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Chapter 3: Problem 12

Image charges for two planes : A point charge \(q\) is located between two parallel infinite conducting planes, a distance \(d\) from one and \(\ell-d\) from the other. Where should image charges be located so that the electric field is everywhere perpendicular to the planes?

Short Answer

Expert verified
Image charges of alternating polarity should be located at every multiple of \(\ell\), beginning from \(-d\) for the first negative image charge. The sequence of charges will be \(q, -q, q, -q, \ldots\) at \(d, \ell-d, \ell+d, 2\ell-d, \ldots\) respectively.

Step by step solution

01

Identify the Image Charges

To make the electric field always perpendicular to the conducting planes, we put an image charge on each plane. An image charge must be identical in magnitude, but opposite in polarity, to the original charge. Thus, the image charges should be \(-q\). Assuming the original point is at \(+d\), these charges will fall at positions -d and \(\ell + d\). This results from imagining that each conducting plane as a mirror, 'reflecting' the point charge, with the reflected (image) points being equidistant from the plane as the original.
02

The Second Set of Image Charges

For each image charge, image charges will be created in the other conductor. After placing the original image charges at positions -d and \(\ell + d\), a second pair of image charges of size \(+q\) at distance \(2\ell + d\) and \(2\ell -d\), are created. They are identical in magnitude to the original charge, but equal in polarity, due to the double mirror effect.
03

Infinite Series of Image Charges

This process is repeated in both directions, which generates an infinite series of charges with alternating polarity: \((-1)^n q\), where n is an integer. The series starts with \(n=0\) for the original charge.

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Key Concepts

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

Image Charges
Image charges are a clever method used in electrostatics to simplify complex problems involving conductors. Essentially, image charges are theoretical charges placed to simulate the effect of conductors on electric fields. This method helps us solve problems without needing to directly calculate the interactions with the conducting surfaces. Imagine a simple scenario: when a real charge is near a conducting surface, the surface influences the electric field due to charge redistribution on itself. To handle this complex interaction, we treat it as if an equal but opposite charge (an image charge) is placed on the other side of the conductor. By doing this, the resulting field outside the conductor remains realistic.

This strategy is vital in ensuring that electric fields are perpendicular to conductive surfaces, which is a necessary condition in electrostatics. Imagine placing a charge between two parallel conducting planes; image charges are used to maintain perpendicularity of electric fields near those planes.
Conducting Planes
Conducting planes are surfaces that can easily transfer electric charge, and they are very effective at modifying electric fields. Key to understanding how charging and electric fields work with conductors is acknowledging that they redistribute their charges to counteract external electric fields. When a point charge is placed near a perfectly conductive plane, the plane redistributes its charges so that the field inside cancels out completely.

In the method of image charges, conducting planes are treated as mirrors that reflect not light, but electric charges. This principle aids in imagining where image charges need to be placed. For an infinite conducting plane, the redistribution leads to a scenario equivalent to placing an image charge somewhere relative to the original charge. The role of conducting planes in problems like the one you've seen is to enforce conditions (like perpendicularity) on electric fields at their surfaces.
Point Charge
In electrostatics, a point charge refers to an idealized model of a particle with electric charge that is concentrated at a single point in space. This simplification is used because it makes calculations tractable and helps us understand how real charges behave in different environments.
  • A single point charge creates an electric field that radiates uniformly in all directions.
  • The field strength decreases with the square of the distance from the charge per Coulomb's Law.
When dealing with conducting surfaces and the method of image charges, the location of the point charge is crucial. For example, in the original problem, the position of the point charge governs the placement of the image charges so that the resulting electric field maintains perpendicularity relative to the conducting planes.

The interaction between point charges and conductive planes is often addressed by "placing" negative image charges across the conducting planes, thus eliminating any resultant electric field inside the conductor.
Electric Field Perpendicularity
Perpendicular electric fields are essential when dealing with conducting surfaces because they ensure the stability of the field within and around the conductor. When the electric field at the surface of a conductor is perpendicular, it indicates that charge redistribution on the conductor's surface has reached a stable configuration.

Why does perpendicularity matter? If the field were not perpendicular, free charges within the conductor would move 鈥� creating currents 鈥� until the field becomes perpendicular. Thus, the condition of perpendicularity is not just mathematical; it reflects the physical reality. In problems involving image charges, ensuring perpendicular fields means placing those charges so that the resulting electric field lines end perpendicularly on the conducting surfaces.

This condition ultimately leads to stable solutions in electrostatic problems and ensures no energy is wasted in unnecessary charge movement.

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Most popular questions from this chapter

Capacitance of a spheroid * Here is the exact formula for the capacitance \(C\) of a conductor in the form of a prolate spheroid of length \(2 a\) and diameter \(2 b\) : \(C=\frac{8 \pi \epsilon_{0} a \epsilon}{\ln \left(\frac{1+\epsilon}{1-\epsilon}\right)}, \quad\) where \(\quad \epsilon=\sqrt{1-\frac{b^{2}}{a^{2}}}\) First verify that the formula reduces to the correct expression for the capacitance of a sphere if \(b \rightarrow a\). Now imagine that the spheroid is a charged water drop. If this drop is deformed at constant volume and constant charge \(Q\) from a sphere to a prolate spheroid, will the energy stored in the electric field increase or decrease? (The volume of the spheroid is \((4 / 3) \pi a b^{2}\).)

Two ways of calculating energy \(* *\) * A capacitor consists of two arbitrarily shaped conducting shells, with one inside the other. The inner conductor has charge \(Q\), the outer has charge \(-Q\). We know of two ways of calculating the energy \(U\) stored in this system. We can find the electric field \(E\) and then integrate \(\epsilon_{0} E^{2} / 2\) over the volume between the conductors. Or. if we know the potential difference \(\phi\), we can write \(U=Q \phi / 2\) (or equivalently \(U=C \phi^{2} / 2\) ). (a) Show that these two methods give the same energy in the case of two concentric shells. (b) By using the identity \(\nabla \cdot(\phi \nabla \phi)=(\nabla \phi)^{2}+\phi \nabla^{2} \phi\), show that the two methods give the same energy for conductors of any shape.

Capacitance coefficients for shells : ? A capacitor consists of two concentric spherical shells. Label the inner shell, of radius \(b\), as conductor 1 ; and label the outer shell, of radius \(a\), as conductor \(2 .\) For this two-conductor system, find \(C_{11}\), \(C_{22}\), and \(C_{12}\)

Maximum field from a passing charge In a colliding beam storage ring an antiproton going east passes a proton going west, the distance of closest approach being \(10^{-10} \mathrm{~m}\). The kinetic energy of each particle in the lab frame is \(93 \mathrm{GeV}\), corresponding to \(\gamma=100\). In the rest frame of the proton, what is the maximum intensity of the electric field at the proton due to the charge on the antiproton? For about how long, approximately, does the field exceed half its maximum intensity?

Field from a filament * On a nylon filament \(0.01 \mathrm{~cm}\) in diameter and \(4 \mathrm{~cm}\) long there are \(5.0 \cdot 10^{8}\) extra electrons distributed uniformly over the surface. What is the electric field strength at the surface of the filament: (a) in the rest frame of the filament? (b) in a frame in which the filament is moving at a speed \(0.9 c\) in a direction parallel to its length?Field from a filament * On a nylon filament \(0.01 \mathrm{~cm}\) in diameter and \(4 \mathrm{~cm}\) long there are \(5.0 \cdot 10^{8}\) extra electrons distributed uniformly over the surface. What is the electric field strength at the surface of the filament: (a) in the rest frame of the filament? (b) in a frame in which the filament is moving at a speed \(0.9 c\) in a direction parallel to its length?
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