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

Want to see more full solutions like this?

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.

Physics for Scientists and Engineers: Foundations and Connections, 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.

Want to see more full solutions like this?

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

Students have asked these similar questions

No chatgpt pls will upvote

air is pushed steadily though a forced air pipe at a steady speed of 4.0 m/s. the pipe measures 56 cm by 22 cm. how fast will air move though a narrower portion of the pipe that is also rectangular and measures 32 cm by 22 cm

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.

Physics for Scientists and Engineers: Foundations and Connections, 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.

Want to see more full solutions like this?

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

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.

Physics for Scientists and Engineers: Foundations and Connections, 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.

Want to see more full solutions like this?

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

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.

Physics for Scientists and Engineers: Foundations and Connections, 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.

Want to see more full solutions like this?

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

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.

Physics for Scientists and Engineers: Foundations and Connections, 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.

Physics for Scientists and Engineers: Foundations and Connections, 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.

Physics for Scientists and Engineers: Foundations and Connections, 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

Concept explainers

Videos

Answer to Problem 51PQ

Explanation of Solution

Answer to Problem 51PQ

Explanation of Solution

Want to see more full solutions like this?

Chapter 24 Solutions