Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Textbook Question
Chapter 25, Problem 47PQ

The infinite sheets in Figure P25.47 are both positively charged. The sheet on the left has a uniform surface charge density of 48.0 μC/m2, and the one on the right has a uniform surface charge density of 24.0 μC/m2.

  1. a. What are the magnitude and direction of the net electric field at points A, B, and C?
  2. b. What is the force exerted on an electron placed at points A, B, and C?

Chapter 25, Problem 47PQ, The infinite sheets in Figure P25.47 are both positively charged. The sheet on the left has a

FIGURE P25.47

(a)

Expert Solution
Check Mark
To determine

The magnitude and direction of the net electric fields at points A, B and C.

Answer to Problem 47PQ

The magnitude and direction of the net electric fields at points A, B and C are 4.07×106i^N/C_, +1.36×106i^N/C_ and +4.07×106i^N/C_ respectively.

Explanation of Solution

Write the expression for finding the electric field due to first sheet.

    E1=σ12ε0                                                                                                         (I)

Here, σ1 is the surface charge density of the first sheet and ε0 is a constant.

Write the expression for finding the electric field due to second sheet.

    E2=σ22ε0                                                                                                        (II)

Here, σ2 is the surface charge density of the second sheet and ε0 is a constant.

The following figure shows the direction and the sum of the fields at points A, B and C.

Physics for Scientists and Engineers: Foundations and Connections, Chapter 25, Problem 47PQ

Write the expression to find the electric field at point A.

    EA=E1i^E2i^                                                                                             (III)

Write the expression to find the electric field at point B.

    EB=+E1i^E2i^                                                                                             (IV)

Write the expression to find the electric field at point C.

    EC=+E1i^+E2i^                                                                                              (V)

Conclusion:

Substitute 48.0×106C/m2 for σ1 and 8.85×1012C2/Nm2 for ε0 in equation (I).

    E1=48.0×106C/m22(8.85×1012C2/Nm2)=2.71×106N/C

Substitute 24.0×106C/m2 for σ1 and 8.85×1012C2/Nm2 for ε0 in equation (II).

    E2=24.0×106C/m22(8.85×1012C2/Nm2)=1.36×106N/C

Substitute 2.71×106N/C for E1 and 1.36×106N/C for E2 in equation (III).

    EA=2.71×106N/Ci^1.36×106N/Ci^=4.07×106i^N/C

Substitute 2.71×106N/C for E1 and 1.36×106N/C for E2 in equation (IV).

    EB=+2.71×106N/Ci^1.36×106N/Ci^=+1.36×106i^N/C

Substitute 2.71×106N/C for E1 and 1.36×106N/C for E2 in equation (IV).

    EC=+2.71×106N/Ci^+1.36×106N/Ci^=+4.07×106i^N/C

Therefore, the magnitude and direction of the net electric fields at points A, B and C are 4.07×106i^N/C_, +1.36×106i^N/C_ and +4.07×106i^N/C_ respectively.

(b)

Expert Solution
Check Mark
To determine

The force exerted on an electron placed at points A, B and C.

Answer to Problem 47PQ

The force exerted on an electron placed at points A, B and C are 6.51×1013i^N_, 2.17×1013i^N_ and 6.51×1013i^N_ respectively.

Explanation of Solution

Write the expression for the force exerted on the electron.

    F=eE                                                                                                         (VI)

Here, e is the charge of the electron.

Conclusion:

Substitute 4.07×106i^N/C for EA and 1.6×1019C for e in equation (VI).

    FA=eEA=1.6×1019C(4.07×106i^N/C)=6.51×1013i^N

Substitute +1.36×106i^N/C for EB and 1.6×1019C for e in equation (VI).

    FB=eEB=1.6×1019C(+1.36×106i^N/C)=2.17×1013i^N

Substitute +4.07×106i^N/C for EC and 1.6×1019C for e in equation (VI).

    FC=eEC=1.6×1019C(+4.07×106i^N/C)=6.51×1013i^N

Therefore, the force exerted on an electron placed at points A, B and C are 6.51×1013i^N_, 2.17×1013i^N_ and 6.51×1013i^N_ respectively.

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Chapter 25 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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