Problem 1 - University Physics with Modern Physics

Question

22.1. A flat sheet of paper of area 0.250 \(\mathrm{m}^{2}\) is oriented so that the normal to the sheet is at an angle of \(60^{\circ}\) to a uniform electric field of magnitude 14 \(\mathrm{N} / \mathrm{C}\) (a) Find the magnitude of the electric flux through the sheet. (b) Does the answer to part (a) depend on the shape of the sheet? Why or why not?(c) For what angle \(\phi\) between the normal to the sheet and the electric field is the magnitude of the flux through the sheet (i) largest and (ii) smallest? Explain your answers.

Step-by-Step Solution

Answer

(a) 1.75 Nm²/C. (b) No, it doesn't. (c) Largest at 0°, smallest at 90°.

Step 1: Understanding the Concepts

Electric flux through a surface is defined by the equation \( \Phi = E \cdot A \cdot \cos(\theta) \), where \( E \) is the electric field strength, \( A \) is the area of the surface, and \( \theta \) is the angle between the normal to the surface and the electric field. Our goal is to apply this formula to solve the problem.

Step 2: Calculate Electric Flux

Given \( E = 14 \, \mathrm{N/C} \), \( A = 0.250 \, \mathrm{m^2} \), and \( \theta = 60^{\circ} \), we plug these values into the flux formula: \[ \Phi = 14 \, \mathrm{N/C} \times 0.250 \, \mathrm{m^2} \times \cos(60^{\circ}) \]\( \cos(60^{\circ}) = 0.5 \), therefore: \[ \Phi = 14 \, \mathrm{N/C} \times 0.250 \, \mathrm{m^2} \times 0.5 = 1.75 \, \mathrm{N} \cdot \mathrm{m^2/C} \].

Step 3: Does Shape Affect Flux?

The answer to part (a) does not depend on the shape of the sheet. Electric flux depends only on the area, the electric field, and the angle between the field and the normal; it does not matter how the area is distributed.

Step 4: Determine Angles for Maximum and Minimum Flux

(i) The electric flux is largest when \( \theta = 0^{\circ} \). This is because \( \cos(0^{\circ}) = 1 \), so \( \Phi = E \cdot A \cdot 1 = E \cdot A \), giving the maximum possible flux.(ii) The electric flux is smallest when \( \theta = 90^{\circ} \) because \( \cos(90^{\circ}) = 0 \), resulting in zero flux through the sheet.

Key Concepts

Electric Field

The electric field is a vector field that represents the force that a positively charged particle would feel at any given point in space. This force is expressed in Newtons per Coulomb (N/C).
In exercises like the one mentioned, the electric field (\( E = 14 \, \mathrm{N/C} \)) influences everything concerning the flux through a surface.

• The stronger the field, the greater the force exerted on charges within it.
• The direction of the electric field is crucial, affecting how it interacts with surfaces like the paper described.

Recognizing the characteristics of the electric field can help you predict and calculate interactions with different materials and objects in its path.

Surface Area

The surface area is a measure of the size of an object's surface. With regards to the electric flux, the surface area\( A = 0.250 \, \mathrm{m^2} \)) specifically determines how much of the field passes through.
* Area only affects the amount of field interacting with the surface, not its shape.* It proves crucial in the formula for electric flux,\( \Phi = E \cdot A \cdot \cos(\theta) \).In application, surface area acts as a gateway for the field lines. The greater the area, the more field lines can traverse.

Angle of Incidence

The angle of incidence, denoted as \( \theta \), represents the inclination between the normal to the surface and the electric field. In this example, the angle is \(60^{\circ}\). Through the \( \cos(\theta) \) term, this angle impacts the volume of field lines penetrating the surface.

• A smaller angle signifies more field lines entering, maximizing flux.
• When \( \theta = 90^{\circ} \), no field lines pass through, making flux zero.

Recognizing these effects helps optimize the orientation of materials for specific exposure to electric fields.

Flux Calculation

Flux calculation ties all these factors together. By formula, \( \Phi = E \cdot A \cdot \cos(\theta) \), where each term - electric field, surface area, and angle - plays a distinct role.
* The given parameters, \( E = 14 \, \mathrm{N/C} \), \( A = 0.250 \, \mathrm{m^2} \), and \( \theta = 60^{\circ} \) lead to the flux \( \Phi = 1.75 \, \mathrm{N} \cdot \mathrm{m^2/C} \).* Changes in any parameter directly alter the flux.Understanding how to calculate flux enables you to predict electric field behavior relative to surfaces, which is critical in many fields such as electronics and physics.