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Chapter 1: Problem 70

Field from two sheets : Two infinite plane sheets of surface charge, with densities \(3 \sigma_{0}\) and \(-2 \sigma_{0}\), are located a distance \(\ell\) apart, parallel to one another. Discuss the electric field of this system. Now suppose the two planes, instead of being parallel, intersect at right angles. Show what the field is like in each of the four regions into which space is thereby divided.

Short Answer

Expert verified
The electric field between the two parallel planes is \(\frac{\sigma_{0}}{2\epsilon_{0}}\), directed from the positively charged plane to the negatively charged one. Outside the planes, the field is zero. For the case of perpendicular planes, the magnitude of the electric field in each quadrant is \(\frac{\sigma_{0}}{\epsilon_{0}}\), with different directions determined by the vector sum of the fields due to each plane.

Step by step solution

01

Calculate Electric Field Between Two Parallel Planes

The electric field due to an infinite plane sheet of surface charge is given by \(E = \frac{\sigma}{2\epsilon_{0}}\), where \(\sigma\) is the surface charge density and \(\epsilon_{0}\) is the permittivity of free space. As the distance between the sheets does not matter for infinite sheets, for the plane with charge density \(3 \sigma_{0}\), the field is \(E_{1} = \frac{3 \sigma_{0}}{2\epsilon_{0}}\) in the direction away from the plane. For the plane with charge density \(-2 \sigma_{0}\), the field is \(E_{2} = \frac{-2 \sigma_{0}}{2\epsilon_{0}}\) in the direction away from the plane. The total field between the sheets is hence \(E_{total} = E_{1} + E_{2} = \frac{\sigma_{0}}{2\epsilon_{0}}\).
02

Discuss Direction of Electric Field

The direction of the electric field is from the positively charged plane to the negatively charged plane. Inside the region between the two planes, the fields due to each plane add up as they are in the same direction, yielding the total field. Outside, the fields are in opposite directions and cancel out, hence the electric field is zero.
03

Calculate Electric Field for Perpendicular Planes

For the planes intersecting at right angles, their respective electric fields will be created in the spaces on either side of them. In each quadrant, the resulting electric field will be a vector sum of the fields due to each plane. This gives four different regions with different field directions, but the same magnitude given by \(|\vec{E}| = \sqrt{E_{1}^2 + E_{2}^2} = \frac{\sigma_{0}}{\epsilon_{0}}\).

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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 an invisible force field created around charged objects. It's a vector field, meaning it has both a magnitude and a direction. The magnitude of the electric field \(E\) generated by a charged surface is proportional to the surface charge density \(\sigma\), and inversely proportional to the permittivity of free space \(\epsilon_{0}\).

For a plane with uniform surface charge density, the electric field can be calculated using the formula \(E = \frac{\sigma}{2\epsilon_{0}}\). Here, the factor of 2 arises because an infinite plane extends equally in both directions, thus affecting the field distribution.

  • The direction of the electric field is always away from a positively charged surface.
  • For a negatively charged surface, the direction is towards the plane.
In this problem, two charged planes with densities \(3\sigma_{0}\) and \(-2\sigma_{0}\) produce fields that add together between the planes but cancel out outside their mutual region.
Surface Charge Density
Surface charge density \(\sigma\) is a measure of how much electric charge is distributed over a surface area. It is typically measured in coulombs per square meter (C/m虏). This density is crucial in determining the electric field produced by a charged surface.

Given the relation \(E = \frac{\sigma}{2\epsilon_{0}}\), you can see that the greater the surface charge density, the stronger the resulting electric field. In simple terms:
  • A high positive surface charge density results in strong outward electric fields.
  • Conversely, a strong negative density results in strong electric fields directed towards the surface.
In the exercise, the densities are given as \(3\sigma_{0}\) and \(-2\sigma_{0}\), meaning one sheet attracts more charges towards it than the other repels, giving rise to the net field calculation.
Permittivity of Free Space
Permittivity of free space, also known as the electric constant, is a fundamental physical constant denoted by \(\epsilon_{0}\). It describes how electric fields interact with a vacuum (or free space) and is instrumental in calculating electric forces and fields.

The value of \(\epsilon_{0} = 8.854 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2)\), serves as a baseline for understanding how materials affect the electric field around them. When applied in the electric field formula, \(E = \frac{\sigma}{2\epsilon_{0}}\), it portrays a measure of how easily the electric field can spread in space without being hindered.

  • The larger the permittivity, the weaker the electric field for a given surface charge density.
  • It helps relate electric phenomena in theoretical situations by setting a standard reference point.
In the context of two charged planes, \(\epsilon_{0}\) aids in quantifying the strength and interaction of fields generated. Understanding how permittivity interacts with charge distributions helps solve problems involving complex field geometries.

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Most popular questions from this chapter

Energy around a sphere * A sphere of radius \(R\) has a charge \(Q\) distributed uniformly over its surface. How large a sphere contains 90 percent of the energy stored in the electrostatic field of this charge distribution?

Oscillating in a ring A ring with radius \(R\) has uniform positive charge density \(\lambda\). A particle with positive charge \(q\) and mass \(m\) is initially located at the center of the ring and is then given a tiny kick. If it is constrained to move in the plane of the ring, show that it undergoes simple harmonic motion (for small oscillations), and find the frequency. Hint: Find the potential energy of the particle when it is at a (small) radius, \(r\), by integrating over the ring, and then take the negative derivative to find the force. You will need to use the law of cosines and also the Taylor series \(1 / \sqrt{1+\epsilon} \approx 1-\epsilon / 2+3 \epsilon^{2} / 8\)

Hole in a shell \(*\) Figure \(1.52\) shows a spherical shell of charge, of radius \(a\) and surface density \(\sigma\), from which a small circular piece of radius \(b \ll a\) has been removed. What is the direction and magnitude of the field, at the midpoint of the aperture? There are two ways to get the answer. You can integrate over the remaining charge distribution, to sum the contributions of all elements to the field at the point in question. Or, remembering the superposition principle, you can think about the effect of replacing the piece removed, which itself is practically a little disk. Note the connection of this result with our discussion of the force on a surface charge - perhaps that is a third way in which you might arrive at the answer.

Maximum field from a ring ** A charge \(Q\) is distributed uniformly around a thin ring of radius \(b\) that lies in the \(x y\) plane with its center at the origin. Locate the point on the positive \(z\) axis where the electric field is strongest.

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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