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Chapter 18: Problem 49

A vertical wall \((5.9 \mathrm{~m} \times 2.5 \mathrm{~m})\) in a house faces due east. A uniform electric field has a magnitude of \(150 \mathrm{~N} / \mathrm{C}\). This field is parallel to the ground and points \(35^{\circ}\) north of east. What is the electric flux through the wall?

Short Answer

Expert verified
The electric flux through the wall is approximately \(1268.49 \, \text{N} \cdot \text{m}^2/\text{C}\).

Step by step solution

01

Calculate the Area of the Wall

The first step is to calculate the area of the wall. Use the formula for area \( \text{Area} = \text{Height} \times \text{Width} \). Substitute the values to find the area: \[ \text{Area} = 5.9 \, \text{m} \times 2.5 \, \text{m} = 14.75 \, \text{m}^2. \]
02

Determine the Component of the Electric Field Perpendicular to the Wall

The electric flux depends on the component of the electric field that is perpendicular to the surface. Since the electric field is 35掳 north of east, calculate the component perpendicular to the eastward direction. Use the cosine function for this: \[ E_\perp = E \cos(90^\circ - 35^\circ) = 150 \, \text{N/C} \times \cos(55^\circ). \]
03

Calculate the Value of Perpendicular Electric Field

Calculate the value from Step 2. First, find \( \cos(55^\circ) \) and then compute: \[ E_\perp = 150 \, \text{N/C} \times 0.5736 = 86.04 \, \text{N/C}. \]
04

Calculate the Electric Flux Through the Wall

The electric flux \( \Phi \) is given by the product of the perpendicular electric field component and the area of the surface: \( \Phi = E_\perp \cdot A \). Substitute the values we have: \[ \Phi = 86.04 \, \text{N/C} \times 14.75 \, \text{m}^2 = 1268.49 \, \text{N} \cdot \text{m}^2/\text{C}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field
An electric field is a region around a charged particle where a force would be exerted on other charged particles. For our exercise, the electric field has a magnitude of 150 N/C and is parallel to the ground, pointing 35掳 north of east. This means that the field not only has a strength (magnitude) but also a specific direction. Understanding both of these aspects is key in calculating related physical concepts, such as electric flux.
Electric Flux Calculation
Electric flux is a measure of the quantity of electric field lines passing through a surface. It helps us understand how much of the field interacts with a given area. In this exercise, after determining the perpendicular component of the electric field, the flux is calculated using the formula:
  • \( \Phi = E_\perp \cdot A \)
Where \( E_\perp \) is the component of the electric field perpendicular to the wall, and \( A \) is the area of the wall. By multiplying these, we get a measure of the flux through the wall.
Perpendicular Component
The perpendicular component of the electric field is crucial when calculating flux. Since the field is not directly perpendicular to the wall, we need to find the component that is. This involves using trigonometry:
  • The given field is at an angle of 35掳 north of east.
  • To find the component perpendicular to the eastward-facing wall, calculate \( E_\perp = E \cos(90^\circ - 35^\circ) \).
  • This simplifies to \( 150 \, \text{N/C} \times \cos(55^\circ) \), resulting in 86.04 N/C.
This component helps us understand how much of the field actually interacts with the wall for flux calculation.
Wall Area
The area of the wall is an essential factor in determining the electric flux. To calculate it:
  • Use the formula \( \text{Area} = \text{Height} \times \text{Width} \).
  • In our exercise, this is \( 5.9 \, \text{m} \times 2.5 \, \text{m} = 14.75 \, \text{m}^2 \).
Knowing the area helps us apply the electric field value over a specific surface, making the flux calculation possible. Larger areas generally result in more flux, assuming the electric field component remains constant.

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