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The square surface shown in the figure measures 3.5 mm on each side. It is immersed in a uniform electric field with magnitude E= 1400 N/C and with field lines at an angle of 35° with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. Normal 4

The square surface shown in the figure measures 3.5 mm on each side. It is immersed in a uniform electric field with magnitude E= 1400 N/C and with field lines at an angle of 35° with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. Normal 4

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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### Electric Flux Through a Surface The square surface shown in the figure measures 3.5 mm on each side. It is immersed in a uniform electric field with magnitude \( E = 1400 \) N/C and with field lines at an angle of \( 35^\circ \) with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. #### Explanation: 1. **Square Surface**: Appears in the middle of the diagram, designated in green, with a side length of 3.5 mm. 2. **Normal Line**: Represented by a downward arrow perpendicular to the square surface. 3. **Electric Field Lines**: Indicated by slanted lines forming an angle with the normal line. 4. **Angle (θ)**: Defined as the angle between the normal and the electric field lines, given as \( 35^\circ \). #### Calculation: To calculate the electric flux (\( \Phi_E \)) through the surface: \[ \Phi_E = E \cdot A \cdot \cos(\theta) \] Where: - \( E \) is the electric field magnitude. - \( A \) is the area of the surface. - \( \theta \) is the angle between the normal to the surface and the electric field. 1. Calculate the area (\( A \)) of the square surface: \[ A = \text{side}^2 = (3.5 \, \text{mm})^2 = (3.5 \times 10^{-3} \, \text{m})^2 = 12.25 \times 10^{-6} \, \text{m}^2 \] 2. Insert values into the electric flux equation: \[ \Phi_E = (1400 \, \text{N/C}) \times (12.25 \times 10^{-6} \, \text{m}^2) \times \cos(35^\circ) \] 3. Calculate \( \cos(35^\circ) \): \[ \cos(35^\circ) \approx 0.819 \] 4. Continue with the calculation: \[ \Phi_E = 1400 \times 12.25 \times 10^{-6} \times 0.819 \] \[ \Phi_E =
Transcribed Image Text:### Electric Flux Through a Surface The square surface shown in the figure measures 3.5 mm on each side. It is immersed in a uniform electric field with magnitude \( E = 1400 \) N/C and with field lines at an angle of \( 35^\circ \) with a normal to the surface, as shown. Take that normal to be "outward," as though the surface were one face of a box. Calculate the electric flux through the surface. #### Explanation: 1. **Square Surface**: Appears in the middle of the diagram, designated in green, with a side length of 3.5 mm. 2. **Normal Line**: Represented by a downward arrow perpendicular to the square surface. 3. **Electric Field Lines**: Indicated by slanted lines forming an angle with the normal line. 4. **Angle (θ)**: Defined as the angle between the normal and the electric field lines, given as \( 35^\circ \). #### Calculation: To calculate the electric flux (\( \Phi_E \)) through the surface: \[ \Phi_E = E \cdot A \cdot \cos(\theta) \] Where: - \( E \) is the electric field magnitude. - \( A \) is the area of the surface. - \( \theta \) is the angle between the normal to the surface and the electric field. 1. Calculate the area (\( A \)) of the square surface: \[ A = \text{side}^2 = (3.5 \, \text{mm})^2 = (3.5 \times 10^{-3} \, \text{m})^2 = 12.25 \times 10^{-6} \, \text{m}^2 \] 2. Insert values into the electric flux equation: \[ \Phi_E = (1400 \, \text{N/C}) \times (12.25 \times 10^{-6} \, \text{m}^2) \times \cos(35^\circ) \] 3. Calculate \( \cos(35^\circ) \): \[ \cos(35^\circ) \approx 0.819 \] 4. Continue with the calculation: \[ \Phi_E = 1400 \times 12.25 \times 10^{-6} \times 0.819 \] \[ \Phi_E =
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