/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 A flat sheet is in the shape of ... [FREE SOLUTION] | 91Ó°ÊÓ

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A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude 90.0 N/C that is directed at 20\(^\circ\) from the plane of the sheet (\(\textbf{Fig. E22.2}\)). Find the magnitude of the electric flux through the sheet.

Short Answer

Expert verified
The electric flux through the sheet is approximately 7.39 N·m²/C.

Step by step solution

01

Determine the Area of the Sheet

The area of a rectangle is calculated using the formula \( A = l \times w \), where \( l \) is the length and \( w \) is the width. For this rectangular sheet, substitute the given values: \( l = 0.400 \, \text{m} \) and \( w = 0.600 \, \text{m} \). So, \( A = 0.400 \, \text{m} \times 0.600 \, \text{m} = 0.240 \, \text{m}^2 \).
02

Understand the Concept of Electric Flux

Electric flux (\( \Phi_E \)) through a surface due to an electric field is given by the formula \( \Phi_E = E \cdot A \cdot \cos\theta \), where \( E \) is the electric field strength, \( A \) is the area through which the field lines pass, and \( \theta \) is the angle between the electric field and the normal (perpendicular) to the surface.
03

Determine the Angle in the Formula

In this problem, the angle given is 20 degrees from the plane of the sheet. Therefore, to find \( \theta \) in the electric flux formula, which is the angle with the normal, calculate \( \theta = 90^\circ - 20^\circ = 70^\circ \).
04

Calculate the Electric Flux

Substitute the known values into the electric flux formula. Given \( E = 90.0 \, \text{N/C} \), \( A = 0.240 \, \text{m}^2 \), and \( \theta = 70^\circ \), calculate:\[ \Phi_E = 90.0 \, \text{N/C} \times 0.240 \, \text{m}^2 \times \cos(70^\circ) \]Using a calculator, \( \cos(70^\circ) \approx 0.3420 \). Thus, \[ \Phi_E = 90.0 \times 0.240 \times 0.3420 = 7.3896 \, \text{N}\cdot\text{m}^2/\text{C} \].

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in physics and is crucial for understanding electric phenomena. An electric field is a region around a charged object where a force would be exerted on other charged objects. It is represented by the symbol \( E \) and is measured in newtons per coulomb (N/C).
Electric fields are generated by electric charges and spread throughout space, affecting other charges/particles within their influence. A uniform electric field means that the field strength and direction are constant regardless of location.
In practical terms, a uniform electric field can be visualized using parallel lines representing the direction and magnitude of the field. These lines are equally spaced, showing that every part of the field has the same strength.
  • The magnitude of the electric field, \( E \), determines the force experienced by a charge placed in the field.
  • The direction of the electric field is taken as the direction in which a positive test charge would move.
  • Uniformity in an electric field implies uniform impact across an immersed body or surface, such as the rectangular sheet in our exercise.
Understanding electric fields is fundamental for topics ranging from electrostatics to applications in electromagnetic devices.
Angle Calculation
When dealing with the intersection of fields and surfaces, understanding angles in these interactions is key. The angle \( \theta \) in this context represents the deviation between the direction of the electric field and the perpendicular line to the surface (known as the "normal").
If an angle is given relative to the plane of a surface, as is often the case for objects partially submerged in electric or magnetic fields, you must convert this angle to one relative to the surface's normal for calculations.
Given: the field is at an angle of 20° from the plane of the sheet. Thus:
  • Consider the normal, a line perpendicular to the sheet: it makes a 90° angle with the plane of the sheet.
  • Calculate the angle with the normal as \( \theta = 90° - \text{angle from the plane} \).
  • For our exercise, \( \theta = 90° - 20° = 70° \).
This conversion is essential because most physics equations involving flux, including our electric flux formula, use the angle between the field and the normal. Recognizing these angle transformations simplifies understanding and solving related problems.
Area of Rectangle
The area of a rectangle is one of the basic geometric properties used to determine the extent of a surface in plane geometry. This calculation is necessary when assessing surfaces over which forces like electric fields act.
To find the area of a rectangle, use the formula \( A = l \times w \), where:
  • \( l \) is the length of the rectangle, and
  • \( w \) is the width of the rectangle.
In our scenario, with values \( l = 0.400 \, \text{m} \) and \( w = 0.600 \, \text{m} \), we calculate:
\[ A = 0.400 \, \text{m} \times 0.600 \, \text{m} = 0.240 \, \text{m}^2 \]
Simply multiply the side lengths together to find the area, which provides the two-dimensional measure needed to evaluate processes like flux through a surface. This area value is then used in conjunction with the angle measurement and field strength to determine the total effect, i.e., the electric flux.
Cosine Function
The cosine function is a trigonometric function and an essential tool when working with angles involving components of vectors and fields. In physics, the cosine function \( \cos(\theta) \) helps translate angles into scalar components for calculations such as those involving forces and flux.
The cosine of an angle, \( \theta \), gives the ratio of the adjacent side to the hypotenuse in a right triangle. This property is particularly important in projecting vector quantities along specific known directions.
  • The cosine function takes an angle and returns a value between -1 and 1.
  • For angles in the first quadrant (0° to 90°), the cosine equals the horizontal line component of a given angle.
  • In the provided exercise, \( \cos(70°) \approx 0.3420 \), which we use to determine how much of the electric field interacts directly along the normal to the surface.
The relevance of \( \cos(\theta) \) in the electric flux formula \( \Phi_E = E \cdot A \cdot \cos(\theta) \) is to assist in correcting for the field interaction angle, affecting how much field passes directly through the surface. The presence of the cosine term helps adjust for the field's relationship with the surface, presenting an accurate depiction of the influence a field has across a particular orientation.

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Most popular questions from this chapter

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area \(\sigma =\) 5.00 \(\times\) 10\(^{-6}\) C/m\(^2\). (a) A small sphere of mass \(m =\) 8.00 \(\times\) 10\(^{-6}\) kg and charge \(q\) is placed 3.00 cm above the sheet of charge and then released from rest. (a) If the sphere is to remain motionless when it is released, what must be the value of \(q\)? (b) What is \(q\) if the sphere is released 1.50 cm above the sheet?

A very small object with mass 8.20 \(\times\) 10\(^{-9}\) kg and positive charge 6.50 \(\times\) 10\(^{-9}\) C is projected directly toward a very large insulating sheet of positive charge that has uniform surface charge density 5.90 \(\times\) 10\(^{-8}\) C/m2. The object is initially 0.400 m from the sheet. What initial speed must the object have in order for its closest distance of approach to the sheet to be 0.100 m?

A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho\). (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\). (b) What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r = R\). (d) Graph the electric-field magnitude as a function of r from \(r = 0\) to \(r = 3R\).

At time \(t =\) 0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet with speed 9.70 \(\times\) 10\(^2\) m/s. The sheet has uniform surface charge density 2.34 \(\times\) 10\(^{-9}\) C/m2. What is the speed of the proton at \(t =\) 5.00 \(\times\) 10\(^{-8}\) s?

A solid conducting sphere carrying charge \(q\) has radius \(a\). It is inside a concentric hollow conducting sphere with inner radius \(b\) and outer radius \(c\). The hollow sphere has no net charge. (a) Derive expressions for the electricfield magnitude in terms of the distance \(r\) from the center for the regions \(r < a, a < r < b, b < r < c\), and \(r > c\). (b) Graph the magnitude of the electric field as a function of \(r\) from \(r =\) 0 to \(r =\) 2c. (c) What is the charge on the inner surface of the hollow sphere? (d) On the outer surface? (e) Represent the charge of the small sphere by four plus signs. Sketch the field lines of the system within a spherical volume of radius 2\(c\).

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