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Intersecting sheets ** (a) Figure \(1.49\) shows the cross section of three infinite sheets intersecting at equal angles. The sheets all have surface charge density \(\sigma .\) By adding up the fields from the sheets, find the electric field at all points in space. (b) Find the field instead by using Gauss's law. You should explain clearly why Gauss's law is in fact useful in this setup. (c) What is the field in the analogous setup where there are \(N\) sheets instead of three? What is your answer in the \(N \rightarrow \infty\) limit? This limit is related to the cylinder in Exercise 1.68.

Short Answer

Expert verified
The electric field at all points in space due to three sheets intersecting at equal angles with surface charge density \(\sigma\) is \(\frac{\sigma}{2 \epsilon_0}\). This result matches with the finding by Gauss's law. For \(N\) sheets, the electric field is \(\frac{N \sigma}{2 \epsilon_0}\). In the case of \(N \rightarrow \infty\), this setup behaves similar to a perfect conductor and the electric field would be infinity.

Step by step solution

01

Find the electric field by adding up fields

At any point in space, the electric field due to each sheet is given by \(\frac{\sigma}{2 \epsilon_0}\) directed away from the sheet. Since all three sheets intersect at equal angles, which means \(\frac{2\pi}{3}\) for each, the total electric field at the intersection point is zero. However, besides the intersection point since all three fields are not equal in direction, their vector sum is not zero. Hence the field away from the intersection point is \(\frac{\sigma}{2 \epsilon_0}\) determined by direction at other points.
02

Apply Gauss's law to find the field

Gauss's law is applicable as these sheets are infinite, producing symmetrical electric fields. Gauss's law states that the electric flux out of any closed surface is equal to the charge enclosed by that surface divided by \(\epsilon_0\). By considering a Gaussian cylindrical surface with two ends parallel to the sheets, it can be seen that, for each sheet that intercepts the Gaussian surface, an equal surface charge \(\sigma\) enters and leaves the surface. This makes the total enclosed charge zero, and so the total electric flux is zero. The electric field is therefore \(\frac{\sigma}{2 \epsilon_0}\) at all points away from the sheets.
03

Calculate the field for \(N\) sheets and the \(N \rightarrow \infty\) limit

The electric field for \(N\) sheets would be determined by the scalar sum of electric fields due to each sheet. It results in \(\frac{N \sigma}{2 \epsilon_0}\). In the case of \(N \rightarrow \infty\), the electric field would be infinity, indicating that this configuration behaves equivalently to a perfect conductor.

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Key Concepts

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

Gauss's Law
Understanding Gauss's law is essential for calculating the electric field produced by symmetrical charge distributions. In essence, Gauss's law relates the electric flux flowing out of a closed surface to the charge enclosed by that surface. The law can be summarized by the equation \[ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} \] where \( \Phi_E \) is the electric flux, \( Q_{\text{enc}} \) is the enclosed electric charge, and \( \epsilon_0 \) is the vacuum permittivity.
When dealing with infinite sheets of charge, this law simplifies the problem because the symmetry of the field allows for an easy choice of the Gaussian surface. For example, in the given exercise, a cylindrical Gaussian surface is chosen such that the caps are parallel to the sheets, simplifying the calculations since the electric field is constant over these caps. Understanding and applying Gauss's law to these situations can reveal both the magnitude and direction of the electric field at any point in space.
Electric Charge Density
Electric charge density is a measure of the amount of electrical charge per unit area or volume. For surface charge density, often denoted as \( \sigma \), it tells us how much charge is distributed over a certain area.
In problems involving infinite sheets with charge, the surface charge density entails how the space around them is influenced. In the exercise given, the sheets have a surface charge density \( \sigma \) which is uniform, meaning that each point on the sheet has the same amount of charge. This uniform charge distribution is a critical factor that simplifies the calculation of the electric field at any point in space, especially using Gauss's law.
Electric Flux
Electric flux, denoted by \( \Phi_E \) represents the number of electric field lines passing through a surface. It is a measure of the distribution of the electric field over an area and it is calculated as the electric field \( \vec{E} \) multiplied by the area \( A \) over which it is spread, and cos\( \theta \), where \( \theta \) is the angle between the electric field and the normal to the surface: \[ \Phi_E = \vec{E} \cdot \vec{A} = E A \cos\theta \] This concept is integral to Gauss's law and is key to understanding how charges create electric fields in the surrounding space. When the surface is parallel to the field lines, like in our exercise with infinite sheets, \( \theta = 0 \) and the electric flux simplifies to just the product of the electric field and the area.
Surface Charge Density
Surface charge density is specifically the amount of charge per unit area on a surface. Represented by \( \sigma \), this quantity allows us to determine the electric field that's generated by surfaces such as the infinite sheets in our exercise.

Surface charge density is related to the electric field in a directly proportional way, meaning as you increase or decrease the charge density, the electric field correspondingly increases or decreases. This is clearly seen from the equation \[ E = \frac{\sigma}{2\epsilon_0} \] used in the exercise, showing that the electric field is directly proportional to the surface charge density.
Electric Field Calculation
The electric field calculation for distributions of charge involves understanding the nature of the electric field and how charge configurations affect it. In our example, the electric field due to each infinite sheet is computed using both direct summation and Gauss's law.
Direct summation involves vector addition of the fields due to each individual sheet. Because the sheets intersect at symmetrical angles, their fields at any point in space can be added vectorially.
Gauss's law simplifies this process by taking advantage of symmetry and easily chosen Gaussian surfaces. For an infinite sheet of charge, or multiple sheets, the field is uniform and perpendicular to the surface, simplifying the calculation to a matter of determining the electric field due to a single sheet and then multiplying by the number of sheets, as seen when dealing with the situation of multiple intersecting sheets.

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Most popular questions from this chapter

Uniform field strength * We know from the example in Section \(1.11\) that the electric field inside a solid sphere with uniform charge density is proportional to \(r\). Assume instead that the charge density is not uniform, but depends only on \(r\). What should this dependence be so that the magnitude of the field at points inside the sphere is independent of \(r\) (except right at the center, where it isn't well defined)? What, should the dependence be in the analogous case where we have a cylinder instead of a sphere?

Find a geometrical arrangement of one proton and two electrons such that the potential energy of the system is exactly zero. How many such arrangements are there with the three particles on the same straight line? You should find that the ratio of two of the distances involved is the golden ratio.

\mathrm{~ F o r c e s ~ o n ~ t h r e e ~ s h e e t s ~ * ?}\( Consider three charged sheets, \)A, B\(, and \)C .\( The sheets are parallel with \)A\( above \)B\( above \)C\(. On each sheet there is surface charge of uniform density: \)-4 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\( on \)A, 7 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\( on \)B\(, and \)-3 \cdot 10^{-5} \mathrm{C} / \mathrm{m}^{2}\( on \)C$. (The density given includes charge on both sides of the sheet.) What is the magnitude of the electrical force per unit area on each sheet? Check to see that the total force per unit area on the three sheets is zero.

Zero field in a sphere In Fig. \(1.51\) a sphere with radius \(R\) is centered at the origin, an, infinite cylinder with radius \(R\) has its axis along the \(z\) axis, and an infinite slab with thickness \(2 R\) lies between the planes \(z=-R\) and \(z=R\). The uniform volume densities of these objects are \(\rho_{1}, \rho_{2}\) and \(\rho_{3}\), respectively. The objects are superposed on top of each other; the densities add where the objects overlap. How should the three densities be related so that the electric field is zero everywhere throughout the volume of the sphere? Hint: Find a vector expression for the field inside each object, and then use superposition.

Force between two strips ** (a) The two strips of charge shown in Fig. \(1.47\) have width \(b\), infinite height, and negligible thickness (in the direction perpendicular to the page). Their charge densities per unit area are \(\pm \sigma .\) Find the magnitude of the electric field due to one of the strips, a distance \(x\) away from it (in the plane of the page). (b) Show that the force (per unit height) between the two strips equals \(\sigma^{2} b(\ln 2) / \pi \epsilon_{0}\). Note that this result is finite, even though you will find that the field due to a strip diverges as you get close to it

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