Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d = 25.0 cm. a. What is the electric field at the location of the sphere with charge +2.00 nC? b. What is the total electric force exerted on the sphere with charge +2.00 nC by the other three spheres? FIGURE P24.51

Question

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Answer 1

Textbook Question

Chapter 24, Problem 51PQ

Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d = 25.0 cm.

a. What is the electric field at the location of the sphere with charge +2.00 nC?

b. What is the total electric force exerted on the sphere with charge +2.00 nC by the other three spheres?

Chapter 24, Problem 51PQ, Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d =

FIGURE P24.51

(a)

Expert Solution

To determine

The electric field at the location of the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 24, Problem 51PQ

Write the equation for the electric field of a charged particle.

E→=kqr2r^ (I)

Here, E→E is the electric field of a charged particle, k is the Coulomb’s constant, r is the distance and r^ is the surface unit vector in the direction for r.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge +2.00 nC.

E→=kq1r12r^1+kq3r32r^3+kq4r42r^4 (II)

Here, q1 is the is the first charge, q4 is the is the third charge, q3 is the is the third charge, r1 is the distance of the first charge to the +2.00 nC charge, r3 is the distance of the third charge to the +2.00 nC charge, r4 is the distance of the fourth charge to the +2.00 nC charge, r^1 is the unit vector in the direction of r1, r^3 is the unit vector in the direction of r3 and r^4 is the unit vector in the direction of r4.

Conclusion:

Substitute 8.99×109 N⋅m2/C2 for k, 1.00 nC for q1, 4.00 nC for q3 and 3.00 nC for q4, d for r1, 2d for r3, d for r4, i^ for r^1, (cos45.0i^+sin45.0j^) for r^3 and j^ for r^4 in equation (II) to find E→.

E→=−k[1.00 nCd2i^−4.00 nC(2d)2(cos45.0i^+sin45.0j^)−3.00 nCd2j^]=−k[(1.00 nC)(10−9 C1 nC)d2i^−(4.00 nC)(10−9 C1 nC)(2d)2(cos45.0i^+sin45.0j^)−(3.00 nC)(10−9 C1 nC)d2j^]=8.99×109 N⋅m2/C2((25.0 cm)(10−2 m1 cm))2(10−9 C)[(−1−2cos45.0°)i^+j^]=(−347i^−635j^) N/C

Thus, the electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

(b)

Expert Solution

To determine

The total electric forceon the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N

Explanation of Solution

Write the equation for the electric force.

F→=qE→ (III)

Here, F→ is the electric force.

Conclusion:

Substitute 2.00 nC for q and (−347i^−635j^) N/C for E→ in equation (III) to find F→.

F→=(2.00 nC)((−347i^−635j^) N/C)=((2.00 nC)(10−9 C1 nC))((−347i^−635j^) N/C)=(−6.94×10−7i^−1.27×10−6j^) N

Thus, the total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N.

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Students have asked these similar questions

A positively charged rod of length L = 0.250 m with linear charge density λ = 2.33 mC/m lies along the x axis (Fig. P24.23). Find the electric field at the position P a distance 0.375 m away from the origin.

A uniformly charged disk with radius R = 40.0 cm and uniform charge density σ = 8.20 ✕ 10−3 C/m2 lies in the xy-plane, with its center at the origin. What is the electric field (in MN/C) due to the charged disk at the following locations? (a) z = 5.00 cm MN/C (b) z = 10.0 cm MN/C (c) z = 50.0 cm MN/C (d) z = 200 cm MN/C

Two infinite, nonconducting sheets of charge are parallel to each other as shown in Figure P24.56. The sheet on the left has a uniform sur- face charge density σ, and the one on the right has a uniform charge density -σ. Calculate the electric field at points (a) to the left of, (b) in between, and (c) to the right of the Figure P24.56 two sheets. (d) What If? Find the electric fields in all three regions if both sheets have positive uniform surface charge densities of value σ.

Answer 2

Textbook Question

Chapter 24, Problem 51PQ

Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d = 25.0 cm.

a. What is the electric field at the location of the sphere with charge +2.00 nC?

b. What is the total electric force exerted on the sphere with charge +2.00 nC by the other three spheres?

Chapter 24, Problem 51PQ, Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d =

FIGURE P24.51

(a)

Expert Solution

To determine

The electric field at the location of the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 24, Problem 51PQ

Write the equation for the electric field of a charged particle.

E→=kqr2r^ (I)

Here, E→E is the electric field of a charged particle, k is the Coulomb’s constant, r is the distance and r^ is the surface unit vector in the direction for r.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge +2.00 nC.

E→=kq1r12r^1+kq3r32r^3+kq4r42r^4 (II)

Here, q1 is the is the first charge, q4 is the is the third charge, q3 is the is the third charge, r1 is the distance of the first charge to the +2.00 nC charge, r3 is the distance of the third charge to the +2.00 nC charge, r4 is the distance of the fourth charge to the +2.00 nC charge, r^1 is the unit vector in the direction of r1, r^3 is the unit vector in the direction of r3 and r^4 is the unit vector in the direction of r4.

Conclusion:

Substitute 8.99×109 N⋅m2/C2 for k, 1.00 nC for q1, 4.00 nC for q3 and 3.00 nC for q4, d for r1, 2d for r3, d for r4, i^ for r^1, (cos45.0i^+sin45.0j^) for r^3 and j^ for r^4 in equation (II) to find E→.

E→=−k[1.00 nCd2i^−4.00 nC(2d)2(cos45.0i^+sin45.0j^)−3.00 nCd2j^]=−k[(1.00 nC)(10−9 C1 nC)d2i^−(4.00 nC)(10−9 C1 nC)(2d)2(cos45.0i^+sin45.0j^)−(3.00 nC)(10−9 C1 nC)d2j^]=8.99×109 N⋅m2/C2((25.0 cm)(10−2 m1 cm))2(10−9 C)[(−1−2cos45.0°)i^+j^]=(−347i^−635j^) N/C

Thus, the electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

(b)

Expert Solution

To determine

The total electric forceon the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N

Explanation of Solution

Write the equation for the electric force.

F→=qE→ (III)

Here, F→ is the electric force.

Conclusion:

Substitute 2.00 nC for q and (−347i^−635j^) N/C for E→ in equation (III) to find F→.

F→=(2.00 nC)((−347i^−635j^) N/C)=((2.00 nC)(10−9 C1 nC))((−347i^−635j^) N/C)=(−6.94×10−7i^−1.27×10−6j^) N

Thus, the total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N.

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Answer 3

Textbook Question

Chapter 24, Problem 51PQ

Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d = 25.0 cm.

a. What is the electric field at the location of the sphere with charge +2.00 nC?

b. What is the total electric force exerted on the sphere with charge +2.00 nC by the other three spheres?

Chapter 24, Problem 51PQ, Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d =

FIGURE P24.51

(a)

Expert Solution

To determine

The electric field at the location of the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 24, Problem 51PQ

Write the equation for the electric field of a charged particle.

E→=kqr2r^ (I)

Here, E→E is the electric field of a charged particle, k is the Coulomb’s constant, r is the distance and r^ is the surface unit vector in the direction for r.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge +2.00 nC.

E→=kq1r12r^1+kq3r32r^3+kq4r42r^4 (II)

Here, q1 is the is the first charge, q4 is the is the third charge, q3 is the is the third charge, r1 is the distance of the first charge to the +2.00 nC charge, r3 is the distance of the third charge to the +2.00 nC charge, r4 is the distance of the fourth charge to the +2.00 nC charge, r^1 is the unit vector in the direction of r1, r^3 is the unit vector in the direction of r3 and r^4 is the unit vector in the direction of r4.

Conclusion:

Substitute 8.99×109 N⋅m2/C2 for k, 1.00 nC for q1, 4.00 nC for q3 and 3.00 nC for q4, d for r1, 2d for r3, d for r4, i^ for r^1, (cos45.0i^+sin45.0j^) for r^3 and j^ for r^4 in equation (II) to find E→.

E→=−k[1.00 nCd2i^−4.00 nC(2d)2(cos45.0i^+sin45.0j^)−3.00 nCd2j^]=−k[(1.00 nC)(10−9 C1 nC)d2i^−(4.00 nC)(10−9 C1 nC)(2d)2(cos45.0i^+sin45.0j^)−(3.00 nC)(10−9 C1 nC)d2j^]=8.99×109 N⋅m2/C2((25.0 cm)(10−2 m1 cm))2(10−9 C)[(−1−2cos45.0°)i^+j^]=(−347i^−635j^) N/C

Thus, the electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

(b)

Expert Solution

To determine

The total electric forceon the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N

Explanation of Solution

Write the equation for the electric force.

F→=qE→ (III)

Here, F→ is the electric force.

Conclusion:

Substitute 2.00 nC for q and (−347i^−635j^) N/C for E→ in equation (III) to find F→.

F→=(2.00 nC)((−347i^−635j^) N/C)=((2.00 nC)(10−9 C1 nC))((−347i^−635j^) N/C)=(−6.94×10−7i^−1.27×10−6j^) N

Thus, the total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N.

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Answer 4

Textbook Question

Chapter 24, Problem 51PQ

Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d = 25.0 cm.

a. What is the electric field at the location of the sphere with charge +2.00 nC?

b. What is the total electric force exerted on the sphere with charge +2.00 nC by the other three spheres?

Chapter 24, Problem 51PQ, Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d =

FIGURE P24.51

(a)

Expert Solution

To determine

The electric field at the location of the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 24, Problem 51PQ

Write the equation for the electric field of a charged particle.

E→=kqr2r^ (I)

Here, E→E is the electric field of a charged particle, k is the Coulomb’s constant, r is the distance and r^ is the surface unit vector in the direction for r.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge +2.00 nC.

E→=kq1r12r^1+kq3r32r^3+kq4r42r^4 (II)

Here, q1 is the is the first charge, q4 is the is the third charge, q3 is the is the third charge, r1 is the distance of the first charge to the +2.00 nC charge, r3 is the distance of the third charge to the +2.00 nC charge, r4 is the distance of the fourth charge to the +2.00 nC charge, r^1 is the unit vector in the direction of r1, r^3 is the unit vector in the direction of r3 and r^4 is the unit vector in the direction of r4.

Conclusion:

Substitute 8.99×109 N⋅m2/C2 for k, 1.00 nC for q1, 4.00 nC for q3 and 3.00 nC for q4, d for r1, 2d for r3, d for r4, i^ for r^1, (cos45.0i^+sin45.0j^) for r^3 and j^ for r^4 in equation (II) to find E→.

E→=−k[1.00 nCd2i^−4.00 nC(2d)2(cos45.0i^+sin45.0j^)−3.00 nCd2j^]=−k[(1.00 nC)(10−9 C1 nC)d2i^−(4.00 nC)(10−9 C1 nC)(2d)2(cos45.0i^+sin45.0j^)−(3.00 nC)(10−9 C1 nC)d2j^]=8.99×109 N⋅m2/C2((25.0 cm)(10−2 m1 cm))2(10−9 C)[(−1−2cos45.0°)i^+j^]=(−347i^−635j^) N/C

Thus, the electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

(b)

Expert Solution

To determine

The total electric forceon the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N

Explanation of Solution

Write the equation for the electric force.

F→=qE→ (III)

Here, F→ is the electric force.

Conclusion:

Substitute 2.00 nC for q and (−347i^−635j^) N/C for E→ in equation (III) to find F→.

F→=(2.00 nC)((−347i^−635j^) N/C)=((2.00 nC)(10−9 C1 nC))((−347i^−635j^) N/C)=(−6.94×10−7i^−1.27×10−6j^) N

Thus, the total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The electric field at the location of the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 24, Problem 51PQ

Write the equation for the electric field of a charged particle.

E→=kqr2r^ (I)

Here, E→E is the electric field of a charged particle, k is the Coulomb’s constant, r is the distance and r^ is the surface unit vector in the direction for r.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge +2.00 nC.

E→=kq1r12r^1+kq3r32r^3+kq4r42r^4 (II)

Here, q1 is the is the first charge, q4 is the is the third charge, q3 is the is the third charge, r1 is the distance of the first charge to the +2.00 nC charge, r3 is the distance of the third charge to the +2.00 nC charge, r4 is the distance of the fourth charge to the +2.00 nC charge, r^1 is the unit vector in the direction of r1, r^3 is the unit vector in the direction of r3 and r^4 is the unit vector in the direction of r4.

Conclusion:

Substitute 8.99×109 N⋅m2/C2 for k, 1.00 nC for q1, 4.00 nC for q3 and 3.00 nC for q4, d for r1, 2d for r3, d for r4, i^ for r^1, (cos45.0i^+sin45.0j^) for r^3 and j^ for r^4 in equation (II) to find E→.

E→=−k[1.00 nCd2i^−4.00 nC(2d)2(cos45.0i^+sin45.0j^)−3.00 nCd2j^]=−k[(1.00 nC)(10−9 C1 nC)d2i^−(4.00 nC)(10−9 C1 nC)(2d)2(cos45.0i^+sin45.0j^)−(3.00 nC)(10−9 C1 nC)d2j^]=8.99×109 N⋅m2/C2((25.0 cm)(10−2 m1 cm))2(10−9 C)[(−1−2cos45.0°)i^+j^]=(−347i^−635j^) N/C

Thus, the electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

(b)

Expert Solution

To determine

The total electric forceon the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N

Explanation of Solution

Write the equation for the electric force.

F→=qE→ (III)

Here, F→ is the electric force.

Conclusion:

Substitute 2.00 nC for q and (−347i^−635j^) N/C for E→ in equation (III) to find F→.

F→=(2.00 nC)((−347i^−635j^) N/C)=((2.00 nC)(10−9 C1 nC))((−347i^−635j^) N/C)=(−6.94×10−7i^−1.27×10−6j^) N

Thus, the total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N.

Answer 8

(a)

Expert Solution

To determine

The electric field at the location of the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 24, Problem 51PQ

Write the equation for the electric field of a charged particle.

E→=kqr2r^ (I)

Here, E→E is the electric field of a charged particle, k is the Coulomb’s constant, r is the distance and r^ is the surface unit vector in the direction for r.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge +2.00 nC.

E→=kq1r12r^1+kq3r32r^3+kq4r42r^4 (II)

Here, q1 is the is the first charge, q4 is the is the third charge, q3 is the is the third charge, r1 is the distance of the first charge to the +2.00 nC charge, r3 is the distance of the third charge to the +2.00 nC charge, r4 is the distance of the fourth charge to the +2.00 nC charge, r^1 is the unit vector in the direction of r1, r^3 is the unit vector in the direction of r3 and r^4 is the unit vector in the direction of r4.

Conclusion:

Substitute 8.99×109 N⋅m2/C2 for k, 1.00 nC for q1, 4.00 nC for q3 and 3.00 nC for q4, d for r1, 2d for r3, d for r4, i^ for r^1, (cos45.0i^+sin45.0j^) for r^3 and j^ for r^4 in equation (II) to find E→.

E→=−k[1.00 nCd2i^−4.00 nC(2d)2(cos45.0i^+sin45.0j^)−3.00 nCd2j^]=−k[(1.00 nC)(10−9 C1 nC)d2i^−(4.00 nC)(10−9 C1 nC)(2d)2(cos45.0i^+sin45.0j^)−(3.00 nC)(10−9 C1 nC)d2j^]=8.99×109 N⋅m2/C2((25.0 cm)(10−2 m1 cm))2(10−9 C)[(−1−2cos45.0°)i^+j^]=(−347i^−635j^) N/C

Thus, the electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The electric field at the location of the sphere with charge +2.00 nC.

Answer 13

Answer to Problem 51PQ

The electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Answer 14

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Bundle: Physics For Scientists And Engineers: Foundations And Connections, Volume 2, Loose-leaf Version + Webassign Printed Access Card For Katz's ... And Connections, Single-term Courses, Chapter 24, Problem 51PQ

Write the equation for the electric field of a charged particle.

E→=kqr2r^ (I)

Here, E→E is the electric field of a charged particle, k is the Coulomb’s constant, r is the distance and r^ is the surface unit vector in the direction for r.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge +2.00 nC.

E→=kq1r12r^1+kq3r32r^3+kq4r42r^4 (II)

Here, q1 is the is the first charge, q4 is the is the third charge, q3 is the is the third charge, r1 is the distance of the first charge to the +2.00 nC charge, r3 is the distance of the third charge to the +2.00 nC charge, r4 is the distance of the fourth charge to the +2.00 nC charge, r^1 is the unit vector in the direction of r1, r^3 is the unit vector in the direction of r3 and r^4 is the unit vector in the direction of r4.

Conclusion:

Substitute 8.99×109 N⋅m2/C2 for k, 1.00 nC for q1, 4.00 nC for q3 and 3.00 nC for q4, d for r1, 2d for r3, d for r4, i^ for r^1, (cos45.0i^+sin45.0j^) for r^3 and j^ for r^4 in equation (II) to find E→.

E→=−k[1.00 nCd2i^−4.00 nC(2d)2(cos45.0i^+sin45.0j^)−3.00 nCd2j^]=−k[(1.00 nC)(10−9 C1 nC)d2i^−(4.00 nC)(10−9 C1 nC)(2d)2(cos45.0i^+sin45.0j^)−(3.00 nC)(10−9 C1 nC)d2j^]=8.99×109 N⋅m2/C2((25.0 cm)(10−2 m1 cm))2(10−9 C)[(−1−2cos45.0°)i^+j^]=(−347i^−635j^) N/C

Thus, the electric field at the location of the sphere with charge +2.00 nC is (−347i^−635j^) N/C.

Answer 15

(b)

Expert Solution

To determine

The total electric forceon the sphere with charge +2.00 nC.

Answer to Problem 51PQ

The total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N

Explanation of Solution

Write the equation for the electric force.

F→=qE→ (III)

Here, F→ is the electric force.

Conclusion:

Substitute 2.00 nC for q and (−347i^−635j^) N/C for E→ in equation (III) to find F→.

F→=(2.00 nC)((−347i^−635j^) N/C)=((2.00 nC)(10−9 C1 nC))((−347i^−635j^) N/C)=(−6.94×10−7i^−1.27×10−6j^) N

Thus, the total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The total electric forceon the sphere with charge +2.00 nC.

Answer 20

Answer to Problem 51PQ

The total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N

Answer 21

Explanation of Solution

Write the equation for the electric force.

F→=qE→ (III)

Here, F→ is the electric force.

Conclusion:

Substitute 2.00 nC for q and (−347i^−635j^) N/C for E→ in equation (III) to find F→.

F→=(2.00 nC)((−347i^−635j^) N/C)=((2.00 nC)(10−9 C1 nC))((−347i^−635j^) N/C)=(−6.94×10−7i^−1.27×10−6j^) N

Thus, the total electric force on the sphere with charge +2.00 nC is (−6.94×10−7i^−1.27×10−6j^) N.

Answer 22

Ch. 24.2 - In a few sentences, explain how you know that...Ch. 24.2 - What is the magnitude of the electric field due to...Ch. 24.3 - Which lines in Figure 24.7 cannot represent an...Ch. 24.4 - Figure 24.10 shows a source that consists of two...Ch. 24.4 - A water molecule is made up of two hydrogen atoms...Ch. 24.5 - a. Figure 24.22A shows a rod of length L and...Ch. 24 - The terms electrostatic force and electrostatic...Ch. 24 - Prob. 2PQ Ch. 24 - A sphere has a charge of 89.5 nC and a radius of...Ch. 24 - Prob. 4PQ

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Answer to Problem 51PQ

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Answer to Problem 51PQ

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