Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 19, Problem 65P

(a)

To determine

The magnitude of total electric force exerted on the charge at the lower left corner due to the other three charges.

(a)

Expert Solution
Check Mark

Answer to Problem 65P

The magnitude of total electric force exerted on the charge at the lower left corner due to the other three charges is 40.8N_.

Explanation of Solution

Figure 1 represents four charges of equal magnitude of 10.0μC.

Principles of Physics: A Calculus-Based Text, Chapter 19, Problem 65P

Write the expression for angle θ from Figure 1,

    θ=tan1(rABrAD)        (I)

Here, θ is the angle, rAB is the distance between the charges A and B, and rAD is the distance between the charges A and D.

Write the expression for magnitude of electric force due to charge 1.

    F1=keq1q4r142        (II)

Here, F1 is the electric force, ke is the coulomb’s constant, q1, q4 are the charges, and r14 is the distance between the charges q1 and q4.

Write the expression for magnitude of electric force due to charge 2.

    F2=keq2q4r242        (III)

Here, F2 is the electric force, ke is the coulomb’s constant, q2, q4 are the charges, and r24 is the distance between the charges q2, and q4.

Write the expression for magnitude of electric force due to charge 1.

    F3=keq3q4r342        (IV)

Here, F3 is the electric force, ke is the coulomb’s constant, q2, q4 are the charges, and r34 is the distance between the charges q2, and q4.

Write the expression for horizontal component of the resultant force.

    Fx=F3F2cosθ        (V)

Write the expression for vertical component of the resultant force.

    Fy=F1F2sinθ        (VI)

Write the expression for the resultant force.

    Fnet=Fx2+Fy2        (VII)

Conclusion:

Substitute 15.0m for rAB, and 60.0m for rAD in equation (I), to find θ.

  θ=tan1(15.0m60.0m)=14.0°

Substitute 8.99×109N.m2/C2 for ke, 10.0μC for q1, 10.0μC for q4, and 0.150m for rAB, in equation (II) to find F1.

    F1=(8.99×109N.m2/C2)(10.0μC×1C106μC)(10.0μC×1C106μC)(0.150m)2=40.0N

Substitute 8.99×109N.m2/C2 for ke, 10.0μC for q2, 10.0μC for q4, and 0.618m for rAB, in equation (III) to find F2.

    F2=(8.99×109N.m2/C2)(10.0μC×1C106μC)(10.0μC×1C106μC)(0.618m)2=2.35N

Substitute 8.99×109N.m2/C2 for ke, 10.0μC for q3, 10.0μC for q4, and 0.600m for rAB, in equation (IV) to find F3.

    F3=(8.99×109N.m2/C2)(10.0μC×1C106μC)(10.0μC×1C106μC)(0.600m)2=2.50N

Substitute 2.50N for F3, 2.35N for F2, and 14.0° for θ in equation (V), to find Fx.

    Fx=2.50N2.35Ncos14.0°=4.78N

Substitute 40.0N for F1, 2.35N for F2, and 14.0° for θ in equation (VI), to find Fy.

    Fy=40.0N2.35Nsin14.0°=40.5N

Substitute 4.78N for Fx, and 40.5N for Fy in equation (VII), to find Fnet.

    Fnet=(4.78N)2+(40.5N)2=40.8N

Therefore, the magnitude of total electric force exerted on the charge at the lower left corner due to the other three charges is 40.8N_.

(b)

To determine

The direction of total electric force exerted on the charge at the lower left corner due to the other three charges.

(b)

Expert Solution
Check Mark

Answer to Problem 65P

The direction of total electric force exerted on the charge at the lower left corner due to the other three charges is 263°_.

Explanation of Solution

Write the expression for angle between the vertical component and the horizontal component of the resultant electric force from Figure 1.

    φ=tan1(FyFx)        (VIII)

Here, φ is the angle between the vertical component and the horizontal component of the resultant electric force.

Conclusion:

Substitute 4.78N for Fx, and 40.5N for Fy in equation (VIII), to find φ.

    φ=tan1(40.5N4.78N)=263°

Therefore, the direction of total electric force exerted on the charge at the lower left corner due to the other three charges is 263°_.

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Chapter 19 Solutions

Principles of Physics: A Calculus-Based Text

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