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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

Negative charge \(-Q\) is distributed uniformly over the surface of a thin spherical insulating shell with radius R. Calculate the force (magnitude and direction) that the shell exerts on a positive point charge \(q\) located a distance (a) \(r > R\) from the center of the shell (outside the shell); (b) \(r < R\) from the center of the shell (inside the shell).

(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 30.0 cm in diameter to produce an electric field of magnitude 1390 N/C just outside the surface of the sphere? (b) What is the electric field at a point 10.0 cm outside the surface of the sphere?

A solid conducting sphere with radius \(R\) carries a positive total charge \(Q\). The sphere is surrounded by an insulating shell with inner radius \(R\) and outer radius 2\(R\). The insulating shell has a uniform charge density \(\rho\). (a) Find the value of \(\rho\) so that the net charge of the entire system is zero. (b) If \(\rho\) has the value found in part (a), find the electric field \(\overrightarrow{E}\) (magnitude and direction) in each of the regions 0 \(< r < R, R < r < 2R\), and \(r > 2R\). Graph the radial component of \(\overrightarrow{E}\) as a function of r. (c) As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in part (b) agree with this rule.

A nonuniform, but spherically symmetric, distribution of charge has a charge density \(\rho\)(\(r\)) given as follows: $$\rho(r) = \rho_0 \bigg(1 - \frac{r}{R}\bigg) \space \space \space \mathrm{for} \space r \leq R$$ $$\rho(r) = 0 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \mathrm{for} \space r \leq R$$ where \(\rho_0 = 3Q/{\pi}R^3\) is a positive constant. (a) Show that the total charge contained in the charge distribution is \(Q\). (b) Show that the electric field in the region \(r \geq R\) is identical to that produced by a point charge \(Q\) at \(r =\) 0. (c) Obtain an expression for the electric field in the region \(r \leq R\). (d) Graph the electric-field magnitude \(E\) as a function of \(r\). (e) Find the value of \(r\) at which the electric field is maximum, and find the value of that maximum field.

A very long uniform line of charge has charge per unit length 4.80 \(\mu\)C/m and lies along the \(x\)-axis. A second long uniform line of charge has charge per unit length -2.40 \(\mu\)C/m and is parallel to the x-axis at \(y =\) 0.400 m. What is the net electric field (magnitude and direction) at the following points on the \(y\)-axis: (a) \(y =\) 0.200 m and (b) \(y =\) 0.600 m?

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