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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

Potential energy in a one-dimensional crystal * * Calculate the potential energy, per ion, for an infinite 1 D ionic crystal with separation \(a\); that is, a row of equally spaced charges of magnitude \(e\) and alternating sign. Hint: The power-series expansion of \(\ln (1+x)\) may be of use.

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.

Escaping field lines * Charges \(2 q\) and \(-q\) are located on the \(x\) axis at \(x=0\) and \(x=a\), respectively. (a) Find the point on the \(x\) axis where the electric field is zero, and make a rough sketch of some field lines. (b) You should find that some of the field lines that start on the \(2 q\) charge end up on the \(-q\) charge, while others head off to infinity. Consider the field lines that form the cutoff between, these two cases. At what angle (with respect to the \(x\) axis) do these lines leave the \(2 q\) charge? Hint: Draw a wisely chosen Gaussian surface that mainly follows these lines.

Field from a spherical shell, right and wrong ** The electric field outside and an infinitesimal distance away from a uniformly charged spherical shell, with radius \(R\) and surface charge density \(\sigma\), is given by Eq. (1.42) as \(\sigma / \epsilon_{0}\). Derive this in the following way. (a) Slice the shell into rings (symmetrically located with respect to the point in question), and then integrate the field contributions from all the rings. You should obtain the incorrect result of \(\sigma / 2 \epsilon_{0}\) (b) Why isn't the result correct? Explain how to modify it to obtain the correct result of \(\sigma / \epsilon_{0} .\) Hint: You could very well have performed the above integral in an effort to obtain the electric field an infinitesimal distance inside the shell, where we know the field is zero. Does the above integration provide a good description of what's going on for points on the shell that are very close to the point in question?

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.
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