$E→tot=E→1+E→2+E→3+E→4+E→5$ (II)

Conclusion:

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→1$ .

$E→1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (III)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→2$ .

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

Substitute $Q5$ for $q$ and $l$ for $r$ in equation (I) to find $E→3$ .

$E→3=k(Q5)l2j^=kQ5l2j^$ (V)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→4$ .

$E→4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^$ (VI)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→5$ .

$E→5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find $E→tot$ .

$E→tot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^$

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

(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

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 $E→tot=kQ2l2j^$ .

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 $E→tot$ .

$E→tot=kQl1(l2+l2)j^=kQ2l2j^$

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

(c)

To determine

Compare the approximate result with the exact result.

(c)

Expert Solution

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.

$E→tot,aE→tot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888$

Conclusion:

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

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Answer 2

Question

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

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 $E→tot=(0.628)kQl2j^$ .

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.

$E→tot=E→1+E→2+E→3+E→4+E→5$ (II)

Conclusion:

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→1$ .

$E→1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (III)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→2$ .

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

Substitute $Q5$ for $q$ and $l$ for $r$ in equation (I) to find $E→3$ .

$E→3=k(Q5)l2j^=kQ5l2j^$ (V)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→4$ .

$E→4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^$ (VI)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→5$ .

$E→5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find $E→tot$ .

$E→tot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^$

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

(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

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 $E→tot=kQ2l2j^$ .

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 $E→tot$ .

$E→tot=kQl1(l2+l2)j^=kQ2l2j^$

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

(c)

To determine

Compare the approximate result with the exact result.

(c)

Expert Solution

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.

$E→tot,aE→tot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888$

Conclusion:

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

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Answer 3

Question

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

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 $E→tot=(0.628)kQl2j^$ .

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.

$E→tot=E→1+E→2+E→3+E→4+E→5$ (II)

Conclusion:

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→1$ .

$E→1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (III)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→2$ .

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

Substitute $Q5$ for $q$ and $l$ for $r$ in equation (I) to find $E→3$ .

$E→3=k(Q5)l2j^=kQ5l2j^$ (V)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→4$ .

$E→4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^$ (VI)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→5$ .

$E→5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find $E→tot$ .

$E→tot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^$

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

(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

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 $E→tot=kQ2l2j^$ .

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 $E→tot$ .

$E→tot=kQl1(l2+l2)j^=kQ2l2j^$

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

(c)

To determine

Compare the approximate result with the exact result.

(c)

Expert Solution

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.

$E→tot,aE→tot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888$

Conclusion:

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

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

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

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 $E→tot=(0.628)kQl2j^$ .

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.

$E→tot=E→1+E→2+E→3+E→4+E→5$ (II)

Conclusion:

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→1$ .

$E→1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (III)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→2$ .

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

Substitute $Q5$ for $q$ and $l$ for $r$ in equation (I) to find $E→3$ .

$E→3=k(Q5)l2j^=kQ5l2j^$ (V)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→4$ .

$E→4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^$ (VI)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→5$ .

$E→5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find $E→tot$ .

$E→tot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^$

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

(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

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 $E→tot=kQ2l2j^$ .

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 $E→tot$ .

$E→tot=kQl1(l2+l2)j^=kQ2l2j^$

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

(c)

To determine

Compare the approximate result with the exact result.

(c)

Expert Solution

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.

$E→tot,aE→tot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888$

Conclusion:

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

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Answer 5

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

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 $E→tot=(0.628)kQl2j^$ .

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.

$E→tot=E→1+E→2+E→3+E→4+E→5$ (II)

Conclusion:

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→1$ .

$E→1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (III)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→2$ .

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

Substitute $Q5$ for $q$ and $l$ for $r$ in equation (I) to find $E→3$ .

$E→3=k(Q5)l2j^=kQ5l2j^$ (V)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→4$ .

$E→4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^$ (VI)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→5$ .

$E→5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find $E→tot$ .

$E→tot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^$

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

(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

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 $E→tot=kQ2l2j^$ .

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 $E→tot$ .

$E→tot=kQl1(l2+l2)j^=kQ2l2j^$

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

(c)

To determine

Compare the approximate result with the exact result.

(c)

Expert Solution

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.

$E→tot,aE→tot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888$

Conclusion:

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

Answer 6

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

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 $E→tot=(0.628)kQl2j^$ .

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.

$E→tot=E→1+E→2+E→3+E→4+E→5$ (II)

Conclusion:

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→1$ .

$E→1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (III)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→2$ .

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

Substitute $Q5$ for $q$ and $l$ for $r$ in equation (I) to find $E→3$ .

$E→3=k(Q5)l2j^=kQ5l2j^$ (V)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→4$ .

$E→4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^$ (VI)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→5$ .

$E→5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find $E→tot$ .

$E→tot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^$

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

Answer 7

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.

Answer 8

(a)

Expert Solution

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 $E→tot=(0.628)kQl2j^$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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.

$E→tot=E→1+E→2+E→3+E→4+E→5$ (II)

Conclusion:

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→1$ .

$E→1=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (III)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→2$ .

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

Substitute $Q5$ for $q$ and $l$ for $r$ in equation (I) to find $E→3$ .

$E→3=k(Q5)l2j^=kQ5l2j^$ (V)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→4$ .

$E→4=k(Q5)(l2+(l2)2)2ll2+(l2)2j^=k(Q5)(54)l225j^=8kQ255l2j^$ (VI)

Substitute $Q5$ for $q$ , $l2+l2$ for $r$ and $ll2+l2$ for $sinθ$ in equation (I) to find $E→5$ .

$E→5=k(Q5)(l2+l2)2ll2+l2j^=kQ102l2j^$ (VII)

Substitute equations (III), (IV), (V), (VI) and (VII) in equation (II) to find $E→tot$ .

$E→tot=kQ102l2j^+8kQ255l2j^+kQ5l2j^+8kQ255l2j^+kQ102l2j^=2kQ10l2j^+16kQ255l2j^+kQ5l2=kQl2(210+16255+15)=(0.628)kQl2j^$

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

Answer 13

(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

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 $E→tot=kQ2l2j^$ .

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 $E→tot$ .

$E→tot=kQl1(l2+l2)j^=kQ2l2j^$

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

Answer 14

(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.

Answer 15

(b)

Expert Solution

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 $E→tot=kQ2l2j^$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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 $E→tot$ .

$E→tot=kQl1(l2+l2)j^=kQ2l2j^$

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

Answer 20

(c)

To determine

Compare the approximate result with the exact result.

(c)

Expert Solution

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.

$E→tot,aE→tot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888$

Conclusion:

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

Answer 21

(c)

To determine

Compare the approximate result with the exact result.

Answer 22

(c)

Expert Solution

Answer to Problem 79PQ

The approximate result is $0.888$ times the exact result.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

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

$E→tot,aE→tot,b=(0.628)kQl2j^kQ2l2j^=(0.628)(2)=0.888$

Conclusion:

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

Answer 27

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

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

Answer to Problem 79PQ

Explanation of Solution

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

Answer to Problem 79PQ

Explanation of Solution

Compare the approximate result with the exact result.

Answer to Problem 79PQ

Explanation of Solution

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