College Physics
College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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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.
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**Problem Statement:** The figure below shows a uniform electric field of magnitude \( E = 480 \, \text{N/C} \) making an angle of \( \phi = 66.5^\circ \) with a flat surface of area \( A = 3.10 \, \text{m}^2 \). What is the magnitude of the electric flux through this surface (in \( \text{N} \cdot \text{m}^2/\text{C} \))? **Diagram Explanation:** - The diagram depicts a flat rectangular surface with lines representing a uniform electric field, denoted as \( \vec{E} \). - The electric field lines intersect the surface at an angle, \( \phi = 66.5^\circ \), from the normal (perpendicular) to the surface. - The angle is clearly marked between the normal to the surface and the direction of the electric field \( \vec{E} \). To calculate the electric flux (\( \Phi \)), use the formula: \[ \Phi = E \cdot A \cdot \cos(\phi) \] where - \( E = 480 \, \text{N/C} \) - \( A = 3.10 \, \text{m}^2 \) - \( \phi = 66.5^\circ \)
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Transcribed Image Text:**Problem Statement:** The figure below shows a uniform electric field of magnitude \( E = 480 \, \text{N/C} \) making an angle of \( \phi = 66.5^\circ \) with a flat surface of area \( A = 3.10 \, \text{m}^2 \). What is the magnitude of the electric flux through this surface (in \( \text{N} \cdot \text{m}^2/\text{C} \))? **Diagram Explanation:** - The diagram depicts a flat rectangular surface with lines representing a uniform electric field, denoted as \( \vec{E} \). - The electric field lines intersect the surface at an angle, \( \phi = 66.5^\circ \), from the normal (perpendicular) to the surface. - The angle is clearly marked between the normal to the surface and the direction of the electric field \( \vec{E} \). To calculate the electric flux (\( \Phi \)), use the formula: \[ \Phi = E \cdot A \cdot \cos(\phi) \] where - \( E = 480 \, \text{N/C} \) - \( A = 3.10 \, \text{m}^2 \) - \( \phi = 66.5^\circ \)
Expert Solution
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Step 1: Formula for electric flux through a surface

The electric flux (straight capital phi subscript E) through a surface is given by the formula:

straight capital phi subscript E equals E times A times cos invisible function application not stretchy left parenthesis theta not stretchy right parenthesis

where:
E is the magnitude of the electric field,
A is the area of the surface, and
theta is the angle between the electric field and the normal to the surface.

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