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A flat sheet of paper of area 0.250 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 N/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.

Short Answer

Expert verified
(a) Flux is 1.75 Nm²/C. (b) No, it doesn't depend on shape. (c) Max when \(\theta = 0^\circ\); min when \(\theta = 90^\circ\).

Step by step solution

01

Write Down Known Values and Formulas

The area of the sheet is given as \(A = 0.250\, \text{m}^2\), the electric field magnitude is \(E = 14\, \text{N/C}\), and the angle between the normal to the sheet and the electric field is \(\theta = 60^\circ\). We can use the formula for electric flux, given by \(\Phi_E = EA \cos\theta\), where \(\Phi_E\) is the electric flux.
02

Calculate the Electric Flux

Substitute the known values into the formula. \[\Phi_E = (14\, \text{N/C}) \times (0.250\, \text{m}^2) \times \cos(60^\circ)\]Since \(\cos(60^\circ) = 0.5\), we have:\[\Phi_E = 14 \times 0.250 \times 0.5 = 1.75\, \text{Nm}^2/\text{C}\]Thus, the magnitude of the electric flux through the sheet is \(1.75\, \text{Nm}^2/\text{C}\).
03

Determine Dependency on Shape

The magnitude of the electric flux depends on the orientation of the surface relative to the electric field, not on the shape of the sheet. Since we use a uniform electric field and account for its interaction with the area, any shape with the same area and orientation would result in the same electric flux.
04

Determine Angles for Maximum and Minimum Flux

(i) The electric flux is largest when \(\cos\theta = 1\), i.e., \(\theta = 0^\circ\). This means the electric field is perpendicular to the surface.(ii) The electric flux is smallest when \(\cos\theta = 0\), i.e., \(\theta = 90^\circ\). In this case, the electric field is parallel to the surface.

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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 object where other charges experience a force. Imagine it like an invisible force field created by electric charges. Electric fields are represented by vectors, meaning they have both a direction and a magnitude. The direction of the field is the direction a positive charge would move if placed in the field. The strength or magnitude of the electric field is measured in newtons per coulomb (N/C).
Key points about electric fields:
  • The electric field is stronger where the lines are closer together, and weaker where they are further apart.
  • The field lines never intersect.
  • A uniform electric field has parallel lines that are evenly spaced.
Understanding electric fields is crucial when calculating electric flux, as they determine the force experienced by surfaces placed in the field.
Orientation of Surfaces
The orientation of a surface in an electric field can significantly affect the amount of electric flux passing through it. Imagine a flat sheet of paper inside a steady breeze (the electric field). The angle at which you hold the paper against the breeze determines how much wind hits it, similar to how a surface's angle affects electric flux.
Orientation involves the angle between the surface's normal (a line perpendicular to the surface) and the electric field lines. This angle is crucial in calculating electric flux:
  • A small angle means the field lines hit the surface directly, increasing the flux.
  • A larger angle reduces the effective area exposed to the field, thus reducing the flux.
Understanding the orientation helps comprehend how surfaces interact differently with electric fields under various angles.
Flux Calculation
Electric flux quantifies how much of an electric field passes through a given surface. It is denoted by \( \Phi_E \), and its calculation is akin to measuring how much wind flows through a window. The formula for electric flux through a flat surface is given by \( \Phi_E = EA \cos\theta \), where:
  • \( E \) is the electric field strength
  • \( A \) is the area of the surface
  • \( \theta \) is the angle between the surface normal and the field direction
To calculate \( \Phi_E \), use the cosine of the angle to adjust for the effective area exposed to the field. A direct hit (\
Perpendicular and Parallel Alignment
When discussing electric flux, the terms perpendicular and parallel alignments are key concepts. These refer to the orientation of the surface relative to the electric field, which dramatically affects the resulting flux.- **Perpendicular Alignment**: When the surface is oriented so that the electric field lines hit it at a right angle (90 degrees to the surface's normal), this results in the maximum flux. This is because all field lines are passing directly through the surface, which is observed when \( \theta = 0^\circ \). Mathematically, this is when \( \cos\theta = 1 \), maximizing the flux equation.- **Parallel Alignment**: Here, the surface is in line with the electric field (\

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Most popular questions from this chapter

How many excess electrons must be added to an isolated spherical conductor 26.0 cm in diameter to produce an electric field of magnitude 1150 N/C just outside the surface?

Charge \(Q\) is distributed uniformly throughout the volume of an insulating sphere of radius \(R =\) 4.00 cm. At a distance of \(r =\) 8.00 cm from the center of the sphere, the electric field due to the charge distribution has magnitude \(E =\) 940 N/C. What are (a) the volume charge density for the sphere and (b) the electric field at a distance of 2.00 cm from the sphere's center?

A cube has sides of length \(L =\) 0.300 m. One corner is at the origin (Fig. E22.6). The nonuniform electric field is given by \(\overrightarrow{E} =\) (-5.00 N/C \(\cdot\) m)\(x\hat{\imath}\) + (3.00 N/C \(\cdot\) m)\(z \hat{k}\). (a) Find the electric flux through each of the six cube faces \(S_1, S_2, S_3, S_4, S_5\), and \(S_6\) . (b) Find the total electric charge inside the cube.

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?

Two very long uniform lines of charge are parallel and are separated by 0.300 m. Each line of charge has charge per unit length +5.20 \(\mu\)C/m. What magnitude of force does one line of charge exert on a 0.0500-m section of the other line of charge?

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