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Chapter 3: Problem 1

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?

Short Answer

Expert verified
The electric field in the rest frame of the filament and in a frame which is moving at \(0.9c\) will be different due to the transformation under special relativity. Calculation of the electric field requires the knowledge of total charge and the dimensions of the filament, as well as application of the corresponding electric field equations in the rest and moving frames.

Step by step solution

01

Find the total charge of the filament

First, calculate total charge on the filament. As we know each electron has a charge of -1.6 x \(10^{-19}\) C, the total charge is calculated by multiplying the number of extra electrons by the charge of one electron. So, \(Q = 5.0x10^{8}*(-1.6x10^{-19} C) = -8.0x10^{-11} C\)
02

Determine the electric field in the rest frame

The electric field at the surface of a charged filament (in rest frame) with the surface charge density \(\sigma\) is given by \(\sigma/(2\epsilon_{0})\). First, find \(\sigma\), which is the total charge divided by the surface area of the filament. The surface area of the filament is approximated by \(A = 2\pi rL = 2\pi*(0.01 cm/100 cm/m)*(4 cm)\), where \(r\) and \(L\) are radius and length of the filament respectively. Then, substitute this value and \(\epsilon_{0} = 8.85x10^{-12} C^2/N*m^2\) into the mentioned equation to find the electric field in the rest frame.
03

Find the electric field in a moving frame

In a frame which is moving at a speed relative to the one in which charges are at rest, the electric field E' is given by \(E'=(\gamma E)\), where E is the electric field in the rest frame and \(\gamma\) is the Lorentz factor given by \(\gamma = 1/\sqrt{1- (\upsilon/c)^2}\). Substitute \(E = \sigma/(2\epsilon_{0})\) from Step 2 and \(\upsilon = 0.9c\) into this equation, to get the electric field in the moving frame.

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. Charges come in two types: positive and negative. Like charges repel each other, while opposite charges attract.
To quantify electric charge, we use the unit called "Coulomb," abbreviated as C. One of the smallest units of charge is the charge of an electron, approximately \(-1.6\times 10^{-19}\, C\).
This charge is inherently negative. When dealing with electric fields generated by charges, the overall effect and strength depend on:
  • the amount of charge involved,
  • how the charge is distributed in space.
In our exercise, the total charge on a nylon filament is calculated by considering the charge of individual electrons and the number of extra electrons.
Surface Charge Density
Surface charge density is an important concept when dealing with charges distributed over a surface. It is defined as the amount of electric charge per unit area on a surface.
The symbol for surface charge density is \(\sigma\) and it is calculated as:\[ \sigma = \frac{Q}{A} \]where \(Q\) is the total charge and \(A\) is the surface area.
In this exercise, the surface charge density is found by dividing the total charge on the filament by its surface area, which is calculated using its radius and length. Surface charge density plays a crucial role in determining the electric field strength at points near the charged surface.
Lorentz Factor
The Lorentz factor is a fundamental component when dealing with scenarios involving objects moving at high velocities, especially close to the speed of light.
It is denoted by \(\gamma\) and given by the equation:\[ \gamma = \frac{1}{\sqrt{1 - \left( \frac{v}{c} \right)^2}} \]where \(v\) is the velocity of the moving object, and \(c\) stands for the speed of light in a vacuum.
When the velocity \(v\) approaches the speed of light, the Lorentz factor \(\gamma\) increases significantly. In our context, a filament moves with a speed of \(0.9c\), making \(\gamma\) quite large. This factor helps us understand how the electric field changes due to relativistic effects.
Relativity in Electromagnetism
Relativity in electromagnetism is crucial when objects move at speeds comparable to the speed of light. According to Einstein's theory of relativity, observed electric and magnetic fields can vary between different reference frames.
In our exercise, the filament moves at a high speed parallel to its length. As such, the electric field experienced by an observer in this moving frame is different from that in the rest frame.
This variation is calculated by multiplying the electric field in the rest frame by the Lorentz factor \(\gamma\). Such computations show the interplay between electric and magnetic fields at high velocities, highlighting why relativity is essential in understanding electromagnetic phenomena.

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Most popular questions from this chapter

Synchrotron current * In a 6 gigaelectron-volt \(\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right)\) electron synchrotron, electrons travel around the machine in an approximately circular path 240 meters long. It is normal to have about \(10^{11}\) electrons circling on this path during a cycle of acceleration. The speed of the electrons is practically that of light. What is the current? We give this very simple problem to emphasize that nothing in our definition of current as rate of transport requires the velocities of the charge carriers to be nonrelativistic and that there is no rule against a given charged particle getting counted many times during a second as part of the current.

Two charges and a plane A positive point charge \(Q\) is fixed a distance \(\ell\) above a horizontal conducting plane. An equal negative charge \(-Q\) is to be located somewhere along the perpendicular dropped from \(Q\) to the plane. Where can \(-Q\) be placed so that the total force on it will be zero?

Decreasing velocity Consider the trajectory of a charged particle that is moving with a speed \(0.8 c\) in the \(x\) direction when it enters a large region in which there is a uniform electric field in the \(y\) direction. Show that the \(x\) velocity of the particle must actually decrease. What about the \(x\) component of momentum?

Moving perpendicular to a wire At the end of Section \(5.9\) we discussed the case where a charge \(q\) moves perpendicular to a wire. Figures \(5.25\) and \(5.26\) show qualtatively why there is a nonzero force on the charge, pointing in the positive \(x\) direction. Carry out the calculation to show that the force at a distance \(\ell\) from the wire equals \(q v l / 2 \pi \epsilon_{0} \ell c^{2}\). That is, use Eq. (5.15) to calculate the force on the charge in its own frame, and then divide by \(\gamma\) to transform back to the lab frame. Notes: You can use the fact that in the charge \(q\) 's frame, the speed of the electrons in the \(x\) direction is \(v_{0} / \gamma\) (this comes from the transverse-velocity-addition formula). Remember that the \(\beta\) in Eq. \((5.15)\) is the velocity of the electrons in the charge's frame, and this velocity has two components. Be careful with the transverse distance involved. There are many things to keep track of in this problem, but the integration itself is easy if you use a computer (or Appendix K).

Compressing a sphere ** A spherical conducting shell has radius \(R\) and potential \(\phi\). If you want, you can consider it to be part of a capacitor with the other shell at infinity. You compress the shell down to essentially zero size (always keeping it spherical) while a battery holds the potential constant at \(\phi\). By calculating the initial and final energies stored in the system, and also the work done by (or on) you and the battery, verify that energy is conserved. (Be sure to specify clearly what your conservation-of-energy statement is, paying careful attention to the signs of the various quantities.)
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