The net electric flux through the cube.

Question

Want to see more full solutions like this?

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 Technology Update, 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.

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

A Gaussian surface in the form of a hemisphere of radius R = 5.84 cm lies in a uniform electric field of magnitude E = 2.20 N/C. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. (a) What is the flux through the base? ______________N · m2/C(b) What is the flux through the curved portion of the surface? _______________N · m2/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.

A square that has 26 cm long edges is centered on the x axis in a region where there exists a uniform electric field given by = (2.00 kN/C)î. (a) What is the electric flux of this electric field through the surface of a square if the normal to the surface is in the +x direction? N · m2/C(b) What is the electric flux through the same square surface if the normal to the surface makes a 60° angle with the y axis and an angle of 90° with the z axis? N · m2/C

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 Technology Update, 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.

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

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 Technology Update, 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.

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

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 Technology Update, 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.

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

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 Technology Update, 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 Technology Update, 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 Technology Update, 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.

Concept explainers

Videos

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

Want to see more full solutions like this?

Chapter 24 Solutions