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An infinite sheet with positive charge per unit area \(\sigma\) lies in the \(x y\) -plane. A second infinite sheet with negative charge per unit area \(-\sigma\) lies in the \(y z\) -plane. Find the net electric field at all points that do not lie in either of these planes. Express your answer in terms of the unit vectors \(\hat{i}, \hat{j},\) and \(\hat{k} .\)

Short Answer

Expert verified
The net electric field is \\( -\frac{\sigma}{2\varepsilon_0} \hat{i} + \frac{\sigma}{2\varepsilon_0} \hat{k} \\\).

Step by step solution

01

Identify Charge Configurations and Fields

We have two infinite sheets of charge: one with positive charge density \( \sigma \) on the \( xy \)-plane and one with negative charge density \(-\sigma \) on the \( yz \)-plane. An infinite charged sheet produces an electric field with magnitude \( E = \frac{\sigma}{2\varepsilon_0} \) directed outward from the plane of the sheet if the charge is positive, and inward if negative. Moreover, the field lines are perpendicular to the sheet.
02

Determine Electric Field Due to xy-plane Sheet

For the infinite sheet located in the \( xy \)-plane, the electric field is perpendicular to the plane (along the \( z \)-axis) and is directed away from the plane due to positive charge. Therefore, the electric field at any point not on the sheet can be expressed as \( \vec{E}_{xy} = \frac{\sigma}{2\varepsilon_0} \hat{k} \).
03

Determine Electric Field Due to yz-plane Sheet

For the infinite sheet located in the \( yz \)-plane, the electric field is perpendicular to the sheet (along the \( x \)-axis) and is directed toward the sheet since the charge is negative. Thus, the electric field can be expressed as \( \vec{E}_{yz} = \frac{\sigma}{2\varepsilon_0} (-\hat{i}) = -\frac{\sigma}{2\varepsilon_0} \hat{i} \).
04

Calculate Net Electric Field

The net electric field is a vector sum of the fields due to each sheet, \( \vec{E} = \vec{E}_{xy} + \vec{E}_{yz} \). Substituting the expressions for both fields, the net electric field is \( \vec{E} = \frac{\sigma}{2\varepsilon_0} \hat{k} - \frac{\sigma}{2\varepsilon_0} \hat{i} \).
05

Express Final Answer in Vector Form

Combine the components of the electric field to produce the final expression, \( \vec{E} = \left(-\frac{\sigma}{2\varepsilon_0} \hat{i}\right) + 0\hat{j} + \left(\frac{\sigma}{2\varepsilon_0} \hat{k}\right) = -\frac{\sigma}{2\varepsilon_0} \hat{i} + \frac{\sigma}{2\varepsilon_0} \hat{k} \). This is the net electric field at all points not on the sheets.

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Key Concepts

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

Infinite Charged Sheets
Infinite charged sheets are a fascinating subject in the study of electromagnetism. Imagine an endless plane spread out in all directions. This is what we mean by "infinite" in this context—there's no edge or boundary. Such a scenario can be rare in reality, but it helps us simplify and understand electric fields better. These sheets have a uniform charge distribution, meaning each section of the sheet carries the same amount of charge. When dealing with an infinite charged sheet, the electric field it produces is uniform. This means the strength of the electric field is the same at every point equidistant from the sheet.
  • The field lines are always perpendicular to the sheet.
  • If the charge is positive, the field lines point away from the sheet.
  • If the charge is negative, the field lines point toward the sheet.
Constant electric fields play a significant role in simplifying complex calculations and predictions about charged particles' behavior.
Vector Addition
Vector addition is a crucial concept when dealing with electric fields, especially when multiple fields interact. Vectors are quantities that have both magnitude and direction. In physics, many quantities like force, velocity, and indeed electric fields, are vectors. To find the resultant vector from two or more vectors, you apply vector addition. It's like finding the overall effect of several different forces acting on an object. In the scenario with two charged sheets, each sheet produces an electric field that can be represented as a vector. To determine the overall or net electric field at any point, we add these vectors together.
  • The electric field from the positive sheet has a direction perpendicular to the sheet and points outward.
  • For the negatively charged sheet, the field points inward, towards the sheet.
To perform vector addition, align the vectors' tails or heads depending on the method like "head-to-tail" or by breaking them into components, aligning along the coordinate axis, and summing. The net electric field will be a combination of these components.
Electromagnetism
Electromagnetism is one of the fundamental forces governing how charged particles interact. It explains phenomena related to electric fields, magnetic fields, and their interactions. When discussing problems involving charged sheets, we are delving into the realm of static electricity, a part of electromagnetism. Key principles of electromagnetism include:
  • Charges create electric fields.
  • Electric fields exert forces on other charges within the field.
  • Like charges repel; opposite charges attract.
These principles help us understand the forces experienced by charges in various configurations. Infinite charged sheets illustrate these electromagnetic interactions as their uniform electric fields significantly influence nearby charges' behavior. Understanding electromagnetism allows us to predict the behavior of electric fields in circuits, motors, and even biological systems.
Charge Distribution
Charge distribution refers to how electric charge is spread over an object. In physics problems involving electric fields, the charge distribution dictates the shape, direction, and intensity of the electric field produced. For infinite charged sheets, the distribution is uniform, meaning the charge is evenly spread across the entire sheet. Uniform charge distributions are significant because:
  • They simplify calculations since the electric field is uniform.
  • They make it easier to predict electric forces acting on other charges nearby.
In our specific problem with two infinite sheets, we consider a positive and a negative charge distribution. The positive distribution on one plane and negative on another results in electric fields that interact—leading to the net field we calculate. Understanding charge distribution allows physicists and engineers to design and analyze systems ranging from capacitors to electrocardiograms effectively.

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Most popular questions from this chapter

Electric Field of the Rarth. The earth has a net electric charge that causes a field at points near its surface equal to 150 \(\mathrm{N} / \mathrm{C}\) and directed in toward the center of the earth. (a) What magnitude and sign of charge would a \(60-\mathrm{kg}\) human have to acquire to overcome his or her weight by the force exerted by the earth's electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 \(\mathrm{m} ?\) Is use of the earth's electric field a feasible means of flight? Why or why not?

The ammonia molecule \(\left(\mathrm{NH}_{3}\right)\) has a dipole moment of \(5.0 \times 10^{-30} \mathrm{C} \cdot \mathrm{m} .\) Ammonia molecules in the gas phase are placed in a uniform electric field \(\overrightarrow{\boldsymbol{E}}\) with magnitude \(1.6 \times 10^{6} \mathrm{N} / \mathrm{C}\) . (a) What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to \(\overrightarrow{\boldsymbol{E}}\) from parallel to perpendicular?(b) At what absolute temperature \(\boldsymbol{T}\)is the average translational kinetic energy \(\frac{3}{2} k T\) of a molecule equal to the change in potential energy calculated in part (a)? (Note: Above this temperature, thermal agitation prevents the dipoles from aligning with the electric field.)

Infinite sheet \(A\) carries a positive uniform charge density \(\sigma\) , and sheet \(B\) , which is to the right of \(A\) and parallel to it, carries a uniform negative charge density \(-2 \sigma .\) (a) Sketch the electric field lines for this pair of sheets. Include the region between the sheets as well as the regions to the left of \(A\) and to the right of \(B\) . (b) Repeat part (a) for the case in which sheet \(B\) carries a charge density of \(+2 \sigma .\)

A negative point charge \(q_{1}=-4.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=0.60 \mathrm{m} .\) A second point charge \(q_{2}\) is on the \(x\) -axis at \(x=\) \(-1.20 \mathrm{m} .\) What must the sign and magnitude of \(q_{2}\) be for the net electric field at the origin to be (a) 50.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -direction and \((b) 50.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) -direction?

A straight, nonconducting plastic wire 8.50 \(\mathrm{cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 \(\mathrm{cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 \(\mathrm{cm}\) directly above its center.

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