The vector electric field created by the 6.00 nC and − 3.00 nC charges together at the origin.

Question

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Answer 1

Question

Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the 6.00 nC and −3.00 nC charges together at the origin.

(a)

Expert Solution

Answer to Problem 35P

The vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Explanation of Solution

The following figure represents the forces acting on the charges.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 23, Problem 35P

Figure-(1)

Here, F1 is the electric force acting on the charge of 5.00 nC by the charge of 6.00 nC and F2 is the electric force acting on the charge 5.00 nC by the charge −3.00 nC.

Write the expression for the electric field at origin by the charge q1 in the negative x axis.

E1→=kq1r12(−i^)

Here, E1→ is the electric field at origin by the charge q1, r1 is the distance of the charge q1 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative x axis direction.

Write the constant value for the Coulomb constant k.

k=8.9876 ×109 N.m2/C2

Write the expression for the electric field at origin by the charge q2 in the negative y axis.

E2→=kq2r22(−j^)

Here, E2→ is the electric field at origin by the charge q2, r2 is the distance of the charge q2 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative y axis direction.

Write the expression for the total vector electric field at origin by the charges q1 and q2.

E→=E1→+E2→ (I)

Here, the E1→ is the electric field at origin by the charge q1 and E2→ is the electric field at origin by the charge q2.

Write the expression to convert the units for charge q1 from nC to C.

q1=6.00 nC (II)

Write the expression to convert the units for charge q2 from nC to C.

q2=|−3.00 nC| (III)

Substitute kq1r12(−i^) for E1→ and kq2r22(−j^) for E2→ in the equation (I) to calculate E→.

E→=kq1r12(−i^)+kq2r22(−j^) (IV)

Conclusion:

Convert the units for the charge q1 from nC to C in equation (II).

q1=6.00 nC(1 C1×109 nC)=6.00×10−9 C

Convert the units for the charge q2 from nC to C in equation (III).

q2=−3.00 nC(1 C1×109 nC)=−3.00×10−9 C

Substitute 6.00×10−9 C for q1, −3.00×10−9 C for q2, 0.300 m for r1 and 0.100 m for r2 and 8.9876×109 N⋅m2/C2 for k in equation (IV) to solve E→.

E→=[(8.9876×109 N⋅m2/C2)(6.00×10−9 C)(0.300 m)2(−i^)+(8.9876×109 N⋅m2/C2)(3.00×10−9 C)(0.100 m)2(−j^)]=[599.17(−i^)+2696.28(−j^)] N/C=[0.599×103(−i^)+2.70×103(−j^)] N/C

Further simplify the above equation.

E→=[−0.599×103i^−2.70×103j^] N/C=[−0.599×103i^−2.70×103j^] N/C(10−3kN/C1 N/C)=[−0.599i^−2.70j^] kN/C

Therefore, the vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

(b)

To determine

The vector force on the 5.00 nC charge.

(b)

Expert Solution

Answer to Problem 35P

The vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

F→=qE→ (IV)

Here, q is the charge on the origin, E→ is the electric field at the origin due to the charges 6.00 nC and −3.00 nC.

Write the expression to convert the units for charge q1 from nC to C.

q=5.00 nC (V)

Conclusion:

Convert the units for the charge q from nC to C in equation (V).

q=5.00 nC(1 C1×109 nC)=5.00×10−9 C

Substitute 5.00×10−9 C for q and [−0.599i^−2.70j^]×103 N/C for E→ in equation (IV) to calculate F→.

F→=5.00×10−9 C×[−0.599i^−2.70j^]×103 N/C=[−(5.00×10−9)×(0.599i^)−(5.00×10−9 C)×(2.70j^)]×103 N=[(−2.99i^−13.48j^)×10−6 N](1 μN1×10−6N)≈(−3.00i^−13.5j^) μN

Therefore, the vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

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Answer 2

Question

Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the 6.00 nC and −3.00 nC charges together at the origin.

(a)

Expert Solution

Answer to Problem 35P

The vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Explanation of Solution

The following figure represents the forces acting on the charges.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 23, Problem 35P

Figure-(1)

Here, F1 is the electric force acting on the charge of 5.00 nC by the charge of 6.00 nC and F2 is the electric force acting on the charge 5.00 nC by the charge −3.00 nC.

Write the expression for the electric field at origin by the charge q1 in the negative x axis.

E1→=kq1r12(−i^)

Here, E1→ is the electric field at origin by the charge q1, r1 is the distance of the charge q1 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative x axis direction.

Write the constant value for the Coulomb constant k.

k=8.9876 ×109 N.m2/C2

Write the expression for the electric field at origin by the charge q2 in the negative y axis.

E2→=kq2r22(−j^)

Here, E2→ is the electric field at origin by the charge q2, r2 is the distance of the charge q2 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative y axis direction.

Write the expression for the total vector electric field at origin by the charges q1 and q2.

E→=E1→+E2→ (I)

Here, the E1→ is the electric field at origin by the charge q1 and E2→ is the electric field at origin by the charge q2.

Write the expression to convert the units for charge q1 from nC to C.

q1=6.00 nC (II)

Write the expression to convert the units for charge q2 from nC to C.

q2=|−3.00 nC| (III)

Substitute kq1r12(−i^) for E1→ and kq2r22(−j^) for E2→ in the equation (I) to calculate E→.

E→=kq1r12(−i^)+kq2r22(−j^) (IV)

Conclusion:

Convert the units for the charge q1 from nC to C in equation (II).

q1=6.00 nC(1 C1×109 nC)=6.00×10−9 C

Convert the units for the charge q2 from nC to C in equation (III).

q2=−3.00 nC(1 C1×109 nC)=−3.00×10−9 C

Substitute 6.00×10−9 C for q1, −3.00×10−9 C for q2, 0.300 m for r1 and 0.100 m for r2 and 8.9876×109 N⋅m2/C2 for k in equation (IV) to solve E→.

E→=[(8.9876×109 N⋅m2/C2)(6.00×10−9 C)(0.300 m)2(−i^)+(8.9876×109 N⋅m2/C2)(3.00×10−9 C)(0.100 m)2(−j^)]=[599.17(−i^)+2696.28(−j^)] N/C=[0.599×103(−i^)+2.70×103(−j^)] N/C

Further simplify the above equation.

E→=[−0.599×103i^−2.70×103j^] N/C=[−0.599×103i^−2.70×103j^] N/C(10−3kN/C1 N/C)=[−0.599i^−2.70j^] kN/C

Therefore, the vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

(b)

To determine

The vector force on the 5.00 nC charge.

(b)

Expert Solution

Answer to Problem 35P

The vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

F→=qE→ (IV)

Here, q is the charge on the origin, E→ is the electric field at the origin due to the charges 6.00 nC and −3.00 nC.

Write the expression to convert the units for charge q1 from nC to C.

q=5.00 nC (V)

Conclusion:

Convert the units for the charge q from nC to C in equation (V).

q=5.00 nC(1 C1×109 nC)=5.00×10−9 C

Substitute 5.00×10−9 C for q and [−0.599i^−2.70j^]×103 N/C for E→ in equation (IV) to calculate F→.

F→=5.00×10−9 C×[−0.599i^−2.70j^]×103 N/C=[−(5.00×10−9)×(0.599i^)−(5.00×10−9 C)×(2.70j^)]×103 N=[(−2.99i^−13.48j^)×10−6 N](1 μN1×10−6N)≈(−3.00i^−13.5j^) μN

Therefore, the vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

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Answer 3

Question

Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the 6.00 nC and −3.00 nC charges together at the origin.

(a)

Expert Solution

Answer to Problem 35P

The vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Explanation of Solution

The following figure represents the forces acting on the charges.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 23, Problem 35P

Figure-(1)

Here, F1 is the electric force acting on the charge of 5.00 nC by the charge of 6.00 nC and F2 is the electric force acting on the charge 5.00 nC by the charge −3.00 nC.

Write the expression for the electric field at origin by the charge q1 in the negative x axis.

E1→=kq1r12(−i^)

Here, E1→ is the electric field at origin by the charge q1, r1 is the distance of the charge q1 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative x axis direction.

Write the constant value for the Coulomb constant k.

k=8.9876 ×109 N.m2/C2

Write the expression for the electric field at origin by the charge q2 in the negative y axis.

E2→=kq2r22(−j^)

Here, E2→ is the electric field at origin by the charge q2, r2 is the distance of the charge q2 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative y axis direction.

Write the expression for the total vector electric field at origin by the charges q1 and q2.

E→=E1→+E2→ (I)

Here, the E1→ is the electric field at origin by the charge q1 and E2→ is the electric field at origin by the charge q2.

Write the expression to convert the units for charge q1 from nC to C.

q1=6.00 nC (II)

Write the expression to convert the units for charge q2 from nC to C.

q2=|−3.00 nC| (III)

Substitute kq1r12(−i^) for E1→ and kq2r22(−j^) for E2→ in the equation (I) to calculate E→.

E→=kq1r12(−i^)+kq2r22(−j^) (IV)

Conclusion:

Convert the units for the charge q1 from nC to C in equation (II).

q1=6.00 nC(1 C1×109 nC)=6.00×10−9 C

Convert the units for the charge q2 from nC to C in equation (III).

q2=−3.00 nC(1 C1×109 nC)=−3.00×10−9 C

Substitute 6.00×10−9 C for q1, −3.00×10−9 C for q2, 0.300 m for r1 and 0.100 m for r2 and 8.9876×109 N⋅m2/C2 for k in equation (IV) to solve E→.

E→=[(8.9876×109 N⋅m2/C2)(6.00×10−9 C)(0.300 m)2(−i^)+(8.9876×109 N⋅m2/C2)(3.00×10−9 C)(0.100 m)2(−j^)]=[599.17(−i^)+2696.28(−j^)] N/C=[0.599×103(−i^)+2.70×103(−j^)] N/C

Further simplify the above equation.

E→=[−0.599×103i^−2.70×103j^] N/C=[−0.599×103i^−2.70×103j^] N/C(10−3kN/C1 N/C)=[−0.599i^−2.70j^] kN/C

Therefore, the vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

(b)

To determine

The vector force on the 5.00 nC charge.

(b)

Expert Solution

Answer to Problem 35P

The vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

F→=qE→ (IV)

Here, q is the charge on the origin, E→ is the electric field at the origin due to the charges 6.00 nC and −3.00 nC.

Write the expression to convert the units for charge q1 from nC to C.

q=5.00 nC (V)

Conclusion:

Convert the units for the charge q from nC to C in equation (V).

q=5.00 nC(1 C1×109 nC)=5.00×10−9 C

Substitute 5.00×10−9 C for q and [−0.599i^−2.70j^]×103 N/C for E→ in equation (IV) to calculate F→.

F→=5.00×10−9 C×[−0.599i^−2.70j^]×103 N/C=[−(5.00×10−9)×(0.599i^)−(5.00×10−9 C)×(2.70j^)]×103 N=[(−2.99i^−13.48j^)×10−6 N](1 μN1×10−6N)≈(−3.00i^−13.5j^) μN

Therefore, the vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

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Answer 4

Question

Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the 6.00 nC and −3.00 nC charges together at the origin.

(a)

Expert Solution

Answer to Problem 35P

The vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Explanation of Solution

The following figure represents the forces acting on the charges.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 23, Problem 35P

Figure-(1)

Here, F1 is the electric force acting on the charge of 5.00 nC by the charge of 6.00 nC and F2 is the electric force acting on the charge 5.00 nC by the charge −3.00 nC.

Write the expression for the electric field at origin by the charge q1 in the negative x axis.

E1→=kq1r12(−i^)

Here, E1→ is the electric field at origin by the charge q1, r1 is the distance of the charge q1 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative x axis direction.

Write the constant value for the Coulomb constant k.

k=8.9876 ×109 N.m2/C2

Write the expression for the electric field at origin by the charge q2 in the negative y axis.

E2→=kq2r22(−j^)

Here, E2→ is the electric field at origin by the charge q2, r2 is the distance of the charge q2 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative y axis direction.

Write the expression for the total vector electric field at origin by the charges q1 and q2.

E→=E1→+E2→ (I)

Here, the E1→ is the electric field at origin by the charge q1 and E2→ is the electric field at origin by the charge q2.

Write the expression to convert the units for charge q1 from nC to C.

q1=6.00 nC (II)

Write the expression to convert the units for charge q2 from nC to C.

q2=|−3.00 nC| (III)

Substitute kq1r12(−i^) for E1→ and kq2r22(−j^) for E2→ in the equation (I) to calculate E→.

E→=kq1r12(−i^)+kq2r22(−j^) (IV)

Conclusion:

Convert the units for the charge q1 from nC to C in equation (II).

q1=6.00 nC(1 C1×109 nC)=6.00×10−9 C

Convert the units for the charge q2 from nC to C in equation (III).

q2=−3.00 nC(1 C1×109 nC)=−3.00×10−9 C

Substitute 6.00×10−9 C for q1, −3.00×10−9 C for q2, 0.300 m for r1 and 0.100 m for r2 and 8.9876×109 N⋅m2/C2 for k in equation (IV) to solve E→.

E→=[(8.9876×109 N⋅m2/C2)(6.00×10−9 C)(0.300 m)2(−i^)+(8.9876×109 N⋅m2/C2)(3.00×10−9 C)(0.100 m)2(−j^)]=[599.17(−i^)+2696.28(−j^)] N/C=[0.599×103(−i^)+2.70×103(−j^)] N/C

Further simplify the above equation.

E→=[−0.599×103i^−2.70×103j^] N/C=[−0.599×103i^−2.70×103j^] N/C(10−3kN/C1 N/C)=[−0.599i^−2.70j^] kN/C

Therefore, the vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

(b)

To determine

The vector force on the 5.00 nC charge.

(b)

Expert Solution

Answer to Problem 35P

The vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

F→=qE→ (IV)

Here, q is the charge on the origin, E→ is the electric field at the origin due to the charges 6.00 nC and −3.00 nC.

Write the expression to convert the units for charge q1 from nC to C.

q=5.00 nC (V)

Conclusion:

Convert the units for the charge q from nC to C in equation (V).

q=5.00 nC(1 C1×109 nC)=5.00×10−9 C

Substitute 5.00×10−9 C for q and [−0.599i^−2.70j^]×103 N/C for E→ in equation (IV) to calculate F→.

F→=5.00×10−9 C×[−0.599i^−2.70j^]×103 N/C=[−(5.00×10−9)×(0.599i^)−(5.00×10−9 C)×(2.70j^)]×103 N=[(−2.99i^−13.48j^)×10−6 N](1 μN1×10−6N)≈(−3.00i^−13.5j^) μN

Therefore, the vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

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Answer 5

Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the 6.00 nC and −3.00 nC charges together at the origin.

(a)

Expert Solution

Answer to Problem 35P

The vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Explanation of Solution

The following figure represents the forces acting on the charges.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 23, Problem 35P

Figure-(1)

Here, F1 is the electric force acting on the charge of 5.00 nC by the charge of 6.00 nC and F2 is the electric force acting on the charge 5.00 nC by the charge −3.00 nC.

Write the expression for the electric field at origin by the charge q1 in the negative x axis.

E1→=kq1r12(−i^)

Here, E1→ is the electric field at origin by the charge q1, r1 is the distance of the charge q1 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative x axis direction.

Write the constant value for the Coulomb constant k.

k=8.9876 ×109 N.m2/C2

Write the expression for the electric field at origin by the charge q2 in the negative y axis.

E2→=kq2r22(−j^)

Here, E2→ is the electric field at origin by the charge q2, r2 is the distance of the charge q2 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative y axis direction.

Write the expression for the total vector electric field at origin by the charges q1 and q2.

E→=E1→+E2→ (I)

Here, the E1→ is the electric field at origin by the charge q1 and E2→ is the electric field at origin by the charge q2.

Write the expression to convert the units for charge q1 from nC to C.

q1=6.00 nC (II)

Write the expression to convert the units for charge q2 from nC to C.

q2=|−3.00 nC| (III)

Substitute kq1r12(−i^) for E1→ and kq2r22(−j^) for E2→ in the equation (I) to calculate E→.

E→=kq1r12(−i^)+kq2r22(−j^) (IV)

Conclusion:

Convert the units for the charge q1 from nC to C in equation (II).

q1=6.00 nC(1 C1×109 nC)=6.00×10−9 C

Convert the units for the charge q2 from nC to C in equation (III).

q2=−3.00 nC(1 C1×109 nC)=−3.00×10−9 C

Substitute 6.00×10−9 C for q1, −3.00×10−9 C for q2, 0.300 m for r1 and 0.100 m for r2 and 8.9876×109 N⋅m2/C2 for k in equation (IV) to solve E→.

E→=[(8.9876×109 N⋅m2/C2)(6.00×10−9 C)(0.300 m)2(−i^)+(8.9876×109 N⋅m2/C2)(3.00×10−9 C)(0.100 m)2(−j^)]=[599.17(−i^)+2696.28(−j^)] N/C=[0.599×103(−i^)+2.70×103(−j^)] N/C

Further simplify the above equation.

E→=[−0.599×103i^−2.70×103j^] N/C=[−0.599×103i^−2.70×103j^] N/C(10−3kN/C1 N/C)=[−0.599i^−2.70j^] kN/C

Therefore, the vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

(b)

To determine

The vector force on the 5.00 nC charge.

(b)

Expert Solution

Answer to Problem 35P

The vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

F→=qE→ (IV)

Here, q is the charge on the origin, E→ is the electric field at the origin due to the charges 6.00 nC and −3.00 nC.

Write the expression to convert the units for charge q1 from nC to C.

q=5.00 nC (V)

Conclusion:

Convert the units for the charge q from nC to C in equation (V).

q=5.00 nC(1 C1×109 nC)=5.00×10−9 C

Substitute 5.00×10−9 C for q and [−0.599i^−2.70j^]×103 N/C for E→ in equation (IV) to calculate F→.

F→=5.00×10−9 C×[−0.599i^−2.70j^]×103 N/C=[−(5.00×10−9)×(0.599i^)−(5.00×10−9 C)×(2.70j^)]×103 N=[(−2.99i^−13.48j^)×10−6 N](1 μN1×10−6N)≈(−3.00i^−13.5j^) μN

Therefore, the vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Answer 6

Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the 6.00 nC and −3.00 nC charges together at the origin.

(a)

Expert Solution

Answer to Problem 35P

The vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Explanation of Solution

The following figure represents the forces acting on the charges.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 23, Problem 35P

Figure-(1)

Here, F1 is the electric force acting on the charge of 5.00 nC by the charge of 6.00 nC and F2 is the electric force acting on the charge 5.00 nC by the charge −3.00 nC.

Write the expression for the electric field at origin by the charge q1 in the negative x axis.

E1→=kq1r12(−i^)

Here, E1→ is the electric field at origin by the charge q1, r1 is the distance of the charge q1 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative x axis direction.

Write the constant value for the Coulomb constant k.

k=8.9876 ×109 N.m2/C2

Write the expression for the electric field at origin by the charge q2 in the negative y axis.

E2→=kq2r22(−j^)

Here, E2→ is the electric field at origin by the charge q2, r2 is the distance of the charge q2 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative y axis direction.

Write the expression for the total vector electric field at origin by the charges q1 and q2.

E→=E1→+E2→ (I)

Here, the E1→ is the electric field at origin by the charge q1 and E2→ is the electric field at origin by the charge q2.

Write the expression to convert the units for charge q1 from nC to C.

q1=6.00 nC (II)

Write the expression to convert the units for charge q2 from nC to C.

q2=|−3.00 nC| (III)

Substitute kq1r12(−i^) for E1→ and kq2r22(−j^) for E2→ in the equation (I) to calculate E→.

E→=kq1r12(−i^)+kq2r22(−j^) (IV)

Conclusion:

Convert the units for the charge q1 from nC to C in equation (II).

q1=6.00 nC(1 C1×109 nC)=6.00×10−9 C

Convert the units for the charge q2 from nC to C in equation (III).

q2=−3.00 nC(1 C1×109 nC)=−3.00×10−9 C

Substitute 6.00×10−9 C for q1, −3.00×10−9 C for q2, 0.300 m for r1 and 0.100 m for r2 and 8.9876×109 N⋅m2/C2 for k in equation (IV) to solve E→.

E→=[(8.9876×109 N⋅m2/C2)(6.00×10−9 C)(0.300 m)2(−i^)+(8.9876×109 N⋅m2/C2)(3.00×10−9 C)(0.100 m)2(−j^)]=[599.17(−i^)+2696.28(−j^)] N/C=[0.599×103(−i^)+2.70×103(−j^)] N/C

Further simplify the above equation.

E→=[−0.599×103i^−2.70×103j^] N/C=[−0.599×103i^−2.70×103j^] N/C(10−3kN/C1 N/C)=[−0.599i^−2.70j^] kN/C

Therefore, the vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Answer 7

Chapter 23, Problem 35P

(a)

To determine

The vector electric field created by the 6.00 nC and −3.00 nC charges together at the origin.

Answer 8

(a)

Expert Solution

Answer to Problem 35P

The vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The following figure represents the forces acting on the charges.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 23, Problem 35P

Figure-(1)

Here, F1 is the electric force acting on the charge of 5.00 nC by the charge of 6.00 nC and F2 is the electric force acting on the charge 5.00 nC by the charge −3.00 nC.

Write the expression for the electric field at origin by the charge q1 in the negative x axis.

E1→=kq1r12(−i^)

Here, E1→ is the electric field at origin by the charge q1, r1 is the distance of the charge q1 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative x axis direction.

Write the constant value for the Coulomb constant k.

k=8.9876 ×109 N.m2/C2

Write the expression for the electric field at origin by the charge q2 in the negative y axis.

E2→=kq2r22(−j^)

Here, E2→ is the electric field at origin by the charge q2, r2 is the distance of the charge q2 from the origin, k is the electrostatic constant. The negative sign indicates that the electric field is in negative y axis direction.

Write the expression for the total vector electric field at origin by the charges q1 and q2.

E→=E1→+E2→ (I)

Here, the E1→ is the electric field at origin by the charge q1 and E2→ is the electric field at origin by the charge q2.

Write the expression to convert the units for charge q1 from nC to C.

q1=6.00 nC (II)

Write the expression to convert the units for charge q2 from nC to C.

q2=|−3.00 nC| (III)

Substitute kq1r12(−i^) for E1→ and kq2r22(−j^) for E2→ in the equation (I) to calculate E→.

E→=kq1r12(−i^)+kq2r22(−j^) (IV)

Conclusion:

Convert the units for the charge q1 from nC to C in equation (II).

q1=6.00 nC(1 C1×109 nC)=6.00×10−9 C

Convert the units for the charge q2 from nC to C in equation (III).

q2=−3.00 nC(1 C1×109 nC)=−3.00×10−9 C

Substitute 6.00×10−9 C for q1, −3.00×10−9 C for q2, 0.300 m for r1 and 0.100 m for r2 and 8.9876×109 N⋅m2/C2 for k in equation (IV) to solve E→.

E→=[(8.9876×109 N⋅m2/C2)(6.00×10−9 C)(0.300 m)2(−i^)+(8.9876×109 N⋅m2/C2)(3.00×10−9 C)(0.100 m)2(−j^)]=[599.17(−i^)+2696.28(−j^)] N/C=[0.599×103(−i^)+2.70×103(−j^)] N/C

Further simplify the above equation.

E→=[−0.599×103i^−2.70×103j^] N/C=[−0.599×103i^−2.70×103j^] N/C(10−3kN/C1 N/C)=[−0.599i^−2.70j^] kN/C

Therefore, the vector electric field that the 6.00 nC and −3.00 nC charges together create at the origin is [−0.599i^−2.70j^] kN/C.

Answer 13

(b)

To determine

The vector force on the 5.00 nC charge.

(b)

Expert Solution

Answer to Problem 35P

The vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

F→=qE→ (IV)

Here, q is the charge on the origin, E→ is the electric field at the origin due to the charges 6.00 nC and −3.00 nC.

Write the expression to convert the units for charge q1 from nC to C.

q=5.00 nC (V)

Conclusion:

Convert the units for the charge q from nC to C in equation (V).

q=5.00 nC(1 C1×109 nC)=5.00×10−9 C

Substitute 5.00×10−9 C for q and [−0.599i^−2.70j^]×103 N/C for E→ in equation (IV) to calculate F→.

F→=5.00×10−9 C×[−0.599i^−2.70j^]×103 N/C=[−(5.00×10−9)×(0.599i^)−(5.00×10−9 C)×(2.70j^)]×103 N=[(−2.99i^−13.48j^)×10−6 N](1 μN1×10−6N)≈(−3.00i^−13.5j^) μN

Therefore, the vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Answer 14

(b)

To determine

The vector force on the 5.00 nC charge.

Answer 15

(b)

Expert Solution

Answer to Problem 35P

The vector force on the 5.00 nC charge is (−3.00i^−13.5j^) μN.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the relation between the electric force and the electric field.

F→=qE→ (IV)

Here, q is the charge on the origin, E→ is the electric field at the origin due to the charges 6.00 nC and −3.00 nC.

Write the expression to convert the units for charge q1 from nC to C.

q=5.00 nC (V)

Conclusion:

Convert the units for the charge q from nC to C in equation (V).

q=5.00 nC(1 C1×109 nC)=5.00×10−9 C

Substitute 5.00×10−9 C for q and [−0.599i^−2.70j^]×103 N/C for E→ in equation (IV) to calculate F→.

F→=5.00×10−9 C×[−0.599i^−2.70j^]×103 N/C=[−(5.00×10−9)×(0.599i^)−(5.00×10−9 C)×(2.70j^)]×103 N=[(−2.99i^−13.48j^)×10−6 N](1 μN1×10−6N)≈(−3.00i^−13.5j^) μN