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Chapter 3: Problem 11

Electron beam A beam of \(9.5\) megaelectron-volt (MeV) electrons \((\gamma \approx 20)\), amounting as current to \(0.05 \mu \mathrm{A}\), is traveling through vacuum. The transverse dimensions of the beam are less than \(1 \mathrm{~mm}\), and there are no positive charges in or near it. (a) In the lab frame, what is the average distance between an, electron and the next one ahead of it, measured parallel to the beam? What approximately is the average electric field strength 1 cm away from the beam? (b) Answer the same questions for the electron rest frame.

Short Answer

Expert verified
The average distance between an electron and the next one ahead of it in the lab frame and electron rest frame is computed using the beam current, charge of the beam, and relativistic energy-momentum relation. The respective electric field strengths 1 cm away from the beam are calculated using the charge density and permittivity of space. Formulas for current, speed of light, and the electric field of an infinite line charge density are employed.

Step by step solution

01

Find the velocity of the electrons

We have the energy given in electronvolts. One electronvolt equals \(1.6 \times 10^{-19} J\). Therefore, the energy of an electron is \(9.5 MeV = 9.5 \times 10^6 eV = 9.5 \times 10^6 \times 1.6 \times 10^{-19} J = 1.52 \times 10^{-12} J\). We use the relativistic energy-momentum relation \((E = \frac{m_0c^2}{\sqrt{1 - v^2/c^2}})\) to find the speed of the electron.
02

Calculate the charge of the beam

The beam current is given as \(0.05 \mu A\). Current \((I)\) is the rate of flow of charge \((Q)\), defined as \(I = \frac{Q}{t}\). Therefore, we can deduce that the charge passing through the beam per second (which is the charge of the beam) is \(Q = I \times t = 0.05 \times 10^{-6} C = 5 \times 10^{-8} C\).
03

Calculate the distance and the electric field in the lab frame

As the beam travels, the electrons are effectively lined up. So, the number of electrons \((n)\) per unit distance \((d)\) can be calculated using the equation \(n = \frac{Q}{ed}\), where \(Q\) is the charge of the beam and \(e\) is the charge of an electron. Solving for \(d\), we get, \(d = \frac{Q}{en} = \frac{5 \times 10^{-8} C}{(1.6 \times 10^{-19} C/electron) \times n}\). The charge density \((\rho)\) can be given by \(\frac{Q}{Ad}\), where \(A\) is the cross-sectional area of the beam (which is negligible here). The electric field \(E\) can be given by \(E = \frac{\rho}{2\pi\epsilon_0}\), where \(\epsilon_0\) is the permittivity of space.
04

Calculate the distance and the electric field in the electron rest frame

In the electron rest frame, the velocity of electrons is 0. The distance remains the same while the electric field changes due to length contraction \((\gamma\) factor) in the moving frame.

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Key Concepts

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

Relativistic Energy-Momentum Relation
When dealing with particles moving at speeds close to that of light, like electrons in an electron beam, the simplifications made in classical physics no longer apply. We use the relativistic energy-momentum relation to find accurate values for their properties. This relation is a core concept in understanding high-speed particles and states that:- The energy of the particle is significantly higher due to its speed.- The equation used is \( E = \frac{m_0c^2}{\sqrt{1 - v^2/c^2}} \), where \( E \) is the total energy, \( m_0 \) is the rest mass, \( c \) is the speed of light, and \( v \) is the velocity of the particle.- As velocity approaches the speed of light, \( v^2/c^2 \) approaches 1, making the energy approach infinity.For the given electron beam with each electron possessing 9.5 MeV of energy, we calculate the speed using this relation. Electrons move extremely fast, hence their energy is affected by relativistic effects making the conventional energy calculations inadequate without using this formula.Understanding this relation helps us compute accurate values for the velocity and hence additional properties like the distance between particles when observing them in laboratory frame or rest frame.
Current and Charge Relationship
In the context of a moving electron beam, understanding the relationship between current and charge is crucial. Current \( (I) \) is essentially the rate at which charge moves through a section in the beam over time. This defines the flow of charge and can be mathematically represented as:- \( I = \frac{Q}{t} \) where \( I \) is the current in amperes, \( Q \) is the total charge in coulombs, and \( t \) is the time in seconds through which the charge flows.- Given the current of the electron beam is 0.05 \( \mu \)A, to find the charge \( Q \), which will help in subsequent calculations, we can rearrange the equation to \( Q = I \times t \).This concept allows us to deduce how much charge a given section of the beam holds per second. Thus, it forms a basis for calculations such as how far apart individual electrons are spaced in the beam as it moves. Knowing the charge is vital for exploring how these electrons interact with each other and potential fields they encounter.
Electric Field Calculation
Calculating the electric field generated by an electron beam lets us understand the force other charges would experience in proximity to the beam. This calculation requires understanding charge distribution and the geometry of the problem:- First, determine the charge density \( (\rho) \) along the beam, which is the total charge per unit length.- The formula for the electric field \( E \) at a distance from a straight line charge is given by \( E = \frac{\rho}{2\pi\epsilon_0} \), where \( \epsilon_0 \) is the permittivity of free space.For the exercise, with the electrons traveling through a small cross-sectional area, the assumption is that the radius is small, simplifying calculation. Despite the negligible radius of the beam, the electric field is non-zero and measurable, impacting any charges that come close to it. Additionally, in the electron rest frame, due to relativistic effects like length contraction, the electric field value can change as it becomes more concentrated.

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Most popular questions from this chapter

Attenuator chain Some important kinds of networks are infinite in extent. Figure \(4.49\) shows a chain of series and parallel resistors stretching off endlessly to the right. The line at the bottom is the resistanceless return wire for all of them. This is sometimes called an attenuator chain, or a ladder network. The problem is to find the "input resistance," that is, the equivalent resistance between terminals \(A\) and \(B\). Our interest in this problem mainly concerns the method of solution, which takes an odd twist and which can be used in other places in physics where we have an iteration of identical devices (even an infinite chain of lenses, in optics). The point is that the input resistance (which we do not yet know - call it \(R\) ) will not be changed by adding a new set of resistors to the front end of the chain to make it one unit longer. But now, adding this section, we see that this new input resistance is just \(R_{1}\) in series with the parallel combination of \(R_{2}\) and \(R\) Use this strategy to determine \(R\). Show that, if voltage \(V_{0}\) is applied at the input to such a chain, the voltage at successive nodes decreases in a geometric series. What should the ratio of the resistors be so that the ladder is an attenuator that halves the voltage at every step? Obviously a truly infinite ladder would not be practical. Can you suggest a way to terminate it after a few sections without introducing any error in its attenuation?

Conductor in a capacitor (a) The plates of a capacitor have area \(A\) and separation \(s\) (assumed to be small). The plates are isolated, so the charges on them remain constant; the charge densities are \(\pm \sigma .\) A neutral conducting slab with the same area \(A\) but thickness \(s / 2\) is initially held outside the capacitor, see Fig. \(3.40 .\) The slab is released. What is its kinetic energy at the moment it is completely inside the capacitor? (The slab will indeed get drawn into the capacitor, as evidenced by the fact that the kinetic energy you calculate will be positive.) (b) Same question, but now let the plates be connected to a battery that maintains a constant potential difference. The charge densities are initially \(\pm \sigma\). (Don't forget to include the work done by the battery, which you will find to be nonzero.)

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?

Unbalanced current \(*\) * As an illustration of the point made in Footnote 13 in Section 4.7, consider a black box that is approximately a \(10 \mathrm{~cm}\) cube with two binding posts. Each of these terminals is connected by a wire to some external circuits. Otherwise, the box is well insulated from everything. A current of approximately \(1 \mathrm{~A}\) flows through this circuit element. Suppose now that the current in and the current out differ by one part in a million. About how long would it take, unless something else happens, for the box to rise in potential by 1000 volts?

Principal radii of curvature * Consider a point on the surface of a conductor. The principal radii of curvature of the surface at that point are defined to be the largest and smallest radii of curvature there. To find the radii of curvature, consider a plane that contains the normal to the surface at the given point. Rotate this plane around the normal, and look at the curve representing the intersection of the plane and the surface. The radius of curvature is defined to be the radius of the circle that locally matches up with the curve. For example, a sphere has its principal radii everywhere equal to the radius \(R\). A cylinder has tone principal radius equal to the cross-sectional radius \(R\), and the other equal to infinity. It turns out that the spatial derivative (in the direction of the\\} normal) of the electric field just outside a conductor can be written in terms of the principal radii, \(R_{1}\) and \(R_{2}\), as follows: $$ \frac{d E}{d x}=-\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) E $$ (a) Verify this expression for a sphere, a cylinder, and a plane. (b) Prove this expression. Use Gauss's law with a wisely chosen pillbox just outside the surface. Remember that near the surface, the electric field is normal to it.
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