An constant electric field, **E** = (3**î** + 4**ĵ**) N/C, goes through a surface with area **A** = (8**î** - 6**k̂**) m². (This surface can also be expressed as an area of 10 m² with the direction of the unit vector (0.8**î** - 0.6**k̂**). What is the magnitude of the electric flux through this area?- **A)** 24 N·m²/C- **B)** 48 N·m²/C- **C)** 0.24 N·m²/C- **D)** 0.48 N·m²/C- **E)** 0

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An constant electric field, E = (31+ 4j) N/C, goes through a surface with area A = (8f – ók) m 2. (This surface can also be expressed as an area of 10 m2 with the direction of the unit vector ( 0.8Î – 0, 6k). What is the magnitude of the electric flux through this area? A) 24 N x m 2/C B) 48 N x m 2/C (c) 0.24 N × m 2/C D) 0.48 N × m ²/C E

An constant electric field, E = (31+ 4j) N/C, goes through a surface with area A = (8f – ók) m 2. (This surface can also be expressed as an area of 10 m2 with the direction of the unit vector ( 0.8Î – 0, 6k). What is the magnitude of the electric flux through this area? A) 24 N x m 2/C B) 48 N x m 2/C (c) 0.24 N × m 2/C D) 0.48 N × m ²/C E

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter6: Gauss's Law

Section: Chapter Questions

Problem 55P: The electric field 10.0 cm from the surface of a copper ball of radius 5.0 cm is directed toward the...

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

An constant electric field, **E** = (3**î** + 4**ĵ**) N/C, goes through a surface with area **A** = (8**î** - 6**k̂**) m². (This surface can also be expressed as an area of 10 m² with the direction of the unit vector (0.8**î** - 0.6**k̂**). What is the magnitude of the electric flux through this area?

- **A)** 24 N·m²/C
- **B)** 48 N·m²/C
- **C)** 0.24 N·m²/C
- **D)** 0.48 N·m²/C
- **E)** 0

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