EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Chapter 23, Problem 40PQ

Three charged metal spheres are arrayed in the xy plane so that they form an equilateral triangle (Fig. P23.40). What is the net electrostatic force on the sphere at the origin?

Chapter 23, Problem 40PQ, Three charged metal spheres are arrayed in the xy plane so that they form an equilateral triangle

Figure P23.40

Expert Solution & Answer
Check Mark
To determine

The net electrostatic force on the sphere at the origin.

Answer to Problem 40PQ

The net electrostatic force on the sphere at the origin is F=(6.01×106N)i^(7.43×106N)j^.

Explanation of Solution

The diagram for the forces on charge on the origin.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 23, Problem 40PQ

Write the expression for Coulomb’s law.

  FE=kq1q2r2r^                                                                                             (I)

Here, FE is the electrostatic force between the spheres, k is the Coulomb’s constant, q1 is the charge of sphere 1, q2 is the charge of sphere 2, r is the distance between the centers of the spheres.

Substitute 8.99×109Nm2/C2 for k, 5.50nC for q1, 3.00nC for q2 and 12.0cm for r to find the magnitude of the force on charge 5.50nC due to 3.00nC.

  F1=(8.99×109Nm2/C2)|(5.50nC)(3.00nC)|(12.0cm)2=(8.99×109Nm2/C2)|(5.50nC109C1.0nC)(3.00nC109C1.0nC)|(12.0cm102m1.0cm)2=1.03×105N

The magnitude of the force on charge 5.50nC due to 3.00nC is in the +x direction.

Substitute 8.99×109Nm2/C2 for k, 5.50nC for q1, 2.50nC for q2 and 12.0cm for r to find the magnitude of the force on charge 5.50nC due to 2.50nC.

  F2=(8.99×109Nm2/C2)|(5.50nC)(2.50nC)|(12.0cm)2=(8.99×109Nm2/C2)|(5.50nC109C1.0nC)(2.50nC109C1.0nC)|(12.0cm102m1.0cm)2=8.58×106N

The magnitude of the force on charge 5.50nC due to 2.50nC is at an angle 60° below to the x direction.

Write the expression to find the net electrostatic force on sphere at origin.

  F=Fxi^+Fyj^                                                                                              (II)

Here, Fx is the x-component of net electrostatic force on sphere at origin, Fy is the y-component of net electrostatic force on sphere at origin, F is the net electrostatic force on sphere at origin.

Write the expression to find the x-component of net electrostatic force on sphere at origin.

  Fx=F2cosθ+F1

Substitute 1.03×105N for F1, 8.58×106N for F2 and 60° for θ to find the x-component of net electrostatic force on sphere at origin.

  Fx=(8.58×106N)cos60°+(1.03×105N)=6.01×106N

Write the expression to find the y-component of net electrostatic force on sphere at origin.

  Fy=F2sinθ

Substitute 8.58×106N for F2 and 60° for θ to find the y-component of net electrostatic force on sphere at origin.

  Fy=(8.58×106N)cos60°=7.43×106N

Conclusion:

Substitute 6.01×106N for Fx and 7.43×106N for Fy in equation (II) to find the net electrostatic force on sphere at origin.

  F=(6.01×106N)i^+(7.43×106N)j^=(6.01×106N)i^(7.43×106N)j^

Thus, the net electrostatic force on the sphere at the origin is F=(6.01×106N)i^(7.43×106N)j^.

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Chapter 23 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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