EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9781305804470
Author: Jewett
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 24, Problem 24.32P

Assume the magnitude of the electric field on each face of the cube of edge L = 1.00 m in Figure P23.32 is uniform and the directions of the fields on each face are as indicated. Find (a) the net electric flux through the cube and (b) the net charge inside the cube. (c) Could the net charge he a single point charge?

Figure P23.32

Chapter 24, Problem 24.32P, Assume the magnitude of the electric field on each face of the cube of edge L = 1.00 m in Figure

(a)

Expert Solution
Check Mark
To determine

The electric flux through the cube.

Answer to Problem 24.32P

The electric flux through the cube is 15.0Nm2/C.

Explanation of Solution

The length of the edge of the cube is 1.00m, the electric field on the each face of the cube is 25.0N/C, 15.0N/C, 35.0N/C, 20.0N/C and 20.0N/C.

The area vector is always out of face. The electric field going into the face of the cube makes 180° with the area vector and electric field going out of the face of the cube makes 0° with the area vector.

Write the expression for total flux through the cube

    ϕE=EAcosθ

Here, ϕE is the net flux, A is the area of the cylinder, E is the electric field and θ is the angle between the electric field lines and the cube.

Write the area of the cube

    A=L2

Here, L is the length of the edge of the cube.

Substitute L2 for A in the above equation.

    ϕE=EL2cosθ=[E1L2cos(180°)+E2L2cos(180°)+E3L2cos(0°)+E4L2cos(0°)+E5L2cos(0°)+E6L2cos(0°)]

Here, E1, E2, E3, E4, E5 and E6 are the flux through the face 1, 2, 3, 4, 5 and 6 of the cube respectively.

Conclusion:

Substitute 1.00m for L, 25.0N/C for E1, 35.0N/C for E2, 15.0N/C for E3, 20.0N/C for E4, 20.0N/C for E5 and 20.0N/C for E6 in above equation to find ϕE.

    ϕE=[(25.0N/C)(1.00m)2cos(180°)+(35.0N/C)(1.00m)2cos(180°)+(15.0N/C)(1.00m)2cos(0°)+(20.0N/C)(1.00m)2cos(0°)+(20.0N/C)(1.00m)2cos(0°)+(20.0N/C)(1.00m)2cos(0°)]={(1.00m)2[(25.0N/C)+(35.0N/C)+(15.0N/C)+(20.0N/C)+(20.0N/C)+(20.0N/C)]=15.0Nm2/C

Therefore, the electric flux through the cube is 15.0Nm2/C.

(b)

Expert Solution
Check Mark
To determine

The net charge inside the cube.

Answer to Problem 24.32P

The net charge inside the cube is 1.33×1010C.

Explanation of Solution

The length of the edge of the cube is 1.00m, the electric field on the each face of the cube is 25.0N/C, 15.0N/C, 35.0N/C, 20.0N/C and 20.0N/C.

Write the expression for flux through the cube by Gauss law

    ϕE=EdA=qclosedε0qclosed=ϕEε0

Here, ε0 is the permittivity of free space and qclosed is the charge enclosed by the cube.

Conclusion:

Substitute 15.0Nm2/C for ϕE and 8.85×1012F/m for ε0 in above equation to find qclosed.

    qclosed=(15.0Nm2/C)(8.85×1012F/m)=1.33×1010C

Therefore, the net charge inside the cube is 1.33×1010C.

(c)

Expert Solution
Check Mark
To determine

The number of charges inside the cube.

Answer to Problem 24.32P

The net charge is not a single point charge.

Explanation of Solution

The electric charge is the physical property of the matter that causes it to experience a force when placed in an electromagnetic field.

The net charge is not a single point charge as it does not produce a uniform field on the face of cubes. The equipotential surface of the single point charge is spherical. So the net charge is not a single point charge.

Conclusion:

Therefore, the net charge is not a single point charge.

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Chapter 24 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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