Chapter 22: Problem 36
In a region of space there is an electric field \(\overrightarrow{E}\) that is in the z-direction and that has magnitude \(E =\) [964 N/(C \(\cdot\) m)]\(x\). Find the flux for this field through a square in the \(xy\)-plane at \(z =\) 0 and with side length 0.350 m. One side of the square is along the \(+x\)-axis and another side is along the \(+y\)-axis.
Short Answer
Step by step solution
Understanding Electric Flux
Calculate Area of the Square
Solve for Electric Field at the Position
Final Electric Flux Calculation
Conclusion and Considerations
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electric Field
- **Directionality**: The field's direction is along the positive z-axis.
- **Magnitude dependence**: It varies linearly with the x-coordinate, demonstrating a spatial change in field strength.
Area Vector
\[ A = (0.350 \, \text{m})^2 = 0.1225 \, \text{m}^2 \]
- **Orientation**: The area vector is crucial for visualizing the interaction with the electric field, especially when establishing the angle \( \theta \) with the field vector.
- **Area Calculation**: The calculation of the area is straightforward for regular shapes like this square: the side length squared.
Dot Product
\[ \Phi_E = \overrightarrow{E} \cdot \overrightarrow{A} = EA \cos\theta \]
- **Angle \( \theta \) Importance**: This angle is between the electric field vector and the area vector. In this exercise, \( \theta = 0^\circ \) since both vectors are along the z-axis, making \( \cos\theta = 1 \).
- **Outcome**: With \( \cos\theta = 1 \), the dot product simplifies to the product of their magnitudes: \( EA \).
Surface Integration
- **Approximations**: For small surfaces or minor field variations, using an average electric field value is sufficient.
- **Integration Necessity**: This becomes crucial when the field greatly varies across the surface or when the shape is complex.