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 23, Problem 72AP

(a)

To determine

The magnitude of the electric force exerted on the charge at q1 (0,0).

(a)

Expert Solution
Check Mark

Answer to Problem 72AP

The magnitude of the electric force exerted on the charge at q1 is 40.85N.

Explanation of Solution

Write the expression for the total electric field.

    E=Ex+Ey                                                                                                                (I)

Here, E is the total electric field, Ex is the electric field resolved in x axis, Ey is the electric field resolved in y axis.

Write the expression for the electric field.

    E=kqr2                                                                                                                      (II)

Here, k is the constant, q is the electric charge, r is the distance from q1.

Rewrite the equation (II) for resolving the forces along x direction.

    Ex=[kq2r22cosθ2+kq3r32cosθ3+kq4r42cosθ4]i^                                                           (III)

Here, q2 is the charge at B, θ2 is the angle made by the q2 with q1, r2 is the distance of q2 form q1, q3 is the charge at C, θ3 is the angle made by the q3 with q1, r3 is the distance of q3 form q1, q4 is the charge at D, θ4 is the angle made by the q4 with q1, r4 is the distance of q4 form q1.

Rewrite the equation (II) for resolving the forces along y direction.

    Ey=[kq2r22sinθ2+kq3r32sinθ3+kq4r42sinθ4]j^                                                            (IV)

Write the expression for the electric force.

    F=qE                                                                                                                     (V)

Conclusion:

The following diagram shows the distance of the charges from (0,0).

Physics for Scientists and Engineers with Modern Physics, Technology Update, Chapter 23, Problem 72AP

Figure-(1)

From the figure-(1) calculate the distance of q3 from the position (0,0).

Consider the triangle ΔCAD.

    tanθ3=15.0cm60.0cmθ3=tan1(0.25)θ3=14.036°

Consider the triangle ΔCAD.

    sinθ3=15.0cmr3r3=15.0cmsin(14.036°)r3=61.847cm

Substitute 9×109Nm2/C2 for k, +10.0μC for q2, 15.0cm for r2, 90° for θ2, +10.0μC for q3, 61.847cm for r3, 14.036° for θ3, +10.0μC for q4, 60.0cm r4, 0° for θ4,in the equation (III) to find Ex.

    Ex=[9×109Nm2/C2(+10.0×106C)(1(15.0cm)2cos90°+1(61.847cm)2cos14.036°+1(60.0cm)2cos0°)]i^=[90000Nm2(1(0.15m)2cos90°+1(0.618cm)2cos14.036°+1(0.60cm)2cos0°)]i^=[90000Nm2(01/m2+2.5401/m2+2.7781/m2)]i^

Calculate further.

    Ex=(9×104Nm2×5.3181/m2)i^=478.62×103i^N

Substitute 9×109Nm2/C2 for k, +10.0μC for q2, 15.0cm for r2, 90° for θ2, +10.0μC for q3, 61.847cm for r3, 14.036° for θ3, +10.0μC for q4, 60.0cm r4, 0° for θ4 in equation (III) to find Ey.

    Ey=[9×109Nm2/C2(+10.0×106C)[1(15.0cm)2sin90°+1(61.847cm)2sin14.036°+1(60.0cm)2sin0°]]j^=[90000Nm2[1(0.15m)2sin90°+1(0.618cm)2sin14.036°+1(0.60cm)2sin0°]]j^=(90000Nm2[44.4441/m2+0.6351/m2+01/m2])j^

Calculate further.

    Ey=(9×104Nm2×45.0791/m2)j^=4057.11×103j^N

Substitute 478.62×103i^N for Ex, 4057.11×103j^N for Ey in the equation (I) to find E.

    E=478.62×103i^N+4057.11×103j^N=(478.62i^+4057.11j^)×103N

Substitute (478.62i^+4057.11j^)×103N for E and +10.0×106C for q in the equation (V) to find F.

    F=[+10.0×106C][(478.62i^+4057.11j^)×103N]=4.786i^+40.57j^NC

Calculate the magnitude of the electric force.

    F=((4.786i^)2+(40.57j^)2)N=22.906+1645.93N=40.85N

Therefore, the magnitude of the electric force exerted on the charge is 40.85N.

(b)

To determine

The direction of the electric force exerted on the charge at q1 (0,0).

(b)

Expert Solution
Check Mark

Answer to Problem 72AP

The direction of the electric force exerted on the charge at q1 is 6.728°.

Explanation of Solution

Write the expression for the direction of the electric force.

    tanα=AB                                                                                                               (VI)

Here, α is the direction of the field, A is the coefficient of the electric force in x axis, B is the coefficient of the electric force in y axis.

Conclusion:

Substitute 4.786 for A, 40.57 for B in the equation (VI) to find α.

    tanα=4.78640.57α=tan1(4.78640.57)α=tan1(0.118)α=6.728°

Therefore, the direction of the electric force exerted on the charge is 6.728°.

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Chapter 23 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

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