The net electric flux through the cube.

Question

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Answer 1

Question

Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

(a)

Expert Solution

Answer to Problem 32P

Net electric flux through the cube is 15.0 N⋅m2/C.

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 32P

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

ϕE=∮E→⋅dA→ (I)

Here, E→ is the electric field, ϕE is the net flux through closed Gaussian surface and A→ is the surface area.

Write the expression for net electric flux through cube.

ϕE=ϕE(top)+ϕE(bottom)+ϕE(right)+ϕE(left)+ϕE(into)+ϕE(out) (II)

Conclusion:

Substitute 25.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(top).

ϕE(top)=EAcos(180°)=−(EA)=−(25.0 N/C)(1.00 m2)=−25.0 N⋅m2/C

Substitute 15.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(bottom)

ϕE(bottom)=EAcos(0°)=EA=(15.0 N/C)(1.00 m2)=15.0 N⋅m2/C

Substitute 35.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(right).

ϕE(right)=EAcos(180°)=−(EA)=−(35.0 N/C)(1.00 m2)=−35.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(left).

ϕE(left)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(into).

ϕE(into)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(out).

ϕE(out)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute −25.0 N⋅m2/C for ϕE(top), 15.0 N⋅m2/C for ϕE(bottom), −35.0 N⋅m2/C for ϕE(right), 20.0 N⋅m2/C for ϕE(left), ϕE(into) and ϕE(out) in equation (II) to calculate ϕE.

ϕE=(−25.0+15.0−35.0+20.0+20.0+20.0) N⋅m2/C=15.0 N⋅m2/C

Therefore, net electric flux through the cube is 15.0 N⋅m2/C.

(b)

To determine

The net charge inside the cube.

(b)

Expert Solution

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

q=ε0ϕE (II)

Here, q is the charge enclosed within the Gaussian surface (cube), ε0 is the permittivity of free space and ϕE is the net electric flux through the cube.

Conclusion:

Substitute 8.85×10−12 C2/N⋅m2 for ε0, 15.0 N⋅m2/C for ϕE in Equation (II) to calculate q.

q=(8.85×10−12 C2/N⋅m2)(15.0 N⋅m2/C)=0.133×10−9 C

Therefore, the charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

(c)

To determine

Whether the net charge could be a point charge.

(c)

Expert Solution

Answer to Problem 32P

No, the net charge can’t be a point charge.

Explanation of Solution

No, single positive charge’s magnitude is q′=1.602×10−19 C, which is not equal to the obtained net charge q=0.133×10−9 C.

Conclusion:

Therefore, the net charge can’t be a point charge.

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Answer 2

Question

Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

(a)

Expert Solution

Answer to Problem 32P

Net electric flux through the cube is 15.0 N⋅m2/C.

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 32P

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

ϕE=∮E→⋅dA→ (I)

Here, E→ is the electric field, ϕE is the net flux through closed Gaussian surface and A→ is the surface area.

Write the expression for net electric flux through cube.

ϕE=ϕE(top)+ϕE(bottom)+ϕE(right)+ϕE(left)+ϕE(into)+ϕE(out) (II)

Conclusion:

Substitute 25.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(top).

ϕE(top)=EAcos(180°)=−(EA)=−(25.0 N/C)(1.00 m2)=−25.0 N⋅m2/C

Substitute 15.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(bottom)

ϕE(bottom)=EAcos(0°)=EA=(15.0 N/C)(1.00 m2)=15.0 N⋅m2/C

Substitute 35.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(right).

ϕE(right)=EAcos(180°)=−(EA)=−(35.0 N/C)(1.00 m2)=−35.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(left).

ϕE(left)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(into).

ϕE(into)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(out).

ϕE(out)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute −25.0 N⋅m2/C for ϕE(top), 15.0 N⋅m2/C for ϕE(bottom), −35.0 N⋅m2/C for ϕE(right), 20.0 N⋅m2/C for ϕE(left), ϕE(into) and ϕE(out) in equation (II) to calculate ϕE.

ϕE=(−25.0+15.0−35.0+20.0+20.0+20.0) N⋅m2/C=15.0 N⋅m2/C

Therefore, net electric flux through the cube is 15.0 N⋅m2/C.

(b)

To determine

The net charge inside the cube.

(b)

Expert Solution

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

q=ε0ϕE (II)

Here, q is the charge enclosed within the Gaussian surface (cube), ε0 is the permittivity of free space and ϕE is the net electric flux through the cube.

Conclusion:

Substitute 8.85×10−12 C2/N⋅m2 for ε0, 15.0 N⋅m2/C for ϕE in Equation (II) to calculate q.

q=(8.85×10−12 C2/N⋅m2)(15.0 N⋅m2/C)=0.133×10−9 C

Therefore, the charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

(c)

To determine

Whether the net charge could be a point charge.

(c)

Expert Solution

Answer to Problem 32P

No, the net charge can’t be a point charge.

Explanation of Solution

No, single positive charge’s magnitude is q′=1.602×10−19 C, which is not equal to the obtained net charge q=0.133×10−9 C.

Conclusion:

Therefore, the net charge can’t be a point charge.

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Answer 3

Question

Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

(a)

Expert Solution

Answer to Problem 32P

Net electric flux through the cube is 15.0 N⋅m2/C.

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 32P

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

ϕE=∮E→⋅dA→ (I)

Here, E→ is the electric field, ϕE is the net flux through closed Gaussian surface and A→ is the surface area.

Write the expression for net electric flux through cube.

ϕE=ϕE(top)+ϕE(bottom)+ϕE(right)+ϕE(left)+ϕE(into)+ϕE(out) (II)

Conclusion:

Substitute 25.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(top).

ϕE(top)=EAcos(180°)=−(EA)=−(25.0 N/C)(1.00 m2)=−25.0 N⋅m2/C

Substitute 15.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(bottom)

ϕE(bottom)=EAcos(0°)=EA=(15.0 N/C)(1.00 m2)=15.0 N⋅m2/C

Substitute 35.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(right).

ϕE(right)=EAcos(180°)=−(EA)=−(35.0 N/C)(1.00 m2)=−35.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(left).

ϕE(left)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(into).

ϕE(into)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(out).

ϕE(out)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute −25.0 N⋅m2/C for ϕE(top), 15.0 N⋅m2/C for ϕE(bottom), −35.0 N⋅m2/C for ϕE(right), 20.0 N⋅m2/C for ϕE(left), ϕE(into) and ϕE(out) in equation (II) to calculate ϕE.

ϕE=(−25.0+15.0−35.0+20.0+20.0+20.0) N⋅m2/C=15.0 N⋅m2/C

Therefore, net electric flux through the cube is 15.0 N⋅m2/C.

(b)

To determine

The net charge inside the cube.

(b)

Expert Solution

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

q=ε0ϕE (II)

Here, q is the charge enclosed within the Gaussian surface (cube), ε0 is the permittivity of free space and ϕE is the net electric flux through the cube.

Conclusion:

Substitute 8.85×10−12 C2/N⋅m2 for ε0, 15.0 N⋅m2/C for ϕE in Equation (II) to calculate q.

q=(8.85×10−12 C2/N⋅m2)(15.0 N⋅m2/C)=0.133×10−9 C

Therefore, the charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

(c)

To determine

Whether the net charge could be a point charge.

(c)

Expert Solution

Answer to Problem 32P

No, the net charge can’t be a point charge.

Explanation of Solution

No, single positive charge’s magnitude is q′=1.602×10−19 C, which is not equal to the obtained net charge q=0.133×10−9 C.

Conclusion:

Therefore, the net charge can’t be a point charge.

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Answer 4

Question

Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

(a)

Expert Solution

Answer to Problem 32P

Net electric flux through the cube is 15.0 N⋅m2/C.

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 32P

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

ϕE=∮E→⋅dA→ (I)

Here, E→ is the electric field, ϕE is the net flux through closed Gaussian surface and A→ is the surface area.

Write the expression for net electric flux through cube.

ϕE=ϕE(top)+ϕE(bottom)+ϕE(right)+ϕE(left)+ϕE(into)+ϕE(out) (II)

Conclusion:

Substitute 25.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(top).

ϕE(top)=EAcos(180°)=−(EA)=−(25.0 N/C)(1.00 m2)=−25.0 N⋅m2/C

Substitute 15.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(bottom)

ϕE(bottom)=EAcos(0°)=EA=(15.0 N/C)(1.00 m2)=15.0 N⋅m2/C

Substitute 35.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(right).

ϕE(right)=EAcos(180°)=−(EA)=−(35.0 N/C)(1.00 m2)=−35.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(left).

ϕE(left)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(into).

ϕE(into)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(out).

ϕE(out)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute −25.0 N⋅m2/C for ϕE(top), 15.0 N⋅m2/C for ϕE(bottom), −35.0 N⋅m2/C for ϕE(right), 20.0 N⋅m2/C for ϕE(left), ϕE(into) and ϕE(out) in equation (II) to calculate ϕE.

ϕE=(−25.0+15.0−35.0+20.0+20.0+20.0) N⋅m2/C=15.0 N⋅m2/C

Therefore, net electric flux through the cube is 15.0 N⋅m2/C.

(b)

To determine

The net charge inside the cube.

(b)

Expert Solution

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

q=ε0ϕE (II)

Here, q is the charge enclosed within the Gaussian surface (cube), ε0 is the permittivity of free space and ϕE is the net electric flux through the cube.

Conclusion:

Substitute 8.85×10−12 C2/N⋅m2 for ε0, 15.0 N⋅m2/C for ϕE in Equation (II) to calculate q.

q=(8.85×10−12 C2/N⋅m2)(15.0 N⋅m2/C)=0.133×10−9 C

Therefore, the charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

(c)

To determine

Whether the net charge could be a point charge.

(c)

Expert Solution

Answer to Problem 32P

No, the net charge can’t be a point charge.

Explanation of Solution

No, single positive charge’s magnitude is q′=1.602×10−19 C, which is not equal to the obtained net charge q=0.133×10−9 C.

Conclusion:

Therefore, the net charge can’t be a point charge.

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Answer 5

Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

(a)

Expert Solution

Answer to Problem 32P

Net electric flux through the cube is 15.0 N⋅m2/C.

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 32P

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

ϕE=∮E→⋅dA→ (I)

Here, E→ is the electric field, ϕE is the net flux through closed Gaussian surface and A→ is the surface area.

Write the expression for net electric flux through cube.

ϕE=ϕE(top)+ϕE(bottom)+ϕE(right)+ϕE(left)+ϕE(into)+ϕE(out) (II)

Conclusion:

Substitute 25.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(top).

ϕE(top)=EAcos(180°)=−(EA)=−(25.0 N/C)(1.00 m2)=−25.0 N⋅m2/C

Substitute 15.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(bottom)

ϕE(bottom)=EAcos(0°)=EA=(15.0 N/C)(1.00 m2)=15.0 N⋅m2/C

Substitute 35.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(right).

ϕE(right)=EAcos(180°)=−(EA)=−(35.0 N/C)(1.00 m2)=−35.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(left).

ϕE(left)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(into).

ϕE(into)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(out).

ϕE(out)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute −25.0 N⋅m2/C for ϕE(top), 15.0 N⋅m2/C for ϕE(bottom), −35.0 N⋅m2/C for ϕE(right), 20.0 N⋅m2/C for ϕE(left), ϕE(into) and ϕE(out) in equation (II) to calculate ϕE.

ϕE=(−25.0+15.0−35.0+20.0+20.0+20.0) N⋅m2/C=15.0 N⋅m2/C

Therefore, net electric flux through the cube is 15.0 N⋅m2/C.

(b)

To determine

The net charge inside the cube.

(b)

Expert Solution

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

q=ε0ϕE (II)

Here, q is the charge enclosed within the Gaussian surface (cube), ε0 is the permittivity of free space and ϕE is the net electric flux through the cube.

Conclusion:

Substitute 8.85×10−12 C2/N⋅m2 for ε0, 15.0 N⋅m2/C for ϕE in Equation (II) to calculate q.

q=(8.85×10−12 C2/N⋅m2)(15.0 N⋅m2/C)=0.133×10−9 C

Therefore, the charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

(c)

To determine

Whether the net charge could be a point charge.

(c)

Expert Solution

Answer to Problem 32P

No, the net charge can’t be a point charge.

Explanation of Solution

No, single positive charge’s magnitude is q′=1.602×10−19 C, which is not equal to the obtained net charge q=0.133×10−9 C.

Conclusion:

Therefore, the net charge can’t be a point charge.

Answer 6

Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

(a)

Expert Solution

Answer to Problem 32P

Net electric flux through the cube is 15.0 N⋅m2/C.

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 32P

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

ϕE=∮E→⋅dA→ (I)

Here, E→ is the electric field, ϕE is the net flux through closed Gaussian surface and A→ is the surface area.

Write the expression for net electric flux through cube.

ϕE=ϕE(top)+ϕE(bottom)+ϕE(right)+ϕE(left)+ϕE(into)+ϕE(out) (II)

Conclusion:

Substitute 25.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(top).

ϕE(top)=EAcos(180°)=−(EA)=−(25.0 N/C)(1.00 m2)=−25.0 N⋅m2/C

Substitute 15.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(bottom)

ϕE(bottom)=EAcos(0°)=EA=(15.0 N/C)(1.00 m2)=15.0 N⋅m2/C

Substitute 35.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(right).

ϕE(right)=EAcos(180°)=−(EA)=−(35.0 N/C)(1.00 m2)=−35.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(left).

ϕE(left)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(into).

ϕE(into)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(out).

ϕE(out)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute −25.0 N⋅m2/C for ϕE(top), 15.0 N⋅m2/C for ϕE(bottom), −35.0 N⋅m2/C for ϕE(right), 20.0 N⋅m2/C for ϕE(left), ϕE(into) and ϕE(out) in equation (II) to calculate ϕE.

ϕE=(−25.0+15.0−35.0+20.0+20.0+20.0) N⋅m2/C=15.0 N⋅m2/C

Therefore, net electric flux through the cube is 15.0 N⋅m2/C.

Answer 7

Chapter 24, Problem 32P

(a)

To determine

The net electric flux through the cube.

Answer 8

(a)

Expert Solution

Answer to Problem 32P

Net electric flux through the cube is 15.0 N⋅m2/C.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Below figure shows the electric field magnitude and its direction in all face’s of the cube.

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 24, Problem 32P

Figure (1)

From Figure (1), it is shown that, the electric fields are perpendicular to the faces of the cube. Therefore, use Gauss law for net flux through the closed Gaussian surface.

Write the expression for net flux through the closed Gaussian surface.

ϕE=∮E→⋅dA→ (I)

Here, E→ is the electric field, ϕE is the net flux through closed Gaussian surface and A→ is the surface area.

Write the expression for net electric flux through cube.

ϕE=ϕE(top)+ϕE(bottom)+ϕE(right)+ϕE(left)+ϕE(into)+ϕE(out) (II)

Conclusion:

Substitute 25.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(top).

ϕE(top)=EAcos(180°)=−(EA)=−(25.0 N/C)(1.00 m2)=−25.0 N⋅m2/C

Substitute 15.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(bottom)

ϕE(bottom)=EAcos(0°)=EA=(15.0 N/C)(1.00 m2)=15.0 N⋅m2/C

Substitute 35.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(right).

ϕE(right)=EAcos(180°)=−(EA)=−(35.0 N/C)(1.00 m2)=−35.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(left).

ϕE(left)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(into).

ϕE(into)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute 20.0 N/C for E, and L2 for A in equation (I) to calculate ϕE(out).

ϕE(out)=EAcos(0°)=EA=(20.0 N/C)(1.00 m2)=20.0 N⋅m2/C

Substitute −25.0 N⋅m2/C for ϕE(top), 15.0 N⋅m2/C for ϕE(bottom), −35.0 N⋅m2/C for ϕE(right), 20.0 N⋅m2/C for ϕE(left), ϕE(into) and ϕE(out) in equation (II) to calculate ϕE.

ϕE=(−25.0+15.0−35.0+20.0+20.0+20.0) N⋅m2/C=15.0 N⋅m2/C

Therefore, net electric flux through the cube is 15.0 N⋅m2/C.

Answer 13

(b)

To determine

The net charge inside the cube.

(b)

Expert Solution

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

q=ε0ϕE (II)

Here, q is the charge enclosed within the Gaussian surface (cube), ε0 is the permittivity of free space and ϕE is the net electric flux through the cube.

Conclusion:

Substitute 8.85×10−12 C2/N⋅m2 for ε0, 15.0 N⋅m2/C for ϕE in Equation (II) to calculate q.

q=(8.85×10−12 C2/N⋅m2)(15.0 N⋅m2/C)=0.133×10−9 C

Therefore, the charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Answer 14

(b)

To determine

The net charge inside the cube.

Answer 15

(b)

Expert Solution

Answer to Problem 32P

The charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression for the charge enclosed within the Gaussian surface (cube) is.

q=ε0ϕE (II)

Here, q is the charge enclosed within the Gaussian surface (cube), ε0 is the permittivity of free space and ϕE is the net electric flux through the cube.

Conclusion:

Substitute 8.85×10−12 C2/N⋅m2 for ε0, 15.0 N⋅m2/C for ϕE in Equation (II) to calculate q.

q=(8.85×10−12 C2/N⋅m2)(15.0 N⋅m2/C)=0.133×10−9 C

Therefore, the charge enclosed within the Gaussian surface (cube) is 0.133×10−9 C.

Answer 20

(c)

To determine

Whether the net charge could be a point charge.

(c)

Expert Solution

Answer to Problem 32P

No, the net charge can’t be a point charge.

Explanation of Solution

No, single positive charge’s magnitude is q′=1.602×10−19 C, which is not equal to the obtained net charge q=0.133×10−9 C.

Conclusion:

Therefore, the net charge can’t be a point charge.

Answer 21

(c)

To determine

Whether the net charge could be a point charge.

Answer 22

(c)

Expert Solution

Answer to Problem 32P

No, the net charge can’t be a point charge.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

No, single positive charge’s magnitude is q′=1.602×10−19 C, which is not equal to the obtained net charge q=0.133×10−9 C.

Conclusion:

Therefore, the net charge can’t be a point charge.

Answer 27

Ch. 24.1 - Suppose a point charge is located at the center of...Ch. 24.2 - If the net flux through a gaussian surface is...Ch. 24 - Prob. 1OQ Ch. 24 - Prob. 2OQ Ch. 24 - Prob. 3OQ Ch. 24 - Prob. 4OQ Ch. 24 - Prob. 5OQ Ch. 24 - Prob. 6OQ Ch. 24 - Prob. 7OQ Ch. 24 - Prob. 8OQ

The net electric flux through the cube.

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The net electric flux through the cube.

Answer to Problem 32P

Explanation of Solution

The net charge inside the cube.

Answer to Problem 32P

Explanation of Solution

Whether the net charge could be a point charge.

Answer to Problem 32P

Explanation of Solution

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