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
Physics for Scientists and Engineers
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ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter23: Continuous Charge Distributions And Gauss's Law
Section: Chapter Questions
Problem 21P: (a) A panicle with charge q is located a distance d from an infinite plane. Determine the electric...
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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 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe462800b-ac16-4590-bafb-82d68c005ec5%2F8063d3eb-ac17-463b-971a-1f0365b617af%2Fayzzz5_processed.jpeg&w=3840&q=75)
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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